Complex Integration: A Compendium of Smart and Little-Known Techniques for Evaluating Integrals and Sums 3031242270, 9783031242274

Table of contents :
Preface
Why Complex Integration?
Goals
Prerequisites
A Note About Nomenclature
Acknowledgements
Prologue
Contents
1 Review of Foundational Concepts
1.1 Sequences and Series
1.2 Integrals
1.3 Two Dimensions
1.4 Analytic Functions
2 Evaluation of Definite Integrals I: The Residue Theorem and Friends
2.1 Poles
2.1.1 Cauchy's Theorem in the Presence of a Pole
2.1.2 The Residue Theorem is a Shortcut
2.2 Branch Points
2.3 Poles and Branch Points
3 Evaluation of Definite Integrals II: Applications to Various Types of Integrals
3.1 Integrands Defined Over [0,infty)
3.2 Integration Over the Unit Circle
3.3 Residue Reduction Contours
3.4 Definite Integrals Evaluated Between Branch Points
3.5 Sums
4 Cauchy Principal Value
4.1 Removable Singularities
4.2 Poles on Contours
4.3 The Hilbert Transform
5 Integral Transforms
5.1 The Fourier Transform
5.2 Laplace Transforms
5.3 Two-Sided Laplace and Mellin Transforms
6 Asymptotic Methods
6.1 Asymptotic Expansions
6.2 The Monotonic h-Transform and a New Complex Integration Technique
6.3 Euler-Maclurin Summation
Appendix Epilogue

Citation preview

Undergraduate Lecture Notes in Physics

Ron Gordon

Complex Integration A Compendium of Smart and Little-Known Techniques for Evaluating Integrals and Sums

Undergraduate Lecture Notes in Physics Series Editors Neil Ashby, University of Colorado, Boulder, CO, USA William Brantley, Department of Physics, Furman University, Greenville, SC, USA Matthew Deady, Physics Program, Bard College, Annandale-on-Hudson, NY, USA Michael Fowler, Department of Physics, University of Virginia, Charlottesville, VA, USA Morten Hjorth-Jensen, Department of Physics, University of Oslo, Oslo, Norway Michael Inglis, Department of Physical Sciences, SUNY Suffolk County Community College, Selden, NY, USA Barry Luokkala , Department of Physics, Carnegie Mellon University, Pittsburgh, PA, USA

Undergraduate Lecture Notes in Physics (ULNP) publishes authoritative texts covering topics throughout pure and applied physics. Each title in the series is suitable as a basis for undergraduate instruction, typically containing practice problems, worked examples, chapter summaries, and suggestions for further reading. ULNP titles must provide at least one of the following: • An exceptionally clear and concise treatment of a standard undergraduate subject. • A solid undergraduate-level introduction to a graduate, advanced, or non-standard subject. • A novel perspective or an unusual approach to teaching a subject. ULNP especially encourages new, original, and idiosyncratic approaches to physics teaching at the undergraduate level. The purpose of ULNP is to provide intriguing, absorbing books that will continue to be the reader’s preferred reference throughout their academic career.

Ron Gordon

Complex Integration A Compendium of Smart and Little-Known Techniques for Evaluating Integrals and Sums

Ron Gordon Northborough, MA, USA

ISSN 2192-4791 ISSN 2192-4805 (electronic) Undergraduate Lecture Notes in Physics ISBN 978-3-031-24227-4 ISBN 978-3-031-24228-1 (eBook) https://doi.org/10.1007/978-3-031-24228-1 © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 This work is subject to copyright. All rights are solely and exclusively licensed by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

This book is dedicated to the memory of my eleventh grade math teacher, Ms. Effie Vellios.

Preface

In real-world maths, computers do almost all the calculating; by contrast, in educational maths, people do almost all the calculating. —Conrad Wolfram, The Math(s) Fix: An Education Blueprint for the AI Age.

Conrad Wolfram, in his book The Math(s) Fix: An Education Blueprint for the AI Age, lays out not only a case for abandoning the traditional math curriculum in which we teach concepts involving hand calculations, but also a proposed all-computer computation curriculum. Wolfram is related to Stephen Wolfram who gifted the computational system Mathematica to us all since the 1980s. You might think this Preface and this book are meant to be a broadside against Wolfram, and I am going to furnish arguments as to why he is dead wrong and we should keep teaching traditional hand computation to everyone. But it isn’t, and I am not. Dr. Wolfram (the author) is largely correct: We need to teach math differently to ready students for dealing with concepts such as machine learning, Kalman filters, graphs, and networks, which have become the backbone of the problems of our society today. So am I saying that a book dedicated to hand calculations is a complete waste of time. No. The opposite, actually. The problems requiring hand calculation have never really gone away. We still generate models of things like imaging detailed objects in turbulent media, or return on European put options, or knocking in an automotive engine. These problems are solved, others will replace them. Mr. Wolfram will tell us simply to fire up Wolfram|Alpha or Mathematica and whatever integral or sum or differential equation that needs to be solved will be solved by the computer. And, again, he isn’t wrong about it. Much of the time, anyway.

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The truth is the systems behind the computers we use as substitutes for hand calculations rely on algorithms written by human beings. Human beings that make mistakes, a lot. Quis custodiet ipsos custodes? So we will always need people who know how to do the hand calculations. I doubt Dr. Wolfram would disagree—otherwise, why would the other Dr. Wolfram need to hire all of those people to keep Mathematica running? I suppose this has always been my calling. When I worked for IBM Microelectronics, I was in a group that did lithographic process design and did so by running interminable simulations of the lithography process on vendor-provided software. (Previously, I worked for such a vendor.) One of my projects was to design test cases that we could model by hand, so we could evaluate the simulation results. This was my favorite part of my job, and we were able to find numerous issues to take to the software vendors. My enthusiasm spilled out to my management, and I was allowed to teach a short course to the engineers running the simulations about how to do simple hand calculations of the lithography models—a sort of Conrad Wolfram-in-reverse, if you will. Even Mathematica itself—and by extension, Wolfram|Alpha—are not perfect. I myself can point to a specific example Mathematica got wrong. The problem was to compute the following sum (which will be considered in the book): ∞  n=1

1 13π 3 √  =− √ 360 2 n 3 sin 2π n 2

13π√ When I checked the result, Mathematica returned− 360 . π squared, not cubed. 2 When I convincingly established that the cube was correct, the good staff at Wolfram Research eventually fixed the error. But it took someone who did hand calculations to fix a presumed bug in their code. Again, this is not to disparage the Dr. Wolfram’s point about math education. In fact, I think he would agree with me that there is a fundamental need to have people who know the basics very well to be making sure the computer models we are running are accurate. But this is all a very long way of answering the question, why do we want to work out these crazy integrals and sums when we have great computer software that can do it for us? But even if there were no need for custodes, even if the likes of Mathematica were perfect and needed no checking, there is a more important, overrriding reason to work out these integrals and sums by hand: It is fun! If you are perusing through this book, it is not likely because you have calculus homework due tomorrow. It is because you have a curiosity as to how really difficult problems in calculus can be solved using complex integration. It is because you want to learn these methods for yourself. There is a power behind them that is addictive. It is like the classic line, “amuse your friends, confound your enemies” taken in a whole new direction.

Preface

ix

Why Complex Integration? I think we have answered the question of why evaluate these integrals and sums by hand in general. But unanswered is the question, why are we focusing on complex integration techniques? The origin of the complex integration techniques is a result of my work on the Math StackExchange website. On this website, there are lots of people asking other people like me how to evaluate various integrals and sums. In fact, there would be several of us working on these problems at once, which would result in very similar, good answers being offered at roughly the same time. Most of the time, these answers would use fundamentally real techniques—“Feynman’s trick” of differentiating a parameter under the integral sign, Taylor expansion, orthogonal function decomposition, for example. I started evaluating integrals using the Residue Theorem just as a way to put down a different answer. It occurred to me, however, that the Residue Theorem as I had learned it in school was not really a great arrow in my quiver of integration techniques I could bring to the table—yet. The examples I had learned in school—and the same examples I was working out in the StackExchange problems—were extremely limited to a few, scientifically selected examples. For example, we are taught to evaluate the following integral using analysis of branch cuts in a multivalued integrand: 1 d x x p (1 − x)q 0

It turns out, however, that the techniques we are taught only work when p + q is an integer. So many times we would be taught to perform inte1 grals such as 0 d x x −1/3 (1 − x)−2/3 , but not something more interesting like 1 −1/3 (1 − x)−2/3 log x. (Note that the factor of log may be treated as the limit 0 dx x of the derivative of x a as a → 0.) The reason why this is so and the evaluation of the more difficult integral using complex integration will be provided in this book. How would we go about attacking these new problems? By doing them correctly! For example, in the branch cut integrals above, the usual way to approach them is to consider something called a “dog-bone” contour formed by the branch cuts above and below the integration interval. We would then need to consider a “residue at infinity” to handle the effect of integrating out to infinity—the residue at infinity was simply a formula derived from an integral about a large circle. But what if the integral about the large circle didn’t converge as the radius became infinite? What if, instead, there was a singular piece that had to cancel another similar piece elsewhere on the contour? By “the contour,” I no longer mean the dog-bone contour but the logical extension of it to the large circle. The cancelations of divergences formed the impetus for attacking whole new classes of problems using complex integration. This is one of the main themes throughout this book.

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Preface

Another theme is the expansion of our quiver of problem-solving techniques. Once we know a more complete form of problems that may be attacked using complex integration techniques, we may attempt to transform difficult integrals into one of those forms. A well-known example of this is the following integral. 1 −1

1 dx x





1+x 1 + 2x + 2x 2 = 4π Cot−1 φ log 2 1−x 1 − 2x + 2x

where φ is the golden ratio. I attacked this problem by transforming the integral into the following form. 1 dx −1

1 x



∞  1 + 2x + 2x 2 1+x = dy p(y) log y log 1−x 1 − 2x + 2x 2 0

where p is a rational function of y. I did this because I knew that integrals of this form, where the poles of p are determined exactly, could be evaluated using the Residue Theorem. In this case, even though the denominator of p was degree eight, I was able to find the poles because of the high degree of symmetry of p.

Goals What are the goals one should expect to achieve by reading this book? Generally, to appreciate whole new types of integrals that may be evaluated using complex integration techniques. Specifically, however, there are several goals. 1. Appreciate that poles and branch points may be treated in a unified manner when it comes to evaluating real integrals using complex integration. 2. Gain perspective on the types of integrals that may be evaluated using complex integration. 3. Learn how to construct even reasonably sophisticated contours in evaluating real integrals using complex integration. 4. Become comfortable with Cauchy principal values and canceling divergences as part of evaluating integrals with combined pole/branch points. 5. Appreciate transform techniques such as Parseval’s Theorem and the Convolution Theorem for evaluating integrals. 6. Gain facility with asymptotic techniques and how they may be used to evaluate real integrals and sums exactly.

Preface

xi

Prerequisites As this is an informal book, there are no real prerequisites for understanding the material other than an enthusiasm for evaluating integrals and sums. In this respect, obviously the reader will have had to have at least two semesters of calculus to appreciate the nature of the problems being solved. That said, I feel that the scope of the book is best focused on examples of solutions of difficult calculus problems via complex integration. Accordingly, more than a passing familiarity with the concept of an analytic function, complex integration, and Laurent series at the level of an undergraduate Complex Analysis course (a la Churchill and Brown) would be the best preparation for the material in the book. Nevertheless, not having such familiarity with this material is hardly a barrier to the enthusiastic student.

A Note About Nomenclature The reader may notice that I place my differentials (“d x”) next to the integral sign rather than after the integrand as is the practice in most places. I do this because of my background in Optics, where we regularly evaluate two-dimensional integrals, and some of us felt that it was less confusing to place the differential next to its own integral. We merely extended the practice to one-dimensional integrals for consistency. I make one exception to this—two-dimensional path integrals where the integrand is of the for Pd x + Qdy in the introductory material. Otherwise, the differentials are next to their integral signs in this book. Northborough, MA, USA

Ron Gordon

Acknowledgements

This book is the product of a lifetime of encouragement, patience, and kindness shown toward me by so many people. I will not attempt to list them all here. I have had some wonderful teachers through the years, and a few in particular deserve my thanks for having a direct impact on this book: Ms. Effie Vellios (Brockton High School, Brockton, MA), Dr. Daniel Adams (Brockton High School), Prof. Sam Holland (University of Massachusetts, Amherst, MA), and Prof. Greg Forbes (University of Rochester, Rochester, NY; Macquarie University, Sydney NSW, Australia). I have dedicated this book to Ms. Vellios, who was the first of my teachers to recognize my affinity for mathematics and effectively lit a candle that burns bright to this very day. High school students who came after my class were sadly robbed of her considerable talent, as she passed away soon after I graduated in 1988. Much of the subject matter discussed here was developed over many years on Math Stack Exchange (math.stackexchange.com). The interactions I have had with the following people have been invaluable and have also had an impact on this book: @robjohn, Mark Viola, @sos440, @O.L, @RandomVariable, Daniel Fischer, @achillehui, and @cody. I am sure I have inadvertently left at least a few people out and I hope that they know who they are. If you are reading this book and are not familiar with the website, I urge you to have a visit and maybe stick around. My editor at Springer, Marina Forlizzi, more or less held my hand through this whole experience, which as someone who has never written a book before I sure as heck needed. Ms. Andrea Feshbach and the Feshbach Family, as well as Dover Books, Inc., provided valuable copyright permissions in the book. I cannot thank them enough. My dear friend, Pete Kasperowicz, took a look at an early version of the book and gave me some great tips even though he doesn’t understand a lick of the math. My parents, Michael and Geri (z”l), were my rocks and encouraged me every step of the way. The set an example for the sort of parent I should be.

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My family has stood behind me every step of the way, even though the writing of this book impacted my two children, Eli and Josh, and my wife Laura most of all. One of my children (Josh) took a Complex Analysis course at his university just, so he would be able to give feedback on some of the problems. Laura never once complained about the time I put into making this book a reality. I am an incredibly lucky nerd. Northborough, MA, USA

Ron Gordon

Prologue

How would you evaluate the following integral? 1 I = −1

1 dx x



 2 2x + 2x + 1 1+x log 1−x 2x 2 − 2x + 1

Well, you could try a number of substitutions. Probably, the one substitution that makes sense from the get-go would be t = 1+x . Some manipulations later, we find 1−x that our integral is equal to the following. ∞ I =2 0

 t −1/2 5 − 2t + t 2 dx log 1 − t2 1 − 2t + 5t 2

Without a further plan, however, it is difficult to know what to do with this. Nevertheless, if we knew that we could transform the integral to a certain form we knew would get us across the finish line, then we would make further substitutions along those lines. This is a major question we will answer in this book: What sorts of integrals can be evaluated exactly using complex integration techniques? In this case, we will see that integrals of the form ∞ dy p(y) log y 0

are perfect candidates for such evaluation. Knowing this form as an end for substitution is a major milestone in solving this problem. We will encounter this form—and others—that may be evaluated using the techniques taught in this book.

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Contents

1 Review of Foundational Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Sequences and Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Two Dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4 Analytic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 1 11 13 18 22

2 Evaluation of Definite Integrals I: The Residue Theorem and Friends . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Poles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1.1 Cauchy’s Theorem in the Presence of a Pole . . . . . . . . . . . . . 2.1.2 The Residue Theorem is a Shortcut . . . . . . . . . . . . . . . . . . . . . 2.2 Branch Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Poles and Branch Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

25 25 26 32 34 39 47

3 Evaluation of Definite Integrals II: Applications to Various Types of Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Integrands Defined Over [0, ∞) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Integration Over the Unit Circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Residue Reduction Contours . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 Definite Integrals Evaluated Between Branch Points . . . . . . . . . . . . . 3.5 Sums . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

49 49 58 75 85 91 99

4 Cauchy Principal Value . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 Removable Singularities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Poles on Contours . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 The Hilbert Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

103 103 117 125 134

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Contents

5 Integral Transforms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 The Fourier Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Laplace Transforms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 Two-Sided Laplace and Mellin Transforms . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

137 139 171 193 202

6 Asymptotic Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1 Asymptotic Expansions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 The Monotonic h-Transform and a New Complex Integration Technique . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3 Euler-Maclurin Summation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

205 205 217 230

Epilogue . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237

Chapter 1

Review of Foundational Concepts

Abstract Before diving into complex integration concepts, it is worth spending some time reviewing material from real calculus. Such material includes sequences and series, Riemann integrals, the Fundamental Theorem, evaluation techniques, double sums and integrals, Green’s Theorem, analytic functions and Cauchy’s Theorem.

1.1 Sequences and Series Sequences The basic building block in any evaluation of integrals is the lowly sequence. Here, we define sequences of real numbers. To realize an integral, we will need to construct sums of sequences, or series. A sequence of real numbers is a mapping from the integers (or a subset thereof) to the real numbers a : Z → R. Sequences have their own notation apart from the usual function notation. For example, a sequence may be denoted as an , where n ∈ Z is denoted as an index. That said, other notations are common, e.g., a[n] or even a(n). The following are examples of sequences defined explicitly: an = n an = n! = (n + 1)    cos 2n−1 arccos (a1 )  |a1 | ≤ 1 an = cosh 2n−1 arccosh(a1 ) |a1 | > 1   an = c1 + c2 n + c3 n 2 2n

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 R. Gordon, Complex Integration, Undergraduate Lecture Notes in Physics, https://doi.org/10.1007/978-3-031-24228-1_1

1

2

1 Review of Foundational Concepts

The following are examples of sequences defined implicitly, or recursively: an+1 = an + 1 a0 = 0 an+1 = nan a0 = 1 an+1 = 2an2 − 1 an+1 = 6an − 12an−1 + 8an−2 Sequences, when arranged in a particular way, represent patterns of numbers. For example, some sequences have numbers arranged in increasing order. Other are arranged in decreasing order. Such sequences are monotone increasing and decreasing, respectively. Some monotone sequences converge to some limit as n increases without bound: 1 1 + n2 π  an = cos 2n an =

Others do not converge but rather diverge, e.g., an = n and an = n! Series We may define a series as a sum over a sequence. A series formed from terms of a sequence form a new sequence; for example consider the sequence An defined as the following series of a sequence ak as follows: An =

n 

ak

(1.1)

k=0

A finite series such as the above geometric series may be referred to as a partial series. For example, you may know of a geometric sequence an = ar n and a partial geometric series n n   r n+1 − 1 (1.2) ak = a rk = a An = r −1 k=0 k=0 We may also define infinite series as the limit of a sequence of partial series, if the limit exists. If the limit exists then we say the infinite sum converges, and if the limit doesn’t exist we say the infinite sum diverges. For example, the geometric series a when |r | < 1. This may be proven above converges and has a limiting value of 1−r as before using the definition of a limit:   n+1        0 and n > N , which means that this inequality holds when N () =

log



a  1−r

log r

 −1

(1.4)

With the limit concept, one may define an infinite series to be the limit of a partial sum as the number of terms increases without limit. Accordingly, as with sequences, series may be convergent or divergent. For example, the above geometric series is convergent when |r | < 1 and divergent otherwise. Another example is the so-called harmonic series: Hn =

n  1 k=1

k

(1.5)

which, using a simple but powerful argument, may be shown to diverge: 1 1 1 1 + + ··· + n + 2 3 2 − 1 2n



1 1 1 1 1 1 1 + + + + + + ··· =1+ + 2 3 4 5 6 7 8

1 1 1 + + · · · + 2n−1 + 1 2n−1 + 2 2n



1 1 1 1 1 1 1 + + + + + + ··· >1+ + 2 4 4 8 8 8 8

1 1 1 + + + · · · 2n 2n 2n  

H2n = 1 +

2n−1 terms

1 1 1 = 1 + + + ··· 2 2 2   n terms

n =1+ 2

(1.6)

i.e., Hn > 1 +

1 log2 n 2

(1.7)

This proof is nifty because not only does it demonstrate that the harmonic series diverges but it also indicates how it diverges, i.e., logarithmically. Along these lines, one may prove that (1.8) Hn ∼ γ + log n, (n → ∞)

4

1 Review of Foundational Concepts

where γ ≈ 0.577216 . . . is the Euler-Mascheroni constant. Fundamental Theorem of Discrete Calculus There are some techniques of evaluating sums that are perfect analogies to the techniques we will use to evaluate integrals, or vice-versa. The first technique we will denote as the Fundamental Theorem of Discrete Calculus. Consider a sequence an and we wish to evaluate a sequence of partial sums: An =

n−1 

ak

(1.9)

k=0

If each term in the sequence may be expressed as the difference between successive terms of another sequence bn —that is, ak = bk+1 − bk , then the series Ak becomes easy to evaluate:

An =

n−1  (bk+1 − bk ) = (b1 − b0 ) + (b2 − b1 ) + · · · + (bn − bn−1 ) k=0

= bn − b0

(1.10)

A sum of differences as above is known as a telescoping sum. In this case, the series value is simply the value of the anti-difference - that is, the sequence bk —at the endpoints of the sum. Accordingly, one way to evaluate a sum exactly is to find the anti-difference sequence of the summand of a series. Let’s go through a few examples. As a first example, let’s compute the following sum. SN =

N  n=1

arctan

2 n2

(1.11)

This old chestnut looks impossible but it is really a simple problem in disguise because of the following convenient fact. arctan

2 = arctan (n + 1) − arctan (n − 1) n2

(1.12)

Then we have a telescoping series and accordingly, S N = arctan (N + 1) + arctan N − arctan 1 − arctan 0 2N +1 3π − arctan 2 = 4 N +N −1 We can also consider the infinite sum by taking the limit as N → ∞.

(1.13)

1.1 Sequences and Series

5

S=

∞ 

arctan

n=1

2 3π = 2 n 4

(1.14)

As another example, consider the less simple problem of evaluating the following sum.     ∞  sin 2πn   S= arctan 2−n (1.15) 2 + cos 2πn n=1 The issue with using the Fundamental Theorem sometimes is that it takes a very good guess to get the antidifference. In this case, we go with the deus ex machina and let     sin 2πn   an = arctan 2−n (1.16) 2 − cos 2πn Then ( ) − sin ( 2π 2n ) ( 2πn ) 221−n −cos ( 2π 2n ) tan (an − an−1 ) = sin ( 2πn ) sin ( 2π 2n ) 1 + 22−n −cos π 221−n −cos 2π ( 2n ) ( 2n ) sin 2πn −n 2 2 −cos

(1.17)

which simplifies somewhat to             1−n sin π21−n + 22 sin π2−n + sin π21−n cos π2−n − sin π2−n cos π21−n             −n 1−n + sin π21−n sin π2−n − 22 cos π21−n + cos π2−n cos π21−n − 22 cos π2−n

−22 23 2

−n

−n

(1.18) The numerator is equal to    1−n    −n sin π 2−n 22 − 22 +1 cos π 2−n + 1

(1.19)

while the denominator is equal to  1−n      −n −n 22 22 − 22 +1 cos π 2−n + 1 + cos π 2−n  1−n    −n 22 − 22 +1 cos π 2−n + 1

(1.20)

Canceling the common factor produces the apparently miraculous result,  an − an−1 = arctan

sin



π  2n

22−n + cos

π  2n

(1.21)

6

1 Review of Foundational Concepts

So an is the antidifference we seek and we now have a telescoping sum. ∞  (an − an−1 ) = a∞ − a0

(1.22)

n=1

where a∞ = lim an

(1.23)

n→∞

Now, sin 2

2−n

π  2n

− cos

π  ∼ 2n

π/2n 1+

log 2 2n

−1

=

π log 2

(1.24)

Note also that a0 = 0. Therefore the sum S is S = arctan

π log 2

(1.25)

The “method” of finding an antidifference works well when we can make a very good guess that an antidifference exists. But in most cases, there is no such antidifference and we will have to find other ways to evaluate the sums we come across. Summation By Parts For example, we may recognize certain sums as not exactly telescoping but having terms that are a product of a first sequence and a difference of a second sequence. This is similar to integration by parts in which the integrand is a product of a first function and a derivative of a second function. Theorem 1.1 (Summation by parts) Let an and bn be sequences of numbers. Then n 

ak (bk+1 − bk ) = an bn+1 − am bm −

k=m

n 

bk (ak − ak−1 )

k=m+1

We begin with the equation ak (bk+1 − bk ) = ak bk+1 − ak−1 bk − bk (ak − ak−1 )

(1.26)

Summing the above from k = m to n produces n  k=m

ak (bk+1 − bk ) =

n  k=m

(ak bk+1 − ak−1 bk ) −

n 

bk (ak − ak−1 )

(1.27)

k=m

Note that the first sum on the RHS is a telescoping sum and is therefore easy to evaluate by the Fundamental Theorem of Discrete Calculus. The first term may also

1.1 Sequences and Series

7

be extracted from the second sum on the RHS; the result is n 

ak (bk+1 − bk ) = an bn+1 − am−1 bm − bm (am − am−1 ) −

k=m

n 

bk (ak − ak−1 )

k=m+1

(1.28) The result follows. As an example, let’s compute the following sum. S=

∞  k=1

Hk k(k + 1)

(1.29)

We rewrite this into a form useful for summation by parts. S=

∞  k=1

Hk

1 1 − k k+1

(1.30)

We may identify ak as Hk and bk as − k1 in the above summation by parts formula. Accordingly, we have ∞  k=1

∞  1 Hk 1 Hk =− − k(k + 1) k+1 k k=1

N  1 1 = − lim − Hk N →∞ k+1 k k=1 ∞

H1  1 HN + + = − lim (Hk − Hk−1 ) N →∞ N + 1 1 k k=2

(1.31)

N = 0. Moreover, it Note that, as N → ∞, HN ∼ γ + log N so that lim N →∞ NH+1 should be clear by the definition of Hk that H1 = 1 and Hk − Hk−1 = k1 . Accordingly,

∞  k=1

∞

 1 Hk π2 =1+ = 2 k(k + 1) k 6 k=2

(1.32)

The above result leads to the following very striking-looking equation. ∞  1+ n=1

+ 13 + · · · + n1 π2 = 1 + 2 + 3 + ··· + n 3 1 2

(1.33)

We will find that many sums involving harmonic numbers may be evaluated using summation by parts because the difference of harmonic numbers is very simple.

8

1 Review of Foundational Concepts

In another example, we look at an exponential sum. S=

∞ 

ka −k

(1.34)

k=1

for a > 1. We can convert this to a sum we can work with by recognizing that a −k = −

 a  −(k+1) a − a −k a−1

(1.35)

Accordingly, ∞  a   −(k+1) k a − a −k a − 1 k=1

∞ 1 a  −k a − + a =− a−1 a a − 1 k=2

1 a 1 1 + = a − 1 a − 1 a 2 1 − a1 1 1 + = a − 1 (a − 1)2 a = (a − 1)2

S=−

(1.36)

Standard Sums We may also recognize some standard sum formulae we will be using along the way. For example, we should all be very familiar with and have already encountered the geometric sum. n−1  1 − rn (1.37) rk = 1−r k=0 For |r | < 1, we may express this as an infinite sum. ∞ 

rk =

k=0

1 1−r

(1.38)

We should also be familiar with derivative forms such as ∞  k=0

kr k =

r (1 − r )2

(1.39)

1.1 Sequences and Series

9

These may be derived by differentiation, summation by parts (above), or (even better) recursion: S = r + 2r 2 + 3r 3 + 4r 4 + . . . r S = r 2 + 2r 3 + 3r 4 + . . . (1 − r )S = r + r 2 + r 3 + r 4 + . . . r = 1−r Geometric sum formulae may apply to complex trigonometric sums; for example, for |r | ≤ 1 ∞ 

1 1 − r eiθ k=0

∞  1 1 − r cos θ r k cos (kθ ) = Re = iθ 1 − re 1 − 2r cos θ + r 2 k=0

∞  r sin θ 1 k = r sin (kθ ) = Im iθ 1 − re 1 − 2r cos θ + r 2 k=0 r k eikθ =

(1.40)

(1.41)

(1.42)

We should also recognize monomial sums. For example, we should be familiar with the following. n  1 k = n(n + 1) (1.43) 2 k=1 We can derive higher powers recursively using the binomial theorem and the Fundamental Theorem of Discrete Calculus. For example, n n n      (k + 1)3 − k 3 = 3 k2 + 3 k+n k=1

k=1

(n + 1)3 − 1 = 3 3

k=1

n 

3 k 2 + n(n + 1) + n 2 k=1

n 

3 5 k 2 = n 3 + 3n 2 + 3n − n 2 − n 2 2 k=1 3 1 = n3 + n2 + n 2 2 1 = n(2n 2 + 3n + 1) 2

(1.44)

10

1 Review of Foundational Concepts

Accordingly, if we define the sum of the mth power as Sm , then S2 =

n  k=1

k2 =

1 n(n + 1)(2n + 1) 6

(1.45)

We can determine Sm again through recursion and the binomial theorem. (n + 1)

m+1



m+1  m + 1

m+1 Sm+1−k −1= Sm + k 1 k=2

(1.46)

Products Let’s briefly discuss products. Products are easily seen to have an equivalence to sums via the logarithm: log

 n 

 (1 + an ) =

k=1

n 

log (1 + an )

(1.47)

k=1

In this case, the convergence of the product as n → ∞ is equivalent to the convergence of the sum over an as n → ∞. That said, there are many cases in which it is advantageous to evaluate a product directly rather than appeal to an equivalent sum. For example, let’s evaluate the following product.

∞  1 n+1 n e2 n − 1 n=2

(1.48)



N  1 n+1 n PN = e2 n − 1 n=2

(1.49)

Consider the partial product

By writing out the terms of PN , we can easily find a simple expression for it: PN = e−2(N −1)

N N −1 (N + 1) N 2(N − 1)!2

(1.50)

Use Stirling’s approximation as well as the definition of e as N → ∞: (N − 1)! ≈



2π(N − 1)(N − 1) N −1 e−(N −1)

(1.51)

Then (N + 1) N ≈ eN N ≈ e2 (N − 1) N . Putting this altogether, we find that the product is e3 /(4π ).

1.2 Integrals

11

1.2 Integrals The integral is the basis of what we will study in this book. At a very high level, an integral is a number we give to a set that is indicative of some measure of the set. The simplest example—and the one we will study here exclusively—is the Riemann integral, which when taken over a simple 1D interval represents the area under a curve representing a function defined on the interval. The Riemann integral, however, may be defined over sets of any number of dimensions and we will study two-dimensional sets in the next section. For now, and for the majority of this book, we will limit ourselves to one-dimensional complexes or manifolds. We define a definite integral of a function over an interval [a, b] as the following limit.

b n b−a  k (1.52) d x f (x) = lim f a + (b − a) n→∞ n k=0 n a

As an example, 1

n 1  2 d x x = lim 3 k n→∞ n k=1 2

0

1 1 n(n + 1)(2n + 1) n→∞ n 3 6 1 = 3

= lim

(1.53)

It will be an exercise to show that 1 dx xm = 0

1 m+1

(1.54)

follows from an induction argument. Another example: 1

n 1  k/n e n→∞ n k=1

d x e x = lim 0

1

1 e1+ n − 1 = lim 1 n→∞ n e n − 1 e−1  = lim  1 n→∞ n en − 1 = e−1

(1.55)

12

1 Review of Foundational Concepts

One more example: 1 0

n dx 1 1 = lim k2 n→∞ n 1 + x2 k=1 1 + n 2 ∞ n 1   (−1)m k 2 m = lim n→∞ n n2 m k=1 m=0

=

∞  m=0

n 1  k2 m n→∞ n n2 m k=1  

(−1)m lim 1 0

d x x m =1/(2 m+1)

∞ 

(−1)m 2m +1 m=0 π = 4

=

(1.56)

Fundamental Theorem of Calculus Recall the Fundamental Theorem of Discrete Calculus: when f k = gk+1 − gk n 

f k = gn+1 − g0

(1.57)

k=0

Equivalently,

n b − a  gk+1 − gk = gn+1 − g0 b−a n k=0 n

(1.58)

If we recognize a differentiable function g such that gn+1 = g(b) and g0 = g(a), the above equation is equivalent to, upon taking the limit as n → ∞, b

d x g  (x) = g(b) − g(a)

(1.59)

a

where g  is the first derivative of g. This is the Fundamental Theorem of Calculus. To evaluate an integral using the FTC, we must know that there is an antiderivative of the integrand. The usefulness of the FTC is limited by the fact that we do not always have such an antiderivative. Integration by parts follows from summation by parts in a similar manner. Note that f n = f (b) and gn+1 = g(b) independent of n.

1.3 Two Dimensions

13

n b − a  gk+1 − gk fk = f (b)g(b) − f (a)g(a) b−a n→∞ n k=0 n

lim

n b − a  f k − f k−1 gk b−a n→∞ n k=1 n

− lim

(1.60)

or, by the definition of the definite integral, b

b



d x f (x)g (x) = f (b)g(b) − f (a)g(a) − a

d x g(x) f  (x)

(1.61)

a

We note that we may define certain improper integrals by setting the lower limit a to −∞ or b to ∞ by taking the respective limits b

b d x f (x) = lim

d x f (x)

a→−∞

−∞

(1.62)

a

∞

b d x f (x) = lim

d x f (x)

b→∞

a

(1.63)

a

Other improper integrals are defined by a weak singularity of the integrand in the region of integration. For example, b I =

d x x −c f (x)

(1.64)

0

where f is finite at x = 0 and 0 < c < 1. We can avoid sampling the integrand at x = 0 by using a substitution x = u −1/c , which transforms the integral into 1 I = c

∞

  du u −1/c f u −1/c

(1.65)

b−c

1.3 Two Dimensions We will frequently come to situations in which we will see double sums of the form S=

n m   k=1 =1

ak

(1.66)

14

1 Review of Foundational Concepts

For such finite sums, we may reverse the order of summation, i.e., S=

n  m 

ak

(1.67)

=1 k=1

When one or both of the upper limits are infinite, then we may still reverse the order of summation under certain conditions of convergence. Reversing the order of summation is an interesting tool to use in evaluating sums. This is true even when n depends on k, as we will see in the next example. This asks that we evaluate S=

∞  k=1

Hk (k + 1)2

(1.68)

One way to attack this sum is to express as a double sum using the definition of Hk , as follows: k ∞   1 1 (1.69) S= 2 (k + 1)  k=1 =1 It is not trivial how one can reverse the order of summation. The solution lies in examining the lattice points sampled by the sum. This lattice is illustrated in Fig. 1.1. Note that, in the above double sum, the sampling of the terms goes to the right for each k. To reverse the order of summation, we sample vertically for each . Accordingly, it should be clear that the above double sum is equal to

Fig. 1.1 Lattice points of summation

1.3 Two Dimensions

15

S= =

∞ ∞  1 =1 ∞  =1



1 (k + 1)2

k= ∞ 

1 1  k=+1 k 2

(1.70)

It turns out that we may express the inner sum as an integral:  ∞  1 ue−u = du k2 eu − 1 k=+1 ∞

(1.71)

0

(The reader is invited to derive this as an exercise.) The sum then becomes  ∞  1

∞

S=

=1



du

0

ue−u eu − 1

(1.72)

For similar reasons as those allowing us to reverse the order of summation, we can also reverse the order of the summation and integration. (Essentially, the sum and integral are linear operators.) The result is ∞ S=

∞

du 0

u  e−u u e − 1 =1 

∞ =−

du ue 0

  1 − e−u 1 − e−u

−u log

(1.73)

We evaluate the integral by substituting x = − log u in the integral to get  S=

1

dx 0

log x log (1 − x) x

(1.74)

This is the sort of integral we will be seeing in later chapters on complex integration. But we can evaluate this integral using familiar techniques we will review later this chapter. As a preview of coming attractions, we will Taylor expand the log (1 − x) term and again reverse the order of integration and summation to get S=−

 ∞  1 1 d x x m−1 log x m 0 m=1

(1.75)

16

1 Review of Foundational Concepts

The reader will be able to integrate by parts and demonstrate that the sum is ∞  1 m3 m=1

(1.76)

Hk = ζ (3) (k + 1)2

(1.77)

S= or

∞  k=1

where ζ is the Riemann zeta function. We may run into similar considerations when doing double integrals over nonrectangular regions. Consider the following example. 1 I =

1 dx

0

x2

x3 dy  1 − y6

(1.78)

We change the order of integration. We can do this by drawing the region of integration and seeing that the integral is just √

1  0

which is 1 4

dy 1 − y6

1 dy  0

or 1 12

1 0



y

dx x3

(1.79)

0

y2 1 − y6

du π = √ 2 24 1−u

(1.80)

(1.81)

While this example does not seem too difficult, it can illustrate how dangerous it can be to simply rely on computer algebra systems (CASs) such as Maple or Mathematica. For example, one version of Maple returned the following result. I = −1/12

2 3 F2 (1/6, 1/2, 1/2; 7/6, 3/2; 1) (5/6)  (2/3) − π 3/2  (5/6)  (2/3)

(1.82)

The interesting part is that the above answer returned by that version of Maple is not incorrect! Rather, it is in such an intimidating form that it might seem impossible to understand or study any further, which would be a shame.

1.3 Two Dimensions

17

In two dimensions, we have another concept: a contour integral in the plane, i.e., an integral of a two-dimensional function F(x, y) = (P(x, y), Q(x, y)) defined on the boundary of and within a region D. If the components P and Q have gradients defined within D, then we may use Green’s Theorem to express an integral over R in terms of an integral over C, the boundary of R. Theorem 1.2 Let P : R2 → R and Q : R2 → R be defined and have continuous partial derivatives throughout an open region D having a boundary C that is oriented in a counterclockwise direction. Then

  ∂Q ∂P − d xd y (1.83) = (Pd x + Qdy) ∂x ∂y C D

One of the immediate joys of Green’s Theorem comes with computing the area of a region bounded by a contour C.  D

1 d xd y = 2

 (xdy − yd x)

(1.84)

C

This formula allows us to compute the area of complex regions such as arbitrary polygons with ease. As an example, consider the following problem. A goat is tied by a rope to a post at the edge of a circular plot of grass. What percentage of the grass will the goat be able to eat if the length of the rope is L and the radius of the plot of grass is R? The situation is shown in Fig. 1.2. The grass that the goat can eat is enclosed within the region D bounded by the thick arcs. The region D subtends an angle 2θ0 from the center of the plot of grass and an angle 2φ0 from the post to which the rope (and goat) is attached. The equation of the arc C R associated with the plot of grass is x = R cos θ , y = R sin θ ; the equation of the arc C L associated with the rope is x = R + L cos φ, y = L sin φ. Plugging this into Eq. (1.84) reveals that the area within the region D is  d xd y = R 2 θ0 + L 2 φ0 − R L sin φ0 .

(1.85)

D

The integration needed to get to this result was trivial once expressions for the angles θ0 and φ0 were determined. To get the area fraction avaailable to the goat, we divide this area by π R 2 . Letting β = L/R, we get the following expression for the area fraction: 



 1 2 1 1 2 1 β β2 1 − β 1− d xd y = arccos 1 − β + β arccos 2 πR π 2 π 2 π 4 D

(1.86)

18

1 Review of Foundational Concepts

Fig. 1.2 Geometry for determining how much grass a goat tied to the edge of a circular plot can eat

For example, if the fraction of grass available to the goat is 0.5, then β ≈ 1.15873, i.e., the length of the rope is about 15.9% longer then the radius of the plot of grass.

1.4 Analytic Functions It is assumed that the reader has had at least a semester of introductory Complex Analysis. Accordingly, we will not be going through analytic function theory in any depth here except to review some interesting results we will need later. A complex function f : C → C of a complex variable z = x + i y may be split into real and imaginary parts as in f (x + i y) = u(x + i y) + iv(x + i y). As for real functions, we may define derivatives and integrals of the complex function f , but these are complicated by the fact that we are considering the behavior of f in the complex plane rather than on a line as in the reals. For example, consider the definition of the derivative of f : f (z + h) − f (z) (1.87) f  (z) = lim h→0 h Here, h is a complex number and is thus defined in the complex plane. It is then unclear from which direction we would take the limit of h in the plane. (The right? The left? Up? Down? From an angle of 23◦ ?)

1.4 Analytic Functions

19

For real functions, we note that when a limit of a function at a value x from the left and the right agree, that function is continuous at x. And when the limit defining the derivative agree from the left and right at x, then the derivative of that function exists at x. (For example, the derivative of |x| does not exist at x = 0 because the limits from the left and right do not agree.) It would then seem, analogously, that the derivative of f should be independent of the direction of h; that is, the derivative of f should be the same regardless of whether h goes to zero parallel to the real or imaginary axis or along whatever line on the complex plane. This would seem to pose a much higher burden on f than in the real case where the agreement is merely between the left and right directions. And this is true: there are only a very small number of functions from all possible functions in the complex plane that satisfy this requirement for its derivatives. These are the analytic functions. To see what analyticity implies, consider partial derivatives of f with respect to the real and imaginary variables. ∂u ∂v ∂f = +i ∂x ∂x ∂x ∂f ∂u ∂v = +i ∂y ∂y ∂y

(1.88) (1.89)

Then if h = h r + i h i f (z + h) − f (z) ≈ h

 ∂u ∂x

   ∂v hi + i ∂∂vx h r + ∂u + i ∂y ∂y hr + i hi

(1.90)

We may rewrite this as ∂u + i ∂∂vx f (z + h) − f (z) ≈ ∂x h 1 + i hhri



hi 1+i hr

∂v ∂y ∂u ∂x

− i ∂u ∂y + i ∂∂vx

 (1.91)

Put this way, it is clear that in order for the derivative to be independent of the slope h i / h r , i.e., the direction of approach in the limit, the Cauchy-Riemann equations must hold: ∂v ∂u = ∂x ∂y ∂u ∂v =− ∂y ∂x

(1.92) (1.93)

20

1 Review of Foundational Concepts

That is, all analytic functions must satisfy the Cauchy-Riemann equations by this very definition.1 We will be dealing mainly with analytic functions in this book. A consequence of a function being analytic at a point z is that it has derivatives of all orders, not just a first derivative, at z. Thus an analytic function may be defined using a Taylor series about the point z, as its derivatives of all orders are defined there. √ Examples of functions that are not analytic at points include, for example, f (z) = z at z = 0, because the derivative at 0 is not defined (or infinite); or f (z) = log z at z = 0 for the same reason. For these functions, those singularities are called branch points as opposed to poles. Integration is also complicated by the fact that integration is in the complex plane and integration between two points can take on an infinite number of paths between the two points, rather than the one path offered by the real line. Again, analytic functions are the key to providing order in the potential chaos of the complex plane. Consider a closed contour in the complex plane defined as a first arbitrary path from z 1 to z 2 , and then a second, different arbitrary path from z 2 back to z 1 . If the value of the integral between z 1 and z 2 is to be independent of path, then the closed loop integral over the net contour C must be zero. Consider the real and imaginary part of the contour integral. 

 dz f (z) = C

(d x + idy)(u + iv) C



 (ud x − vdy) + i

=

(vd x + udy)

(1.94)

C

C

We may now apply Green’s Theorem to these contour integrals and we get the following double integrals over the region D interior to the contour C.  dz f (z) = − C



 d xd y D

∂u ∂v + ∂x ∂y





 +i

d xd y D

∂v ∂u − ∂x ∂y

(1.95)

And lo and behold, by the Cauchy-Riemann equations the integrals are indeed zero. Thus, we can say that, because the paths we used were arbitrary, that for any closed contour C, t e contour integral about C of an analytic function is zero. Actually, this is not quite right because Green’s Theorem requires that the derivatives of u and v be continuous within the region D. This is a stronger condition than what is necessary for an integral of an analytic function defined within and on a closed contour C to vanish. With this weaker condition, we have the fundamental result we will be using in this book: Cauchy’s Theorem (which we will not prove here).

1

This is a necessary but not sufficient condition as noted in P. Morse and H. Feshbach, Methods of Theoretical Physics, Feshbach Publishing LLC (2005), Sec. 4.2.

1.4 Analytic Functions

21

Theorem 1.3 A function analytic on and within a closed contour C satisfies the following.  dz f (z) = 0 (1.96) C

The converse is also true: if an integral of a function over a closed contour vanishes, then the function is analytic; this is Morera’s Theorem and is also not proven here. As an example, note that the real (or imaginary) part of an analytic function is not analytic. For example, 

2π dz x = i

|z|=1

dθ eiθ cos θ 0

2π

2π dθ sin θ cos θ + i

=− 0

dθ cos2 θ 0

= iπ = 0

(1.97)

Of course, we will be more interested in functions that are meromorphic inside of of closed contour C; that is, functions that are almost analytic except at a few isolated poles inside C. For example, consider a function f with exactly one pole inside C. If we wished to maintain the truth of Cauchy’s Theorem, we could deform the contour C into a new contour C  that detoured around the pole so as to define a new region D  in which the pole is excluded. We will be doing this with great frequency throughout the book, and not just in the presence of poles but also branch points. But as we will see with the residue Theorem later, we need not exclude poles from the contour to see interesting results. Consider the Cauchy Integral formula. Theorem 1.4 A function analytic on and within a closed contour C satisfies the following with the complex number (point) a lying inside the contour C. 1 f (a) = i2π

 dz C

f (z) z−a

(1.98)

To see why, deform the contour C into a small circle of radius  about a; the the contour integral is  2π   f (z) = i dθ f a + eiθ dz (1.99) z−a C

0

Because f is analytic inside C and in particular in the vicinity of a, we may evaluate the integral in the limit as  → 0. Cauchy’s Integral Theorem and Cauchy’s Theorem may be used to derive some other fascinating and useful results. For example, consider the Argument Principle.

22

1 Review of Foundational Concepts

Theorem 1.5 A function f meromorphic on and inside of a closed contour C, where f has Nz zeroes and N p poles throughout the interior of C, satisfies the following. 1 i2π

 dz C

f  (z) = Nz − N p f (z)

(1.100)

To see this, consider the function f (z) = (z − z 0 ) Nz (z − z 1 )−N p g(z)

(1.101)

where g is analytic in and on C and have no zeros in and on C. Then Np Nz f  (z) g  (z) = − + f (z) z − z0 z − z1 g(z)

(1.102)

The Argument Principle follows from application of Cauchy’s Theorem to the last two terms and the Cauchy Integral Formula to the first two terms. Note that the analysis with distinct zeroes and poles follows in a similar fashion. The Argument Principle may be used to prove Rouché’s Theorem, which may be stated as follows. Theorem 1.6 If f and g are analytic in and on a contour C, and if |g| < | f | on C, then f and f + g have the same number of zeroes inside of C. A proof of Rouché’s Theorem will not be given here. We will, however, demonstrate its use. For example, the number of roots of the function h(z) = z 8 + 15z 4 + 7z + 1 within |z| < 2 is eight, because the function f (z) = z 8 has an absolute value on |z| = 2 (256) that is greater than the maximum absolute value of the function g(z) = 15z 4 + 7z + 1 (255) on |z| = 2, and obviously z 8 has eight zeroes within |z| = 2.

Problems 1.1 Evaluate the following sum. ∞  k=0

1.2

k k 4 − 3k 2 + 1

(a) Evaluate the following sum. ∞  (2k + 1)Hk  2 k + k2 k=1

1.4 Analytic Functions

23

(b) Can you state and demonstrate a pattern from the above problem and the summation by parts example in the text? 1.3 Evaluate the following sum. ∞  k=0

Hk (k + 1)3

1.4 Use Green’s Theorem to find the area of an arbitrary polygon, specified by N vertices (x1 , y1 ), (x2 , y2 ),…,(x N , y N ). (This is sometimes called the "Shoelace Theorem.")

Chapter 2

Evaluation of Definite Integrals I: The Residue Theorem and Friends

Abstract Many students cite using the residue theorem to evaluate real definite integrals and sums as the main factor in their interest in Complex Analysis. This chapter will review the residue theorem of course. But the residue theorem is only useful in the presence of poles and no branch points. To attack very difficult integrals, we will need to consider integrands with poles and branch points, and occasionally the poles may be essential. Accordingly, we seek a more methodical treatment of applying complex integration to evaluation of real definite integrals and sums. Such a treatment is presented here. We will then present several examples. Ultimately, we will demonstrate a unified approach to using complex integrals to evaluate real definite integrals and sums, regardless of the type of singularities occurring in the complex integrand.

2.1 Poles The point of this section is to develop a unified framework for thinking about the evaluation of contour integrals in the presence of any sort of singularity, whether a pole or a branch point. (We will deal with essential singularities, which are simply poles of arbitrarily large order, in another section.) In a first course on complex analysis, branch points in contour integrals are treated as something weird and alien and, well, other. We can include the poles inside the contour and use the Residue Theorem, but we have to resort to deforming the contour (somehow) to deal with branch points. Here, we are going to take things a bit differently. And slowly. We will slow down the process of integrating in the complex plane so we can learn a little more about the meaning of the Residue Theorem. For example, we will examine a very well-known integral commonly seen in an application of the Residue Theorem in a first treatment. Here, however, we will construct a contour that avoids the pole usually included in the contour so we can learn details about the Residue Theorem that are neglected in a first treatment. These details will not only enable us to understand the mechanisms behind the Residue Theorem, but they will also put branch point singularities into a new focus. The result will be a unified treatment of any singularity using contour integration. © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 R. Gordon, Complex Integration, Undergraduate Lecture Notes in Physics, https://doi.org/10.1007/978-3-031-24228-1_2

25

26

2 Evaluation of Definite Integrals I …

After that, we will not only see old friends in a new light, but we will also attack new problems to which many would not think to apply complex integration techniques.

2.1.1 Cauchy’s Theorem in the Presence of a Pole As promised, we will motivate our discussion of the Residue Theorem from an in-depth treatment of a relatively simple integral. Of course, we will end with the foregone conclusion that contour integrals with poles can be treated with the Residue Theorem. What’s different, however, is that we will take the time to understand the mechanism through an in-depth treatment of a fairly simple integral, one that virtually every student of Complex Analysis has encountered. To begin, we will demonstrate how one may use Cauchy’s Theorem to show that ∞ −∞

dx =π 1 + x2

(2.1)

Yes, this is almost trivially evaluated using the substitution x = tan θ . And applying the Residue Theorem using a semicircular contour in the upper (or lower) half plane is about as trivial. But these are almost robotic, not even semiconscious, evaluations. What else needs to be said? A lot. Rather than draw a simple semicircle around the pole and state the Residue Theorem to evaluate the integral—you know, like zombies, we will exclude the pole at z = i with a detour. We will use Cauchy’s Theorem to evaluate the integral on the LHS of Eq. (2.1). To do so, we will need to express the contour integral in Eq. (2.2) parametrically at each portion of the contour C R . We will find that evaluating definite integrals like that in Eq. (2.1) by complex integration involves the following steps: 1. Determine an analytic integrand based on a contour, and 2. determine a contour based on the analytic integrand, such that the contour and integrand that result produce the definite integral to be evaluated. Easy, right? In this spirit, we should define a contour integral that will produce the integral in Eq. (2.1). In this case, obviously if we use the integrand of the definite integral in Eq. (2.1), simply expanded to the complex plane, then that definite integral falls out if part of the contour coincides with the real axis. On the other hand, we need to exclude any singularities from inside the contour so the integrand is analytic in and on the contour C. Again, rather than simply assume the Residue Theorem is true, we should deform the contour around the singularity, i.e., pole at z = i. Because Cauchy’s Theorem requires a closed contour, the deformation must be continuous.

2.1 Poles

27

Table 2.1 Components of the contour C R Contour segment label

Contour segment

γ1

γ4

x ∈ [−R, − A ]     y ∈ − A , − A + i 1 −  2R −  2A  A A π φ ∈ 3π 2 − arcsin  R , arcsin  R − 2      y ∈  A + i 1 −  2R −  2A ,  A

γ5

x ∈ [ A , R]

γ6

θ ∈ [0, π ]

γ2 γ3

Fig. 2.1 Semicircle contour with a pole interior to the contour excluded

That contour integral is as follows.

CR

dz 1 + z2

(2.2)

where C R is the following contour: To stress, we will be working at a far more basic level, one that is instructive and provides additional insights into the computation. We will be hitting the pause button a lot, looking for those little details that provides whole new perspectives. We begin by defining sections of the contour C in Table 2.1. Let’s pause and assess the situation. In Fig. 2.1, we have defined the detour around the pole more generally than is usually presented. As shown in Fig. 2.1, the detour is defined by, not one, but two parameters:  R , the radius of a circular piece around the pole at z = i, and  A , the half-width of a gap surrounding the origin on the real axis. Note that there is nothing preventing us from using, e.g., en ellipse or square rather than a circle around the pole, but there is no need to introduce unnecessary complexity at this point.

28

2 Evaluation of Definite Integrals I …

In Table 2.1, each segment is defined by a variable of integration having values within the respective interval. Accordingly,

 dz = 1 + z2 k=1 6

CR



dz 1 + z2

γk

From Table 2.1, we may express the integral over each contour segment in terms of its parametrizing variable:  γ1

 γ2

 γ3

 γ4

 γ5

 γ6

dz = 1 + z2

− A −R

dx 1 + x2 √2 2   R − A

(2.3)

1−

dz =i 1 + z2

0

dz = i R 1 + z2

− π2 +arcsin ( A / R )



dφ 3π 2

−arcsin ( A / R )

0 dz =i 1 + z2 √2 R A

dz = iR 1 + z2

eiφ

2 1 + i +  R eiφ

dy 1 + ( A + i y)2

(2.4)

(2.5)

(2.6)

 R − 2A

1−

dz = 1 + z2

dy 1 + (− A + i y)2

dx 1 + x2 π dθ 0

eiθ 1 + R 2 ei2θ

(2.7)

(2.8)

It is useful to understand each of the above parametrizations, including the geometry of the integration limits. Although it seems like we are introducing a plethora of details to solve a simple problem, one should keep in mind that the point is to learn new details about the process of using a contour integral and Cauchy’s Theorem to evaluate a simple definite integral. When we have discovered our sea legs, we will not need so much detail to make progress. Now that we have written out the integrals for each portion of the contour C, we may determine their contributions to the contour integral as R → ∞ and  A → 0. It is important to account for symmetries in the expressions above; these will guide how we combine and manipulate these components of the complex integral in Eq. (2.2). For example, combining the integrals in Eqs. (2.3) and (2.7) produces, in the limit as R → ∞ and  A → 0, the following expression:

2.1 Poles

29

 γ1 +γ5

⎛  ⎞ A ∞ dz d x d x ⎠ = lim ⎝ +  A →0 1 + z2 1 + x2 1 + x2 −∞

(2.9)

A

Perhaps it is unclear why Eq. (2.9) is written that way, rather than that which simply reproduces the integral on the LHS of Eq. (2.1). Of course, in this case the RHS of Eq. (2.9) is equal to the LHS of Eq. (2.1). We will see later, however, that this is not necessarily the case. For example, when the contour C passes through a simple pole, the limit in Eq. (2.9) may exist. In that case, the limit is called the Cauchy principal value of the integral. We will revisit this concept many times in examples that follow. Next, combining the integrals in Eqs.(2.4) and (2.7) produces the following:  γ2 +γ4

√2 2   R − A  dy

1−

dz =i 1 + z2

0

1 1 + ( A − i y)2

−

1



1 + ( A + i y)2

√2 2   R − A y dy

2 1 − y 2 − 4 2A y 2

1−

= −4 A

(2.10)

0

Does the integral in Eq. (2.10) vanish, as we expect, as  A → 0? We suspect it must, although there are a few factors that may make us doubt this. For example, setting  A and  R to zero results in a situation in which we have zero times a singular integral, or an indeterminate quantity. In this case, we can get around this bump in the road by evaluating the integral on the RHS of Eq. (2.10) exactly as is; this exact evaluation produces: 

dz 1 =−  2 γ2 +γ4 1 + z 2 1 +  2A  ⎤ ⎡   δ(2 − δ) + 2 2A + 2 A 1 +  2A 1 + 2 2A − 2 A 1 +  2A ⎥ ⎢  ⎥ log ⎢   ⎦, ⎣ 2 2 2 2 1 + 2 A + 2 A 1 +  A δ(2 − δ) + 2 A − 2 A 1 +  A

(2.11)

 where δ =  2R −  2A . That said, it should be clear that, as  A → 0, the RHS of Eq. (2.11) approaches zero, independent of the value of  R . That is, the integrals along the paths leading to and from the contour portion that encircles the pole is zero. This is not always the case, as we will see later when we discuss branch points.

30

2 Evaluation of Definite Integrals I …

Let’s pause again. Why bother with any analysis of the integrals up and back along the detours when it is clear they will vanish? The answer lies in the idea of unifying the way we treat poles and branch point singularities. While it is clear that, for an isolated pole, the integrals up and back along the detour will cancel each other out, we will see that for branch point singularities, they will not. So, it is prudent to set up the framework we will need to understand how the analysis to follow works for branch points. We may also show that the integral over the contour portion γ6 vanishes in the limit as R → ∞. This is because of the following fact that should have been encountered in a first treatment: Theorem 2.1 ML Inequality        dz f (z) ≤ M L    

(2.12)

γ

where M = max | f (z)| , and z∈γ

L = length(γ ).

In keeping with the informal tone, we will not prove the ML Inequality here although reviewing the proof from that first course is a good idea. (A good way to think about the validity of the ML Inequality is to think about applying the Cauchyover γ6 in Schwartz inequality to the integral over γ6 .) In the case of the integral

2 over γ6 occurs Eq. (2.8), the maximum absolute value of the integrand 1/ 1 + z

at z = i R and is 1/ R 2 − 1 . The length of γ6 is π R. Accordingly,  γ6

dz = iR 1 + z2

π dθ 0

πR , ≤ 2 R −1

eiθ 1 + R 2 ei2θ (2.13)

and the integral over γ6 vanishes as R → ∞. At this point, we have demonstrated that the integrals over γ2 , γ4 , and γ6 vanish as  A → 0 and R → ∞. The contour integral is then equal to the integral over γ1 and γ5 , which together combine to form the original real integral we seek, and over γ3 , which we need to evaluate.

2.1 Poles

31

Per Eq.(2.5):  γ3

dz = i R 1 + z2



− π2 +arcsin ( A / R )

dφ 3π 2

−arcsin ( A / R )

− π2 +arcsin ( A / R )



= i R

dφ 3π 2

1 = 2

−π 2

−arcsin ( A / R )

eiφ i2 R eiφ +  2R ei2φ

+arcsin ( A / R )



dφ 3π 2

eiφ

2 1 + i +  R eiφ

−arcsin ( A / R )

dφ 1 − i 21  R eiφ

(2.14)

We do not set  R → 0 here, or even take the limit as  R → 0. First, in the limit as  A → 0, the ratio  A / R goes to zero. Next, imagine Taylor expanding the integrand in Eq. (2.14) in powers of  R . For the zeroth order term, the integral is equal to 2π . Higher powers, however, are simply integrals of some power of eiφ over a complete cycle, or zero. Accordingly, π

− 2 3π 2

dφ = −2π 1 − i 21  R eiφ

(2.15)

The integral in Eq. (2.15) is negative because the path γ3 is traversed clockwise. Putting together the above results, we find that

C

dz = 1 + z2

∞ −∞

dx −π 1 + x2

(2.16)

Let’s pause once more. We have reduced the original contour integral in Eq. (2.2) to the very nice and simple expression in Eq. (2.16) by taking limits as R → ∞ and  A → 0. Note that we did not need to take  R → 0; rather, the only restriction on  R is that | R | < 1. There are two reasons for this restriction: (1) we need to be able to imagine a Taylor series expansion of the integrand in Eq. (2.15), and (2) otherwise, the geometry would become ridiculous. Does this fact have any useful implications for us? It is not immediately clear right now, but perhaps we will gain more insight from the next example. The final step is, of course, applying Cauchy’s Theorem. Because the contour C enclosed no poles and the integrand is analytic on and inside C, the contour integral is zero. Accordingly, we reproduce the result in Eq. (2.1). Perhaps we set a record for the longest derivation of the arctangent integral in the history of mathematics. Again, as stated earlier, the point is not the result which we

32

2 Evaluation of Definite Integrals I …

knew already but the precise story told on the journey to the result. For example, we learned that the entire contribution to the contour integral in the various limits came from the integral over the circular arc around the pole, γ3 and the integral over the real axis γ1 and γ5 . The contributions from the detour to the arc around the pole, γ2 and γ4 , canceled each other out. Let’s take a look at this in general.

2.1.2 The Residue Theorem is a Shortcut Again, the key step in the above evaluation of the integral in Eq. (2.1) is the cancellation of the integrals over the contour portions γ2 and γ4 in Eqs. (2.4) and (2.6), respectively, as expressed via Eq. (2.11). That is, the detour introduced in the real integral needed to avoid including the pole at z = i did not generate a contribution to the contour integral in Eq. (2.2). This is true in general. Consider a closed contour enclosing a pole of a meromorphic (i.e., analytic except at a finite set of poles) function. To use Cauchy’s Theorem in evaluating the contour integral, we need to exclude the pole which means including a detour to and around the pole. For example, the detour to the pole can include of a pair of lines, each a distance  A from a normal to the contour that includes the pole. An important thing to note is that the integrals over the lines segments of the detour to the pole (e.g., the straight lines in Fig. 2.1) cancel each other out and hence contribute nothing to the contour integral. We will see, however, that this non-contribution to the contour integral of such a detour is only true for poles. For poles, the contribution to the contour integral occurs at the detour around the singularity. This is why the Residue Theorem works. Consider a contour integral over a contour C of a meromorphic function f having poles z k , k ∈ 1, 2, . . . , N inside C. Then because the detours out to the poles contribute nothing to the contour integral, we may write

N  

2π

dz f (z) = i R

dφ eiφ f z k +  R eiφ

(2.17)

k=1 0

C

Because f is meromorphic, it has a Laurent expansion in a neighborhood of the pole at z = z k : f (z) =

∞ 

an(k) (z − z k )n

(2.18)

n=−∞

Note that in Eq. (2.18), there is a separate set of Laurent expansion coefficients an(k) corresponding to each pole z k . That is, the (k) indicates the pole to which the Laurent expansion corresponds. We may write, in the neighborhood of each pole, z − z k =  R eiφ , which brings us to the pièce de résistance:

2.1 Poles

33



N  

2π

dz f (z) = i

dφ

k=1 0

C

=i

∞ N  

∞ 

an(k)  (n+1) ei(n+1)φ R

n=−∞

an(k)  (n+1) R

k=1 n=−∞

2π dφ ei(n+1)φ

(2.19)

0

It should be crystal clear that the integrals on the RHS of Eq. (2.19) are all zero except when n = −1, and we obtain the much-simplified expression

dz f (z) = i2π

N 

(k) a−1

(2.20)

k=1

C

(k) represents the residue Look familiar? It should be apparent that the coefficient a−1 of f at the pole z k . Equation (2.20) is accordingly an expression of the Residue Theorem: (k) Res f (z) = a−1 z=z k

(2.21)

when f has a Laurent expansion as in Eq. (2.18). We may formalize this in a statement of the Residue Theorem. Theorem 2.2 The integral of a function f over a closed contour C is equal to i2π times the sum of the residues of function f at the poles z k that are located in the interior of C. For the case exhaustively analyzed above, the residue of the integrand about the pole inside the contour C at z = i is computed by expanding the integrand in a Laurent series about that pole. Letting ζ = z − i: 1 1 1 1 . = = 2 2 1+z 1 + (i + ζ ) i2ζ 1 − iζ /2 Accordingly, Res z=i

1 1 = . 1 + z2 i2

There is a simple expression for the residue of a function with a simple pole. Let f (z) = p(z)/q(z), with p being analytic in the interior of a contour C and q having a set of zeroes z k , k ∈ 1, 2, ..., N in the interior of C. Then

34

2 Evaluation of Definite Integrals I …

Res z=z k

p (z k ) p(z) =  . q(z) q (z k )

(2.22)

There is a similar, albeit more complicated, formula when there is a double pole. In general, however, experience has shown that for poles that are nonsimple, the residues are best computed by expansion into a Laurent series.

2.2 Branch Points We have demonstrated that the Residue Theorem is in effect a shortcut because of the cancellation of the integrals up and back along the detour to a pole. This cancellation allows us to replace a single contour integral over a contour in which poles are located with small, simple contours about each pole. When the singularities are poles only, the Residue Theorem allows us to focus only on the pieces contributing to the contour integral. Nevertheless, as we have been more than hinting, poles are just one case of a singularity and the cancellation along the detour is only true for poles. Branch point singularities, as well as the cuts defined to keep integrands single-valued, need to be √ kept from the interior of closed contours. But why? Surely z, for example, is finite at z = 0, right? √ As discussed in the previous chapter, z is not analytic at z = 0 because its derivative is not finite there. And, as also discussed, functions are multivalued; e.g., √ −1 can be i when we represent −1 = eiπ or −1 when −1 = e−iπ Multivaluedness is a weird problem that baffles a lot of students. Multivalued √ functions can be represented by Riemann surfaces which in the case of z looks like a two-level Lazy Susan that joins back with itself after two laps around the origin. For functions like z α , α irrational and log z, the Lazy Susan just keeps spiraling around the origin, up or down, forever, unlike city garages. Riemann surfaces are really cool ways of visualizing multivaluedness in general but are utterly useless for our purposes here. Instead, we deal with multivaluedness in a way that is kind of unsatisfying but is perfect for our purposes: a branch cut extending from a branch point out in some direction. Generally, a branch cut can be any curve extending out from a branch point but here we will only need a line. The use of a branch cut is to enforce a single-valuedness to being a multivalued function to analyticity. To go back to the √ z example, if we make the negative real axis a branch cut, then away from that branch cut, arg z ∈ [−π, π ). Note we will only need a single branch cut to define an analytic function outside the branch point and branch cut. With respect to branch point singularities, we will discover a very interesting fact: the integrals up and back along the detour to the singularity on the contour—usually along the branch cut—do not cancel. The difference between those integrals provides the contribution to the contour integral while the detour around the singularity

2.2 Branch Points

35

contributes nothing. Looked at it this way, poles and branch point singularities are a ying and yang, complements to each other. Of course, some singularities are both poles and branch points. We will visit examples of this, but first, let’s recall the evaluation of contour integrals having integrands with poles. Because for poles the detour to a pole contributes nothing to the contour integral, we now have reason in everyday evaluation to simply ignore the detour to the pole. In doing so, we may consider the contribution to the contour integral by integrating a full circle of radius  R about the pole. This artificial construct is merely a shortcut: we do not need to break up a contour for a detour to a pole, as the detour contributes nothing to the contour integral. So, let’s take a look at an example of computing the value of an integral whose complex counterpart has integrand with a branch point singularity. Let’s compute the value of the following integral: ∞ −∞

log 1 + x 2 dx (1 − i x)2

(2.23)

If we had to guess the form of the contour integral from which we may determine the integral in Eq. (2.23), we may try the following:



log 1 + z 2 dz (2.24) (1 − i z)2 C

Note that the integrand in Eq. (2.24) has branch point singularities at z = ±i. So a first guess would have us consider the contour C to be that defined in Fig. 2.1 and Table 2.1. Unfortunately, that contour will not work here. The reason is that the

function log 1 + z 2 has a branch cut over {z : z = it, t ∈ [1, ∞)}: the contour C as defined in Fig. 2.1 and Table 2.1 will cross the branch cut. So let’s approach the branch point along either side of that branch cut; our contour C is the defined in Fig. 2.2 and Table 2.2. In Table 2.2, each segment is defined by a variable of integration having values within the respective interval. Accordingly,

Fig. 2.2 Semicircle contour with a pole interior to the contour excluded

36

2 Evaluation of Definite Integrals I …

Table 2.2 Components of the contour C R Contour segment label

Contour segment

γ1

x ∈ [−R, R]

γ2

γ5

θ ∈ [0, π/2 − arcsin RA ]     y ∈  A + i R,  A + i 1 +  2R −  2A  φ ∈ π2 − arcsin  RA , arcsin  RA − 3 π2      y ∈ − A + i 1 +  2R −  2A , − A + i R

γ6

θ ∈ [π/2 + arcsin

γ3 γ4

CR

A R , π]

6   log 1 + z 2 log 1 + z 2 dz = dz (1 − i z)2 (1 − i z)2 k=1 γk

From Table 2.1, we may express the integral over each contour segment in terms of its parametrizing variable:  γ1

 γ2

 γ3

 γ4

 γ5

 γ6

R log 1 + z 2 log 1 + x 2 dz = dx (1 − i z)2 (1 − i x)2

(2.25)

−R



2

log 1 + z dz (1 − i z)2

= iR 0

log 1 + z 2 dz =i (1 − i z)2

√2 2

  R − A log 1 + ( A + i y)2 dy (1 − i ( A + i y))2 R



2

log 1 + z dz (1 − i z)2



π 2

−arcsin ( A / R )

R =i

√

1+

(2.27)



2 log 1 + i +  R eiφ dφ eiφ

2 (2.28) 1 − i i +  R eiφ

− 3π 2 +arcsin ( A / R )

2

log 1 + z dz (1 − i z)2

(2.26)

1+

log 1 + z 2 dz = i R (1 − i z)2

log 1 + R 2 ei2θ dθ

2 1 − i Reiθ

π/2−arcsin  ( A /R)

dy

log 1 + (− A + i y)2 (1 − i (− A + i y))2

 2R − 2A

log 1 + R 2 ei2θ = iR dθ

2 π/2+arcsin ( A /R) 1 − i Reiθ 

(2.29)

π

(2.30)

2.2 Branch Points

37

The integrals over γ2 and γ6 lead to the following integral when we take  A → 0: π iR 0

log 1 + R 2 ei2θ dθ

2 1 − i Reiθ

(2.31)

We again invoke the ML-Inequality to deduce that the integral in Eq. (2.31) is

bounded by 2π R log R/ R 2 − 1 as R → ∞. Hence, the integrals over γ2 and γ6 do not contribute to the contour integral in this limit. The integral over γ4 —the detour around the branch point—may be evaluated as  A → 0. Note the integration limits—ultimately from −3π/2 to π/2 as a → 0— correspond to two adjacent representations of arg i. Again, because we traverse γ4 in a clockwise manner, the traversal on the upward traversal to the left of the branch cut has argument 2π less than the upward traversal on the right side. Generally speaking, many of the traversals up and back will have this decrease of 2π in the argument. Let’s now evaluate the integral over γ4 :  γ4



2

 − 3π +arcsin ( A / R ) log 1 + i +  R eiφ 2 log 1 + z 2 dz = i R dφ eiφ

2 π (1 − i z)2 −arcsin ( / ) 1 − i i +  R eiφ A R 2  ∼ i R

− 3π 2 π 2

dφ e

iφ



log i2 R eiφ +  2R ei2φ ( A → 0)

2 2 − i R eiφ

(2.32)

In contrast with the situation involving a pole, we take  R → 0 for the branch point singularity so that the contribution, i.e., the integral over γ4 , vanishes. If we do not take  R → 0, then the evaluation of the remaining integrals becomes a bit more complicated than many of us care for. (NB: we could have taken  R → 0 for the pole but we didn’t have to do that.) In the situation with the pole, the integral over γ4 is determined by the residue of the integrand at that pole. With a branch point, in contrast, the integral over a contour around the branch point merely serves as a connection between paths along either side of the associated branch cut. How do we evaluate the integrals over γ3 and γ5 , i.e., the detour to the branch point? We first consider the limit as  A → 0:  γ3

1+

 R log 1 + z 2 log 1 − y 2 dz →i dy (1 − i z)2 (1 + y)2

(2.33)

R

Note that 1 − y 2 < 0 in the integration interval. This is the source of the branch

2 cut along axis above the 2the imaginary

2 branch

point z = i. So on γ3 , log 1 − y = log − y − 1 = log (−1) + log y − 1 . The question is, what is log (−1)? The answer by itself it doesn’t really matter. We could set −1 = eiπ along γ3 or −iπ e or ei3π . What does matter, however, is what log (−1) is on γ5 , because log (−1) on γ5 differs from log (−1) on γ3 by −2π , as mentioned before. The reason for this

38

2 Evaluation of Definite Integrals I …

is that the transition from γ3 to γ5 through γ4 in a clockwise sense. In the limit as  A → 0, the transition forces a change in phase by a full cycle in the clockwise sense, or −2π . Accordingly,  γ3 +γ5

 

∞ log y 2 − 1 − iπ log 1 + z 2 log y 2 − 1 + iπ dz =i dy − (1 − i z)2 (1 + y)2 (1 + y)2 1+ R

∞ = 2π 1

dy (1 + y)2

=π

(2.34)

That is, the contribution to the original contour integral in Eq. (2.24) comes from the detour to the branch point and not from the detour around the branch point. This is the complement of the situation with poles that led to the Residue Theorem. With branch points, however, we do not have an analogy to the Residue Theorem because of the lack of a Laurent expansion with respect to a branch point. As we will see in the following sections, evaluating definite integrals that involve a branch point in the generating contour integral produces—hopefully—other integrals that are easier to evaluate, as happened above. Let’s put the above together to finally evaluate the original integral in Eq. (2.23). We set R → ∞ and let  A → 0 then  R → 0. The result for the contour integral is

dz C

∞ log 1 + z 2 log 1 + x 2 = d x +π (1 − i z)2 (1 − i x)2

(2.35)

−∞

Therefore, by Cauchy’s Theorem: ∞ −∞

log 1 + x 2 dx = −π (1 − i x)2

(2.36)

Of course, this result can be derived in other, real-based ways. But it works in this book as an example of contour integration in the presence of a branch point singularity. There will be other examples of this—integrals that are readily evaluated using real methods that are not quite so readily evaluated using complex methods— presented unapologetically. The point is to explore the possibilities of complex integration rather than find snappy solutions to integral evaluation. The next few sections will go through examples of several types of integrals that may be attacked using the sort of thinking outlined in this section. No, we will not go through analyses as painstaking as those presented here. But we will begin to understand the types of contours—and integrands—that may be used to evaluate some definite integrals and sums.

2.3 Poles and Branch Points

39

2.3 Poles and Branch Points We will occasionally encounter a situation in which the contour integral we need to evaluate a definite integral has both poles and branch points. The subsequent analysis when the poles and branch points are separate is straightforward when the poles and branch points are separated. But it’s when a pole and a branch point occur at the same point, however, that we will develop something new (OK, new to us) and which will amaze our friends and confound our enemies. To begin, in the spirit of the previous branch point example, let’s evaluate the following definite integral: ∞ −∞

log a 2 + x 2 dx b2 + x 2

(2.37)

To evaluate that integral, we introduce the following contour integral:

C

log a 2 + z 2 dz b2 + z 2

(2.38)

where C is the contour shown in Fig. 2.3. Beginning here, we will display less of the contour definition in tables as we did for the integrals in Sec. (2.1); the table-building should be performed by the reader until the reader feels comfortably proficient in defining the integrals. Rather, the more interesting skill lies in simply defining a contour that will produce the real, definite integral whose value we seek. In Eq. (2.37), the integrand has not only branch point at z = ±ia but also poles at z = ±ib. Accordingly, we should nominally design the contour C to detour around each of these singularities as in Fig. 2.3. Along these lines, Fig. 2.3 shows the contour C. The question is, which singularity on the plot is the pole and which is the singularity? That will depend on whether a < b, a = b, or a > b. We will consider each of these cases in turn because each case provides interesting insights in their own right.

Fig. 2.3 Semicircle contour with a pole and branch point interior to the contour excluded

40

2 Evaluation of Definite Integrals I …

a > b In this case, we encounter the branch point first before we encounter the pole. This could make the computation of the integral in Eq. (2.38) a little tricky because we will have to monitor the phase of the integrand carefully as we traverse the pole at z = ib. And while we will soon see circumstances where we will have to perform such monitoring, we have an out: the shortcut that is the Residue Theorem. Accordingly, we instead define the contour C to be that shown in Fig. 2.2 instead of Fig. 2.3. As we did in the previous section, let’s set up the contour integral in detail.  γ1

 γ2

R log a 2 + z 2 log a 2 + x 2 dz = dx b2 + z 2 b2 + x 2

(2.39)

2 π/2 2 i2θ log a 2 + z 2 iθ log a + R e dz = i R dθ e b2 + z 2 b2 + R 2 ei2θ

(2.40)

−R

0

√2 2

  R − A log a 2 + ( A + i y)2 dy (2.41) b2 + ( A + i y)2 γ3 R 

2  −3π/2+arcsin

 ( A / R )  log a 2 + ia +  R eiφ log a 2 + z 2 dz = i R dφ eiφ

2 b2 + z 2 b2 + ia +  R eiφ 

log a 2 + z 2 dz =i b2 + z 2

a+

γ4

π/2−arcsin ( A / R )

 R log a 2 + z 2 dz =i b2 + z 2 √2 γ5 a+

dy

(2.42)

log a 2 + (− A + i y)2

(2.43)

b2 + (− A + i y)2

 R − 2A

2 π  2 i2θ log a 2 + z 2 iθ log a + R e dz = iR dθ e b2 + z 2 b2 + R 2 ei2θ

γ6

(2.44)

π/2

As we did in the previous section (and pretty much from here on in), we consider the limits  A → 0 and R → ∞. In this limit, the sum of the integrals over γ2 and γ6 vanishes as 2π log R/R as R → ∞. The sum of the integrals over γ3 and γ5 is 

∞ i

dy a+ R



∞ log y 2 − a 2 + iπ log y 2 − a 2 − iπ dy − = −2π 2 2 2 2 2 y −b y −b y − b2 a+ R

(2.45) We may now consider the limit as  R → 0. The integral over γ4 , around the branch point, may be shown to vanish as i2π  R log  R /(a 2 − b2 ) in this limit. The contour

2.3 Poles and Branch Points

41

integral is then equal to

C



∞ ∞ log a 2 + z 2 log a 2 + x 2 dy dz = dx − 2π 2 2 2 2 2 b +z b +x y − b2 −∞

(2.46)

a

The contour integral above is also equal to, by the Residue Theorem, the residue of the integrand at the pole z = ib. Evaluating the second integral on the RHS, we have ∞ −∞

  log a 2 + z 2 log a 2 + x 2 a+b π = i2π Res dx − log z=ib b2 + x 2 b a−b b2 + z 2

(2.47)

Because the pole of the integrand within the contour C is simple, the residue computation is straightforward:



log a 2 + z 2 1 Res log a 2 − b2 = 2 2 z=ib b +z i2b

(2.48)

And we have the integral we seek: ∞ −∞

log a 2 + x 2 2π dx = log (a + b) 2 2 b +x b

(2.49)

Actually, this result is only valid when a > b > 0. Clearly, when a and/or b is negative, the result does not change. Therefore we may write, for all real a, b: ∞ −∞

log a 2 + x 2 2π log (|a| + |b|) dx = 2 2 b +x |b|

(2.50)

a b, as the pole comes before the branch point. Now, some wiseguy in the back may retort that we could approach the branch point from the real axis, as in Fig. 2.1. If we could use the contour in Fig. 2.1, then we could apply the residue theorem as we did for the case a > b. Unfortunately, and if that wiseguy paid attention to the previous example, they would know that the integrand of the contour integral has a branch cut along the imaginary axis z = i y for |y| > a. The contour in Fig. 2.1 is away from the branch cut and accordingly that contour is of no use in determining the integral.

42

2 Evaluation of Definite Integrals I …

So we will evaluate the contour integral based on the contour in Fig. 2.3. And as can be imagined at this point, there are a lot of moving parts and it’s gonna get messy:  γ1

R log a 2 + z 2 log a 2 + x 2 dz = dx b2 + z 2 b2 + x 2

(2.51)

−R

 log a 2 + z 2 dz = iR b2 + z 2

log a 2 + R 2 ei2θ dθ e b2 + R 2 ei2θ

π/2−arcsin  ( A /R)

iθ

γ2

0

(2.52)

√ b+  2R − 2A

2   2 log a + z log a 2 + ( A + i y)2 dz =i dy (2.53) b2 + z 2 b2 + ( A + i y)2 γ3 R 

 −π/2+arcsin

2 2 iφ 2  ( A / R )  2 log a + ib +  e R log a + z dz = i R dφ eiφ

2 b2 + z 2 b2 + ib +  R eiφ γ4

π/2−arcsin ( A / R )

√2 2

2    R − A 2 log a + z log a 2 + ( A + i y)2 dz =i dy b2 + z 2 b2 + ( A + i y)2 √2 2 γ5

(2.54)

a+

b−

 γ6

 γ7

 γ8

 γ9

log a 2 + z 2 dz = i R b2 + z 2

π/2−arcsin ( A / R )

log a 2 + z 2 dz =i b2 + z 2 √ a+

log a 2 + z 2 dz = i R b2 + z 2

(2.56) dy

log a 2 + (− A + i y)2 b2 + (− A + i y)2

 2R − 2A

(2.57)



2  log a 2 + ib +  R eiφ dφ eiφ

2 b2 + ib +  R eiφ

−3π/2+arcsin  ( A / R )

−π/2−arcsin ( A / R )

R log a 2 + z 2 dz =i b2 + z 2 √2 b+



2  log a 2 + ia +  R eiφ dφ eiφ

2 b2 + ia +  R eiφ

−3π/2+arcsin  ( A / R )

√ b−  2R − 2A 

 log a 2 + z 2 dz = iR b2 + z 2

γ10

 R − A

(2.55)

dy

log a 2 + ( A + i y)2

 R − 2A

π

π/2+arcsin ( A /R)

b2 + ( A + i y)2

log a 2 + R 2 ei2θ dθ e b2 + R 2 ei2θ iθ

(2.58) (2.59)

(2.60)

2.3 Poles and Branch Points

43

As before, we consider the limits  A → 0 and R → ∞. In this limit, the sum of the integrals over γ2 and γ10 vanishes as 2π R log R/ R 2 − 1 as R → ∞. The sum of the integrals along a first side of the imaginary axis is the sum of the integrals over γ3 and γ5  dz

b+ a+

 R  R log a 2 + z 2 log a 2 − y 2 log a 2 − y 2 = i dy + i dy b2 + z 2 b2 − y 2 b2 − y 2 ∞

γ3 +γ5

b+  R

=i ∞

b− R



2

log y 2 − a + iπ dy +i b2 − y 2

a+  R

b− R



log y 2 − a 2 + iπ dy b2 − y 2

(2.61) Similarly, the sum of the integrals along the opposite side of the imaginary axis is the sum of γ7 and γ9  γ7 +γ9

b−

∞  R log a 2 + z 2 log y 2 − a 2 − iπ log y 2 − a 2 − iπ dz =i dy +i dy b2 + z 2 b2 − y 2 b2 − y 2 a+ R

b+ R

(2.62) Combining Eqs. (2.61) and (2.62), we find the log terms in the integrand disappear and we get a fairly easy integral we can evaluate.  γ3 +γ5 +γ7 +γ9

b−

 R ∞ log a 2 + z 2 dy dy dz = 2π − 2π b2 + z 2 b2 − y 2 y 2 − b2 a+ R

(2.63)

b+ R

Didn’t that happen last time: when adding the contributions up and back along a branch cut, logarithms in the integrand cancel and we are left with a much simpler integral to evaluate. This is not guaranteed, but generally speaking when we add integrals up and back along each side of a branch cut we turn a set of difficult integrals into a much easier one. What we see here is the integration intervals skipping over the pole; the integral near the pole is treated separately. The astute reader will notice that we have not set  R → 0 and we are not ready to do so yet. We can however evaluate the above integrals and get   log a 2 + z 2

 dz γ3 +γ5 +γ7 +γ9

b2 + z 2

=

       2b −  R R b + a + R π π log − log + log b R b − a − R b 2b +  R

=

  (2b −  R )(b − a −  R ) π log b (2b +  R )(b + a +  R )

(2.64)

44

2 Evaluation of Definite Integrals I …

While this portion results in a finite limit as  R → 0, one can see that the intermediate steps produced terms in log  R , which canceled during simplification. This sort of thing where potentially divergent terms are generated and canceled will be a theme later, where the cancellation is less trivial. The integral over γ6 , i.e., around the branch point, clearly vanishes as  R → 0:

log i2 R aeiφ +  2R ei2φ

−3π/2 

dφ e

i R

iφ

b2 − a 2 + i2 R aeiφ +  2R ei2φ

π/2

(2.65)

which vanishes as −i2π  R log  R /(b2 − a 2 ) as  R → 0. Again, an integral immediately around a branch point is zero. Finally, the integral over γ4 and γ8 , about the pole at z = ib, is as follows:  γ4

−π/2 2

 2 i2φ 2 iφ log a 2 + z 2 iφ log a − b + i2 R be +  R e dz = i R dφ e b2 + z 2 i2 R beiφ +  2R ei2φ π/2

log b2 − a 2 − i2 R beiφ −  2R ei2φ + iπ dφ i2b +  R eiφ

−π/2 

=i π/2

=−

 γ8



π π2 log b2 − a 2 − i 2b 2b

( R → 0)

(2.66)

2  −3π/2 2 i2φ 2 iφ log a 2 + z 2 iφ log a − b + i2 R be +  R e dz = i R dφ e b2 + z 2 i2 R beiφ +  2R ei2φ −π/2

log b2 − a 2 − i2 R beiφ −  2R ei2φ − iπ dφ i2b +  R eiφ

−3π/2 

=i −π/2

=−



π2 π log b2 − a 2 + i 2b 2b

( R → 0)

(2.67)

Adding these, we see that the terms ±iπ from either side of the branch cut vanish. Accordingly,



log a 2 + z 2 π (2.68) dz = − log b2 − a 2 2 2 b +z b γ4 +γ8

2.3 Poles and Branch Points

45

and the contour integral results from putting all this together:

C



  ∞

log a 2 + z 2 log a 2 + x 2 b+a π π dz = dx − log − log b2 − a 2 2 2 2 2 b +z b +x b b−a b −∞ ∞

= −∞

log a 2 + x 2 2π log (a + b) dx − 2 2 b +x b

(2.69)

The assumption above was that b > a > 0. To make the result valid for a, b < 0, we replace with absolute values in the final result, by Cauchy’s Theorem: ∞ −∞

log a 2 + x 2 2π log (|a| + |b|) dx = 2 2 b +x |b|

(2.70)

So after all that, we ended up with the exact same result derived for the case a > b. The reason for this is that the phase shifts of the branch cuts cancel. Of course, we will see that this result holds up for a = b, because...well, it’s pretty plain at this point. But since we are math enthusiasts, we are going to work that case on its own separate merits. And in the process, we are going to encounter some new animals in our zoo. a = b In this case, the pole and branch point coincide at z = ia. This will separate the wheat from the chaff. Again, we define the contour C to be that shown in Fig. (2.2). Along these lines,  dz γ1

 γ2

 γ3

R log a 2 + z 2 log a 2 + x 2 = d x a2 + z2 a2 + x 2



2 π/2 2 i2θ log a 2 + z 2 iθ log a + R e dz = iR dθ e a2 + z2 a 2 + R 2 ei2θ

(2.72)



log a 2 + z 2 dz =i a2 + z2

(2.73)

 dz γ4

γ5

 dz

0  a+  2R − 2A



dy



log a 2 + ( A + i y)2

a 2 + ( A + i y)2 

2  −3π/2+arcsin  ( A / R ) log a 2 + ia +  R eiφ dφ eiφ (2.74)

2 a 2 + ia +  R eiφ R



log a 2 + z 2 = i R a2 + z2

 log a 2 + z 2 dz =i a2 + z2

γ6

(2.71)

−R

π/2−arcsin ( A / R )

R dy  a+  2R − 2A

log a 2 + (− A + i y)2 a 2 + (− A + i y)2

2

π 2 i2θ log a 2 + z 2 iθ log a + R e = i R dθ e a2 + z2 a 2 + R 2 ei2θ π/2

(2.75)

(2.76)

46

2 Evaluation of Definite Integrals I …

Again, the integrals over γ2 and γ6 vanish in the limit as R → ∞. We consider the integrals over γ3 and γ5 as  A → 0:

∞ log a 2 + z 2 log y 2 − a 2 + iπ = i dy a2 + z2 y2 − a2

(2.77)

∞ log a 2 + z 2 log y 2 − a 2 − iπ dz = −i dy a2 + z2 y2 − a2

(2.78)

 dz γ3

 γ5

a+ R

a+ R

Accordingly, the contribution to the contour integral from γ3 and γ5 is  γ3 +γ5

  ∞ log a 2 + z 2 2a +  R dy π dz = −2π = − log a2 + z2 y2 − a2 a R

(2.79)

a+ R

Note here that this contribution up and back along wither side of the branch cut diverges in the limit as  R → 0. If we hold any hope that our contour integration technique will produce the expected result, we expect the integral over γ4 to include a divergent term that will cancel the divergent term in Eq. (2.79). While it may seem weird to be working with divergent quantities, we will find that working with divergent portions of the contour integral has certain benefits. The main benefit of working with divergent pieces is that it provides an automatic check on the results: if after evaluation the contour integral we still have divergent pieces, we know there is a mistake somewhere. We expect the cancellation to occur in the evaluation of the integral over γ4 , which we evaluate here in the limit as  A → 0.  γ4

−3π/2

 2 i2φ iφ log a 2 + z 2 iφ log i2a R e +  R e dz = i R dφ e a2 + z2 i2a R eiφ +  2R ei2φ π/2

=

1 2a

1 = 2a

−3π/2 

dφ log i2a R eiφ

π/2 −3π/2 

π/2

 π  dφ i + log (2a R ) + iφ 2

π i 1 π2 − log (2a R ) + = −i 2a a 2a 2 π = − log (2a R ) a



9π 2 π2 − 4 4



(2.80)

2.3 Poles and Branch Points

47

And hopefully it is clear now that the divergences, in fact, do cancel! This happened for us in the case a < b except that the divergences canceled within the integrals along either side of the branch cut, so it was not as apparent. Accordingly, the contour integral is now equal to

C

∞ log a 2 + z 2 log a 2 + x 2 2π log (2a) dz = d x − a2 + z2 a2 + x 2 a

(2.81)

−∞

Applying Cauchy’s Theorem and then adjusting for the case a < 0, we arrive at the expected result: ∞ −∞

log a 2 + x 2 2π log (2|a|) dx = a2 + x 2 |a|

(2.82)

This section, it is hoped, has brought to the fore a couple of concepts that are not completely clear from a first treatment of complex integration techniques of evaluating real definite integrals. • Cauchy’s Theorem does not distinguish between singularites, and neither do we. If we attack problems with poles or branch points in the same way, then using Cauchy’s Theorem to evaluate definite integrals becomes easier to contemplate: all we need do is think about ways to construct closed contours that detour around the singularities. • The Residue Theorem, then, is essentially a shortcut we can take once we understand the structure of the singularities enough to construct a contour. In the case a > b, we were able to use the simpler contour once we determined that the pole could be treated as “inside” the contour. This shortcut works because we realize that there is no contribution to the contour integral along the detour portions of the contour. But by default, contours should detour around all singularities the same. In the chapter that follows, we will apply the teachings of this chapter to explore different types of definite integrals that may be evaluated using complex integration. Many of these will not be encountered on a first treatment of the subject.

Problems 2.1 Evaluate the following integral ∞ dx −∞

log x 2 + a 2 (a − i x)2

48

2 Evaluation of Definite Integrals I …

for the cases (i) a > 0, and (ii) a < 0. In both cases, the contour used should be closed in the upper half-plane. That is, in one of the cases there should be a pole and branch point at the same point. 2.2 Verify the endpoints of the integration intervals in (i) Table 2.1 and (ii) Table 2.2. 2.3 Verify Eq. (2.11). 2.4 Evaluate the following integrals using the contour-based analysis detailed in this chapter. (i)

∞ −∞

log x 2 + a 2 dx

2 x 2 + b2

for the cases (a) a > b, (b) a < b, and (c) a = b, (ii)

∞ −∞

for a > 0,



−3/2 d x x 2 + a2

Chapter 3

Evaluation of Definite Integrals II: Applications to Various Types of Integrals

Abstract In the previous section, we demonstrated that poles and branch points, rather than being two completely different cases that must be treated differently, are actually complementary in the contributions they provide to a contour integral. In this sense, the Residue Theorem is simply a shortcut because the integrals over the detours to the pole cancel. In the presence of branch points, these integrals do not cancel. This point of view allows us to consider cases avoided in first treatments of contour integration, namely, singularities that are both poles and branch points simultaneously. We now get to apply this point of view to some familiar types of integrals. This point of view also enables us to treat integrals that are usually not considered ones that anyone would want to evaluate using contour integration. In fact, the thought of using contour integration on some of these integrals may send chills down the spine. But using results from the previous chapter, we will see how this point of view pays dividends to those who treasure new ways of evaluating integrals. We will also illustrate applications to the evaluation of sums.

3.1 Integrands Defined Over [0, ∞) Many times we are asked to evaluate integrals having the following form: ∞ dx f (x) 0

where f : [0, ∞) → R is integrable over its domain. Note that f is usually not even and is therefore not continued over the negative reals. Accordingly, contours such as those in Figs. 2.2 and 2.3 will not be appropriate for this case. The contour we will be using for all of the integrals evaluated in this section is known as a keyhole contour and is defined as follows. Let’s begin with an example that isn’t particularly difficult but which result of its evaluation will leave us a bit bemused.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 R. Gordon, Complex Integration, Undergraduate Lecture Notes in Physics, https://doi.org/10.1007/978-3-031-24228-1_3

49

50

3 Evaluation of Definite Integrals II: Applications to Various Types of Integrals

Fig. 3.1 Keyhole contour excluding a branch point at the origin

The Mystery Symmetry So let’s evaluate the following integral. ∞ dx

x2

0

xα + 2x cos (πβ) + 1

(3.1)

It should not be difficult to show that α ∈ (−1, 1) for the integral to converge. For reasons that will be appreciated later, we will also restrict β ∈ (−1, 1). We consider the following contour integral over the contour C in Fig. 3.1. We will be assuming that A can be set to zero.  zα dz 2 z + 2z cos (πβ) + 1 C

where  γ1

 γ2

zα = dz 2 z + 2z cos (πβ) + 1

dx R

zα = iR dz 2 z + 2z cos (πβ) + 1

 dz γ3

R

zα = z 2 + 2z cos (πβ) + 1

2π d θ eiθ 0

R dx R

xα x2 + 2x cos (πβ) + 1

R2 ei2θ

Rα eiαθ + 2Reiθ cos (πβ) + 1

xα ei2πα x2 + 2x cos (πβ) + 1

(3.2)

(3.3)

(3.4)

3.1 Integrands Defined Over [0, ∞)

 γ4

zα dz 2 = iR z + 2z cos (πβ) + 1

51

0 d φ eiφ 2π

R2 ei2φ

Rα eiαφ + 2R eiφ cos (πβ) + 1 (3.5)

We could have provided detours for the two poles we will find, but as we have seen it is far easier to take the shortcut the Residue Theorem provides. There is an important point we must make here: we have implicitly defined arg z ∈ [0, 2π ) for all z in a disk that contains the contour C. We set this up when we defined log z such that the branch cut is alone the positive real axis. We could have defined the log term such that the branch cut is along the negative real axis, but we would have had to perform some mathematical gymnastics to get the definite integral we seek. This definition of arg z will have direct implications on the computation of the definite integral. Back to the contour integral. Clearly, the integral over γ4 vanishes as R → 0 because α > −1. Regarding the integral over γ2 , the integral vanishes as R → ∞ because in this limit the integral behaves as Rα−1 , and α < 1. This leaves the integral over γ1 and γ3 , and we get that the contour integral is  C

  zα = 1 − ei2πα dz 2 z + 2z cos (πβ) + 1

∞ dx 0

xα (3.6) x2 + 2x cos (πβ) + 1

This is a lovely, simple expression for the contour integral in terms of the definite integral; it is pretty much what we will always see with a factor of z α in the contour integral. By the Residue Theorem, the contour integral is also equal to i2π times the sum of the residues at the polesinside the  contour C. The poles of the integrand are where z 2 + 2z cos (πβ) + 1 = z + eiπβ z + e−iπβ = 0 Res

z=−eiπβ

Res

z=−e−iπβ

eiπα eiπαβ zα = z 2 + 2z cos (πβ) + 1 −i2 sin (πβ) α eiπα e−iπαβ z = z 2 + 2z cos (πβ) + 1 i2 sin (πβ)

(3.7) (3.8)

It should be clear that we take the minus sign to be eiπ because, as we stated, arg z ∈ [0, 2π ); this accounts for the factor eiπα in both of the residues. By the Residue Theorem, we get the following equality.   1 − ei2πα

∞ dx 0

eiπα sin (π αβ) xα = i2π x2 + 2x cos (πβ) + 1 sin (πβ)

(3.9)

52

3 Evaluation of Definite Integrals II: Applications to Various Types of Integrals

Doing the algebra, we see that ∞ dx 0

xα π sin (π αβ) = 2 x + 2x cos (πβ) + 1 sin (π α) sin (πβ)

(3.10)

What a nice symmetric result! Here’s a question: can the nice result on the RHS symmetric in α and β can be deduced from the integral on the LHS without evaluating the integral? That is, ∞ 0

xα dx 2 = x + 2x cos (πβ) + 1

∞ dx 0

xβ x2 + 2x cos (π α) + 1

(3.11)

Note that α and β are both ∈ (−1, 1) and are accordingly interchangeable. The origin of the symmetry seems be a mystery. It’s log, log, log! For our next integral, we will look at something that outwardly has no branch points, nor any symmetry: ∞ dx 1 + x3 0

Those who have taken a first course in Complex Analysis have likely seen the trick we will be using here. Because the integrand is meromorphic, a consideration of a contour integral using this integrand will not produce anything meaningful over the keyhole contour defined in Fig. 3.1. What we need to do is introduce a factor of log z in the contour integral. As we saw in the previous section, a log in the presence of a branch point effectively becomes a factor of −i2π . Accordingly, we consider the following contour integral:  log z dz (3.12) 1 + z3 C

where again, C is the contour defined in Fig. 3.1, such that  dz γ1

 γ2

log z = 1 + z3

R dx R

log z dz = iR 1 + z3

2π

log x 1 + x3

(3.13)

  log Reiθ dθ e 1 + R3 ei3θ

(3.14)

iθ

0

3.1 Integrands Defined Over [0, ∞)

 γ3

 γ4

53

log z dz = 1 + z3

R dx R

log z dz = iR 1 + z3

log x + i2π 1 + x3

0 dφ e

iφ

2π

  log R eiφ 1 + R3 ei3φ

(3.15)

(3.16)

As R → ∞ and R → 0, the contour integral becomes  C

log z dz = −i2π 1 + z3

∞ 0

dx 1 + x3

(3.17)

By the Residue Theorem, the contour integral is also equal to i2π times the sum of the residues of the poles of the integrand of the contour integral inside the contour   C. In this case, the poles are at z ∈ eiπ/3 , eiπ , ei5π/3 . Before computing the residues, it is important for the reader to understand why we chose the poles to be the set above and not something differing in phase by some multiple of 2π . The reason those values for the phase are used for these poles is because we have already specified that each z on and in the contour C must satisfy arg z ∈ [0, 2π ). Note, for example, that arg z = 0 on γ1 and arg z = 2π on γ3 . The residues at these poles are as follows. log z iπ/3 = i2π/3 1 + z3 3e π = e−iπ/6 9

Res

z=eiπ/3

Res

z=eiπ

Res

z=ei5π/3

log z iπ = i2π 3 1+z 3e π =i 3

log z i5π/3 = i10π/3 3 1+z 3e 5π = − eiπ/6 9

(3.18) (3.19)

(3.20) (3.21)

(3.22) (3.23)

Adding the residues, we arrive at the following result: ∞ 0

√ dx 2 3 π = 1 + x3 9

(3.24)

54

3 Evaluation of Definite Integrals II: Applications to Various Types of Integrals

This result is fairly standard and we should make sure the trick exploiting the multivaluedness of the logarithm in the complex plane is understood. We will study many examples of such an introduction of multivaluedness to evaluate definite integrals. The Universal Polynomials Let’s discuss a class of definite integrals that are easily evaluated from the quantities we have already computed. We begin by evaluating an integral similar to the one we just computed: ∞ log x dx (3.25) 1 + x3 0

We introduce the following contour integral for computing the definite integral:  dz C

log2 z 1 + z3

(3.26)

What should be clear to the reader is that the poles are the same as for the previous integral. In fact, we could write the previous result for the log-less definite integral as ∞ − i2π 0 ∞

=⇒ 0

 dx = i2π Rk log zk 3 1+x 3

k=1

 dx =− Rk log zk 3 1+x 3

(3.27)

k=1

where z1 = eiπ/3 , etc., and Rk = Res z=zk

1 1 + z3

(3.28)

That is, R1 = 13 e−i2π/3 , R2 = 13 , and R3 = 13 e−i4π/3 . Also, log z1 = i π3 , log z2 = iπ , because, again, arg z ∈ [0, 2π ). Using these quantities, we can compute log z3 = i 5π 3 the definite integral with the additional log factor:  C

log2 z dz = 1 + z3

That is,

∞ 0

∞ − i4π 0

 log2 x − (log x + i2π )2 = i2π Rk log2 zk 3 1+x m

dx

log x dx + 4π 2 1 + x3

(3.29)

k=1

∞ 0

 dx = i2π Rk log2 zk 1 + x3 m

k=1

(3.30)

3.1 Integrands Defined Over [0, ∞)

where

∞ − i2π 0

55

 dx = i2π Rk log zk 3 1+x m

(3.31)

k=1

Putting together Eqs. (3.30) and (3.31), we get ∞ 0

log x 1 =− Rk log zk (log zk − i2π ) 3 1+x 2 m

dx

(3.32)

k=1

The reader can use the information we have already derived regarding the poles zk 2 and their residues Rk in Eq. (3.32) to see the sum on the RHS is − 2π . 27 The analysis we did above on the integrals with factors of log in the integrand hopefully gave the reader some ideas about the general integrals of meromorphic functions times a power of log. It should be clear that the form of Eq. (3.32) is independent of the meromorphic function in the integrand. We will demonstrate below and in the problems the following relation: ∞

1  Rk log zk Pm (log zk ) m+1 m

dx f (x) logm x = −

(3.33)

k=1

0

with Rk = Resz=zk f (z). That is, when the poles of f are simple, we need only compute the residue of f at the poles. In contrast, when f has a repeated pole, the sum on the RHS of Eq. (3.33) is complicated by derivatives of f and accordingly there is no universal polynomial with repeated poles. The derivation of the universal polynomials is straightforward and involves generating a recurrence relation for, say, Pm (log zk ) in terms of Pm−1 (log zk ), Pm−2 (log zk ),..., P0 (log zk ). To give a taste: P0 ( ) = 1 P1 ( ) = − i2π

(3.34) (3.35)

P2 ( ) = 2 − i3π − 2π 2

(3.36)

We will leave the derivation of further universal polynomials and their applications to the exercises. Contributions from Infinity In the previous examples we were able to show that any contributions from the large arcs of the contour vanish as a radius R as the large arcs increase without bound. Here is an example in which we need to be super careful about how we treat the contributions of the contour integral as R → ∞.

56

3 Evaluation of Definite Integrals II: Applications to Various Types of Integrals

Moreover, the previous examples did not require a transformation before we introduced a contour integral. In this example, we will be performing transformations on the definite integral before determining the best contour integral representation. Note that this example will not be using the usual contour for this class of integrals. Consider evaluating the following definite integral: ∞ dx  0

x2

x  + 1 (eπx + 1)

(3.37)

Normally, we would consider the following contour integral: 

z log z  dz  2 z + 1 (eπz + 1) C

(3.38)

where C is the keyhole contour from Fig. (3.1). Nevertheless, it turns out that the evaluation of the integral over the large arc of radius R as well as the sum of the residues involves challenges that we will not encounter until the final chapter on asymptotics. Instead, we are going to exploit some symmetry in the integrand to derive an interesting representation of the integral. We begin with the observation that eπz

π

1 1 1 = − tanh z +1 2 2 2

(3.39)

We will be breaking the integral up into odd and even portions so the integral we want to evaluate becomes ⎤ ⎡ R ∞  R π

1 x x x 1  = lim ⎣ x ⎦ dx  2 dx 2 dx 2 − tanh 2 x +1 2 x +1 x + 1 (eπ x + 1) R→∞ 2 0 0 0 ⎤ ⎡ R 

1 x eπ x − 1 ⎥ ⎢ = lim ⎣log R2 + 1 − dx 2 ⎦ 4 R→∞ x + 1 eπ x + 1 −R

We will be using the exponential form of the tanh function for this analysis. Clearly, the log term will diverge as R → ∞; accordingly, we expect the integral to have a similar divergent quantity. Accordingly, consider the following contour integral  z eπz − 1 (3.40) dz 2 z + 1 eπz + 1 C where C is the semicircular contour of radius R in the upper half plane. Accordingly, this contour integral is equal to

3.1 Integrands Defined Over [0, ∞)

R −R

57

x eπx − 1 + iR dx 2 x + 1 eπx + 1

π

eπRe − 1 Reiθ R2 ei2θ + 1 eπReiθ + 1 iθ

d θ eiθ 0

(3.41)

The second integral vanishes as R → ∞; this will be shown in the exercises. The contour integral is then equal to the original integral, and is also equal to i2π times the sum of the residues at the poles within C. This is where things get interesting. The poles of the integrand within C are at zk = i(2k + 1) for k ∈ {0, 1, 2, ...}. The pole for k = 0 is a double pole while the poles for k ≥ 1 are simple. To compute the pole for k = 0, we expand in a series for z = z0 + ζ = i + ζ : eπ(i+ζ ) − 1 (i + ζ )(−2 − π ζ − · · · ) i+ζ = 2 π(i+ζ ) (i + ζ ) + 1 e +1 (i2ζ + ζ 2 )(−π ζ − π2 ζ 2 1 (i + ζ )(1 + π2 ζ + · · · ) = π ζ 2 (1 − i 21 ζ )(1 + π2 ζ + · · · )   1 1 1 − i ζ + ··· = πζ2 2 Accordingly, the residue is the coefficient of ζ −1 : Res z=z0

1 z eπz − 1 = 2 πz z +1e +1 i2π

(3.42)

The simple poles zk , k ≥ 1, have residues z eπz − 1 1 = 2 πz z=i(2k+1) z + 1 e +1 i2π Res



1 1 + k k +1

 (3.43)

The definite integral of interest is then R −R

N (R)  z eπz − 1 x eπx − 1 = i2π dx 2 Res z=i(2k+1) z 2 + 1 eπz + 1 x + 1 eπx + 1

(3.44)

k=0

= 1+

N (R)   k=1

= 2HN (R) +

1 1 + k k +1 1 N (R) + 1

 (3.45) (3.46)

N (R) is the number of poles within the semicircle of radius R, which is N (R) = 1 (R − 1); HN is the N th harmonic number. Recall that 2 HN ∼ log N + γ

(N → ∞)

(3.47)

58

3 Evaluation of Definite Integrals II: Applications to Various Types of Integrals

where γ ≈ 0.577216 is the Euler-Mascheroni constant. Accordingly, R dx −R

x2

x eπx − 1 + 1 eπx + 1

2 (log N (R) + γ ) = −2 (log 2 − γ ) + 2 log (R − 1)

(3.48) (3.49)

and therefore, the original integral is ∞ dx 

x2

0

x + 1 (eπx + 1) 

(3.50)

    1 = lim log R2 + 1 − 2 log(R − 1) + 2 (log 2 − γ ) 4 R→∞

At this point the reader should see that the divergent logs cancel in the limit as R → ∞. This should be fully expected and provides a natural check on the evaluation. Accordingly, ∞ 1 x  = (log 2 − γ ) dx  2 (3.51) πx 2 x + 1 (e + 1) 0

So this particular case led to an alternative to the keyhole contour when we broke the integrand up into odd and even components. In this way, we were able to express the contributions of the residues to the integral, including the divergent piece, using a much simpler sum than what we would have gotten through the keyhole contour. If we are to take any lesson from this book, it is to recognize when the usual way of attacking problems leads to a dead end and be creative.

3.2 Integration Over the Unit Circle Frequently we are asked to evaluate integrals—mainly involving trigonometric functions—over a single cycle. That is, we are looking at integrals of the type 

φ 0 +2π

dx ρ(cos x, sin x) = −i φ0

|z|=1

  z + z −1 z − z −1 dz ρ , z 2 i2

(3.52)

where ρ ∈ Q is a rational function. These types of integrals are usually covered in a first course in complex analysis. We will review an example. Then, we will build on out previous work a little to extend this technique into new areas. We will see, for example, that φ0 can be any value and its importance lies in defining a branch cut in situations where we encounter multivalued functions.

3.2 Integration Over the Unit Circle

59

Review Preview For the review, let’s see how we use the Residue Theorem to evaluate the following definite integral: π 0

dθ a2 cos2 θ + b2 sin2 θ 2π

=

a2 0

= i2



+

|z|=1

b2

dθ   − a2 − b2 cos θ

(3.53)

dz       a2 − b2 z 2 − 2 a2 + b2 z + a2 − b2

The poles of the integrand are easily found to be z1 = (a + b)/(a − b) and z2 = (a − b)/(a + b). We further note that when a > 0 and b > 0, |z1 | > 1 and is accordingly outside the contour; its residue is zero. Nevertheless, |z2 | < 1 so the residue is nonzero:   i2 i2     =      Res  2 z=z2 a − b2 z 2 − 2 a2 + b2 z + a2 − b2 2 a2 − b2 z2 − 2 a2 + b2 = And therefore

π a2 0

cos2

i −ab

π dθ = 2 2 ab θ + b sin θ

(3.54)

This is valid when ab > 0. But what happens when ab < 0? The result in Eq. (3.54) is negative despite the integrand being strictly positive. We could simply declare that the result should be enclosed in an absolute value sign, and such a declaration would be correct. But let’s answer the question a little more definitively. Note that, when a > 0 and b < 0, the pole z1 may be expressed as z1 = (a + b)/(a − b) = (a − |b|)/(a + |b|) < 0; accordingly, z2 is now > 0. So in this case, it is z2 that has zero residue, while   i2 i2     =      Res  2 2 2 2 2 2 2 2 2 z=z1 a −b z −2 a +b z+ a −b 2 a − b z1 − 2 a2 + b2 =

i ab

which produces as the result for the integral −π/(ab), which is positive. Accordingly, we may now state that, for all real a, b,

60

3 Evaluation of Definite Integrals II: Applications to Various Types of Integrals

π a2 0

cos2

π dθ = 2 2 |ab| θ + b sin θ

(3.55)

Going forward, we should be able to short-circuit this thinking. Generally, however, we should make sure we look at our results skeptically before celebrating them. This is where things get fun We are arriving at a place where we address integrals that are not traditionally evaluated using complex analytic techniques. It will turn out that the simplest changes to an integral like that in Eq. (3.54) will bring us to explore new techniques. It is not that the analysis below is groundbreaking. Rather, we will be testing the assumption that there are only a limited class of integrals that may be evaluated using complex techniques. It will turn out that with a little patience we can vastly expand that class of integrals, and we will encounter new insights along the way. We are interested in the following definite integral: π 0

1 θ = dθ 2 2 a cos2 θ + b2 sin2 θ

2π dθ 0

θ p − q cos θ

(3.56)

where, as before, p = a2 + b2 , q = a2 − b2 . This is the simplest case of an integral of a polynomial times a trigonometric rational function. It might seem that the linear term in the numerator of the integrand of this integral might confound the usual substitution z = eiθ that works so well with trigonometric rational functions like that above. For example, the usual substitution would introduce a factor of −i log z in the numerator of the integrand. So let’s consider the following contour integral. −

i 2

 C

dz −i log z

= z p − q z+z−1 2

 dz C

qz 2

log z − 2pz + q

(3.57)

What is C? Nominally, we have been taking C to be the unit circle. Nevertheless, there is a branch point at z = 0 due to the log in the numerator of the integrand. Accordingly, because the definite integral is over θ ∈ [0, 2π ], we take the branch cut to be the positive real axis. That is, C appears to be a keyhole contour but with the outer radius being unity rather than increasing without limit in the previous section. But we still haven’t defined C completely. Note that we also have a pair of poles where the denominator of the integrand is zero. The poles are at  p z± = ± q

p2 −1 q2

(3.58)

3.2 Integration Over the Unit Circle

61

Fig. 3.2 Keyhole contour in unit circle with a branch cut along the positive real axis and a pole on the branch cut

and we note that both poles are real and positive, with |z+ | > 1 and |z− | < 1. So we have a pole on the branch cut within the unit circle. Whoa. As with any other pole, we will need to form a detour—for convenience, circular detours—around the pole at z = z− . The resulting contour C is pictured in Fig. (3.2). Before we proceed, let’s take stock of what just happened. In conventional evaluations of definite integrals using complex integration, we usually select a contour that was taught to us in a first course: a semicircle, a keyhole, a unit circle. We will see other standard contours such as wedges and rectangles. But here, we have designed a custom contour based on the singularities of the integrand of the complex representation of the definite integral on the RHS of Eq. (3.56). Using such a contour construction, the range of definite integrals that may be evaluated using complex integration may be expanded significantly from those learned in a first course. What will happen is that, even though we have derived a contour integral based on the definite integral, the contour integral when broken apart will have other contributions from the parts of the contour that avoid the branch cut, branch point, and pole. Accordingly, we will see that the contour integral is equal to the sum of the definite integral and other contributions from the detours. Let us break apart the contour integral on the RHS of Eq. (3.57). The terms are arranged counterclockwise beginning at θ = 0 on the large circular arc.  γ1

1 log z = dz 2 qz − 2pz + q 2

 dz γ2

log z = qz 2 − 2pz + q

2π dθ 0

z − +R

dx 1

θ p − q cos θ

(3.59)

log x + i2π qx2 − 2px + q

(3.60)

62

3 Evaluation of Definite Integrals II: Applications to Various Types of Integrals

 γ3

log z dz 2 = iR qz − 2pz + q

π

  log z− ei2π + R eiφ dφ e 2    q z− + R eiφ − 2p z− + R eiφ + q iφ

2π

(3.61)  dz γ4

 γ5

 γ6

 γ7

qz 2

log z = − 2pz + q

R dx z− −R

log z dz 2 = iR qz − 2pz + q log z dz 2 = qz − 2pz + q

log x + i2π qx2 − 2px + q

0 dφ 2π

log R + i2π qR2 ei2φ − 2pR eiφ + q

z − −R

dx R

log z = iR dz 2 qz − 2pz + q

0

(3.62)

qx2

log x − 2px + q

(3.63)

(3.64)

  log z− + R eiφ dφ e   2  q z− + R eiφ − 2p z− + R eiφ + q iφ

π

(3.65)  dz γ8

log z = qz 2 − 2pz + q

1 dx z− +R

log x qx2 − 2px + q

(3.66)

Before we combine and simplify the above terms, we should note that we defined the branch cut such that arg z = 2π below the cut and arg z = 0 above. Note that these values of the phase of z are applied to the detours above and below the branch cut. This should be apparent in the third integral where the reader may have been a bit puzzled at the z− ei2π term. That factor is needed because of the log. It will turn out that this factor will be crucial to the evaluation of the definite integral. We have encountered the Cauchy principal value for integrals earlier. Let’s provide a definition: Definition 1 The Cauchy principal value of a definite integral of a function f over an interval in which f has a simple pole at x = x2 is defined as follows. x1 PV

⎞ ⎛ x − 2 R x1 dx f (x) = lim ⎝ dx f (x) + dx f (x)⎠ →0

x0

x0

(3.67)

x2 +R

As we have seen, the Cauchy principal value is very useful and we will be seeing this many more times over the rest of this book. It is useful because it allows us to define integrals in terms of cancellations of divergent quantities, as we will see soon.

3.2 Integration Over the Unit Circle

63

Let’s get the following question out of the way: what does the Cauchy principal value represent? With Definition 1 in mind, let’s combine the integrals along the lines below and above the branch cut, i.e., the second, fourth, sixth, and eighth integrals above as R → 0:  dz γ2 +γ4 +γ6 +γ8

z− +R

log z = qz 2 − 2pz + q

dx

log x + i2π qx2 − 2px + q

dx

log x + i2π qx2 − 2px + q

1

R + z− −R z− −R

+

dx

log x qx2 − 2px + q

dx

log x qx2 − 2px + q

R

1 + z− +R

⎛

⎜ = −i2π ⎝

z− −R R

z− +R

1 = −i2π PV 0

dx + qx2 − 2px + q

1

dx qx2 − 2px + q

⎞ dx ⎟ ⎠ qx2 − 2px + q

(3.68)

As for the integrals around the pole, we may combine them and let R → 0. For example, the integral over γ3 is  dz γ3

  log z− ei2π + R eiφ d φ eiφ   2  q z− + R eiφ − 2p z− + R eiφ + q 2π

−1 i(φ−2π ) π log z− + i2π + log 1 + R z− e iφ    = iR d φ e  2 qz− − 2pz− + q − 2(p − qz− )R eiφ + qR2 ei2φ

log z = iR qz 2 − 2pz + q

π

2π

π log z− + i2π =i 2 p − qz−

(R → 0)

(3.69)

2 Note that z− is a zero of qx2 − 2px + q, so the term qz− − 2pz− + q above vanishes. Similarly, the integral over γ2 is

 γ2

log z = iR dz 2 qz − 2pz + q =i

0

  log z− + R eiφ dφ e 2    q z− + R eiφ − 2p z− + R eiφ + q iφ

π

π log z− 2 p − qz−

(R → 0)

(3.70)

64

3 Evaluation of Definite Integrals II: Applications to Various Types of Integrals

Now let’s evaluate the Cauchy principal value in Eq. (3.68). As usual, we break the integrand up into partial fractions. 1 PV 0

1 dx = PV qx2 − 2px + q q(z+ − z− )



1 dx 0

1 1 − x − z+ x − z−

 (3.71)

Now, the Cauchy principal value of the integral of the first term is simply the usual integral as the pole at x = z+ is outside of the integration interval. 1 PV 0

  z+ − 1 dx = log x − z+ z+

(3.72)

In contrast, the pole at x = z− is in the integration interval so we will have to use the definition of the Cauchy principal value. 1 PV 0

dx x − z−

⎛ z− −R ⎞  1 dx dx ⎠ = lim ⎝ + R →0 x − z− x − z− z− +R 0      R 1 − z− + log = lim log R →0 z− R   1 − z− = log z−

Using the fact that z− z+ = 1, we may conclude that 1 PV 0

dx = log z− qx2 − 2px + q

(3.73)

Accordingly, the contour integral is then  dz C

log z 1 = qz 2 − 2pz + q 2

2π dθ 0

θ π2 i2π log z− iπ log z− − − + p − q cos θ q(z+ − z− ) p − qz− p − qz−

(3.74)

Finally, we determine the value of the integral using Cauchy’s theorem and using the fact that q(z+ − z− ) = 2 p2 − q2 and p − qz− = p2 − q2 . The terms with log z− cancel and we have that 2π dθ 0

θ = p − q cos θ

2π 2 p2 − q2

(3.75)

3.2 Integration Over the Unit Circle

and accordingly

π dθ 0

65

π2 θ = 2|ab| a2 cos2 θ + b2 sin2 θ

(3.76)

where we added in absolute values for similar reasons as for the previous example and the result in Eq. (3.76) is valid for all a, b ∈ R and = 0. Hopefully it is clear that the result is correct at least for a = b. How else could we check the result? The important point of this example is that we are expanding the types of integrals that may be attacked using complex integration. Let’s do an even more challenging example, one that is quite surprising in its difficulty. Along the way, we will meet a new friend. Dilog die! Sometimes polynomial-times-trig-rational integrands are actually a lot simpler than we might expect. And sometimes, we can be led into a rabbit hole—although, isn’t that why this book is so much fun? Consider the following integral. π I1 =

dx 0

x sin x 1 + cos2 x

(3.77)

If one wanted to evaluate this integral without any of the hassle of setting up weird contours as in the previous example, they might think about integration by parts. Or at least until they realize that the result will be an integral over arctan cos x which does not lend itself to the desired complexity avoidance. Nevertheless, there is a nifty trick that leads to a quick evaluation: substitute x → π − x. π I1 =

dx 0

π =π 0

(π − x) sin (π − x) 1 + cos2 (π − x)

sin x − dx 1 + cos2 x

π =π

dx 0

π dx 0

sin x − I1 1 + cos2 x

x sin x 1 + cos2 x (3.78)

66

3 Evaluation of Definite Integrals II: Applications to Various Types of Integrals

and accordingly π I1 = 2

π dx 0

sin x 1 + cos2 x

π = [− arctan (cos x)]π0 2 π2 = 4

(3.79)

Like most of the results we derive using Cauchy’s Theorem or the Residue Theorem, this is an elegant result in which we find the integral by algebraically solving for it. Aren’t these fantastic? So, one should expect that a very similar integral π I2 =

dx 0

x cos x 1 + sin2 x

(3.80)

is just as easily evaluated as I1 . Well, if that were the case, it would not be appearing here. The problem is that cos (π − x) = − cos x so that we do not get the nifty algebraic solution as for I1 ; rather, we just get the trivial statement that 0 = 0. And, as before, integration by parts leaves us with a not-very-nice integrand. So we are after all going to evaluate I2 somehow using complex techniques. And for this we will use a different formulation than converting each factor of x into a log as we did in the previous example. It’s not that the log we used in the previous example won’t work here—it will, as will be seen in the exercises—but it is always good to attack problems like this from multiple angles. So, at the risk of introducing a deus ex machina, consider the following representation of I2 : ⎤ ⎡ π π  ∂ ⎣ x cos x sin (αx) ⎦ = dx dx (3.81) ∂α 1 + sin2 x 1 + sin2 x 0

0

α=1

This is a case of what many refer to as “Feynman’s trick,” which may have originated with a story about how the Nobel-winning physicist Richard Feynman used differentiating under the integral as a nice, simple alternative to complex integration. 1 The idea, of course, is to convert the integral over [0, π ] into an integral over a full cycle so we may integrate over a unit circle in the complex plane. So it makes sense to play around with the following:

1

The irony is not lost on us.

3.2 Integration Over the Unit Circle 2π dx 0

sin (αx) = 1 + sin2 x

π dx 0

67

sin (αx) + 1 + sin2 x

2π dx π

π = (1 + cos (π α))

dx 0

sin (αx) 1 + sin2 x

sin (αx) + sin (π α) 1 + sin2 x

π dx 0

cos (αx) 1 + sin2 x

(3.82) So what we have here is not one but a pair of integrals over the unit circle. The first, π 0

1 cos (αx) = dx 2 2 1 + sin x

π dx −π

cos (αx) 1 + sin2 x

(3.83)

is over θ ∈ [−π, π ]. In this case the associated contour integral will have a branch cut along the negative real axis. The second, 2π dx 0

sin (αx) 1 + sin2 x

(3.84)

is over θ ∈ [0, 2π ]. In this case the associated contour integral will have a branch cut along the positive real axis. By subbing z = eiθ in each of the integrals in Eqs. (3.83) and (3.84), we get the following contour integrals:  dz

i4 C±

z 1+α z 4 − 6z 2 + 1

(3.85)

We will need to determine C+ and C− . We do this by noting the poles inside the √ 2 − 1 . That is, there is a pole along the branch cut in unit circle are at ±z0 = ± either case. Accordingly, C is the contour shown in Fig. 3.2 with the detour around + √ the pole at z = 2 − 1, and C− is C+ reflected about the imaginary axis. Let 2π eiαx (3.86) I (α) = dx 1 + sin2 x 0

Then with regard to the contour C+ , the following are its components (some of which are combined):

68

3 Evaluation of Definite Integrals II: Applications to Various Types of Integrals  i4

dz γ1

 i4

dz

γ2 +γ4 +γ6 +γ8

z4

dz γ3

(3.87)

z 1+α = i4 1 − ei2π α PV 4 2 z − 6z + 1

 i4

z 1+α = I (α) − 6z 2 + 1

z 1+α z 4 − 6z 2 + 1

π = −4R 2π

1 dx 0

x4

x1+α − 6x2 + 1

(3.88)

 i2π 1+α z0 e + R eiφ d φ eiφ   4 2 z0 + R eiφ − 6 z0 + R eiφ + 1

(3.89)  i4

dz γ3

 i4

dz γ5

z 1+α z 4 − 6z 2 + 1

0 = −4R 2π

z 1+α = −4R z 4 − 6z 2 + 1

0 π

 iφ 1+α R e iφ dφ e   4 2 iφ − 6 R eiφ + 1 R e

(3.90)

 1+α z0 + R eiφ d φ eiφ  4 2  z0 + R eiφ − 6 z0 + R eiφ + 1

(3.91) The contour over C+ is equal to i2π times the residue at the pole z = √ integral

2 − 1 , where the negative is taken to be eiπ (why?): −z0 = −  dz

i4 C+

z 1+α z 1+α = i2π(i4) Res z 4 − 6z 2 + 1 z=z0 eiπ z 4 − 6z 2 + 1 −8π eiπα z0α 4z02 − 12 π = √ eiπα z0α 2 =

(3.92)

Applying the residue theorem and rearranging the furniture a little, we get the following expression for I (α):   I (α) = i4 ei2πα − 1 PV

1 dx 0

 2 π x1+α + √ z0α eiπα + 1 4 2 x − 6x + 1 2 2

(3.93)

As we are only interested in the imaginary part of I (α), we may simplify this result for I and obtain 2π dx 0

sin αx = Im [I (α)] 1 + sin2 x 1 = −8 sin2 π αPV

dx 0

x4

x1+α π + √ z0α sin π α (1 + cos π α) − 6x2 + 1 2

(3.94)

3.2 Integration Over the Unit Circle

69

Fig. 3.3 Keyhole contour in unit circle with a branch cut along the negative real axis and a pole on the branch cut

That takes care of one of the integrals we need. Now we need the other: define π J (α) =

dx −π

eiαx 1 + sin2 x

(3.95)

for which we defined the contour C− (Fig. 3.3). We leave a similar analysis of the analogous contour integral and residue at the pole z = z0 for the reader. The result is π dx −π

cos αx = Re [J (α)] 1 + sin2 x 1 = −8 sin π αPV

dx 0

x1+α x4 − 6x2 + 1

(3.96)

π + √ z0α (1 + cos π α) 2 So combining Eqs. (3.82), (3.94), and (3.96), we finally arrive at the following expression: π 0

π π sin αx = √ z0α sin π α − 8 sin2 αPV dx 2 2 1 + sin x 2 2

1 dx 0

x1+α x4 − 6x2 + 1

(3.97)

To get the original integral we want to evaluate, we differentiate both sides with respect to α and set α = 1:

70

3 Evaluation of Definite Integrals II: Applications to Various Types of Integrals

π 0

1

π 2 √ x cos x x2 log x = − dx 2 − 1 − 8PV dx √ x4 − 6x2 + 1 1 + sin2 x 2 2

(3.98)

0

Whew! We have now expressed the original integral in terms of a Cauchy principal value of an integral over a log times a rational. Rest assured, this is progress! It’s just that it is hard to see unless we know that integrals of logs times rationals are usually expressed in terms of dilogarithms. The dilogarithm is not commonly taught at the undergraduate or even graduate level in most coursework. Until recently, the dilogarithm was discussed in hushed tones among only the most devoted integro-philes. But recent studies of the integrals in Bose-Einstein and Fermi-Dirac statistics have provided an appreciation of dilogarithms beyond the esoteric. What we are going to see is that the dilogarithm has some really amazing properties we can exploit to derive new results. Before we discuss those properties, let’s simplify the principal value integral a bit. Decomposing the integrand, we get ⎛ ⎞   1 1 ⎝ 1 8x2 √

− √

⎠ = 1+ √ x4 − 6x2 + 1 2 x− 2+1 x+ 2+1 ⎛ ⎞   1 1 ⎝ 1 √

− √

⎠ − 1− √ 2 x− 2−1 x+ 2−1

(3.99)

Accordingly, the principal value integral may be evaluated based on evaluating the following four integrals. 1 PV

dx

log x √

x− 2+1

(3.100)

dx

log x √

x+ 2+1

(3.101)

dx

log x √

x− 2−1

(3.102)

dx

log x √

x+ 2−1

(3.103)

0

1 PV 0

1 PV 0

1 PV 0

3.2 Integration Over the Unit Circle

71

The only one of the four that actually requires a Cauchy principal value to define the integral is in Eq. (3.102). We will now express these integrals in terms of dilogarithms. To begin, when y ∈ / [0, 1], we have 1 0

log x = Li2 dx x−y

  1 y

(3.104)

where the dilogarithm of z ∈ (0, 1) is defined as 1 Li2 (z) := −

dt

log (1 − zt) t

(3.105)

0

This is but one of many representations of the dilogarithm. The reader might know that the dilogarithm has a series expansion that looks almost like that for the...monologarithm (?) − log (1 − z): Li1 (z) = − log (1 − z) = Li2 (z) = ··· Lis (z) =

∞  zk k=1 ∞  k=1

k zk k2

∞  zk k=1

ks

(3.106)

(3.107)

(3.108)

where Eq. (3.108) defines a polylogarithm. From this it should be clear that dilogs satisfy Li2 (0) = 0 and Li2 (1) = π 2 /6. An interesting thing about the dilogarithm is that there are very few exact results but there are a dizzyingly enormous panoply of identities satisfied by dilogarithms. We will need a couple of those identities to evaluate the sought-after integral. Nevertheless, before we get to those identities, we will need to evaluate the Cauchy principal value of the dilogarithm of a real argument w ∈ / (−∞, 1]. We begin by extending the dilogarithm to arguments outside of (−∞, 1):   1 1 π2 − log2 w − Li2 (3.109) Li2 (w) = 3 2 w We will leave derivation of these results to the exercises. In the meantime, the Cauchy principal value integral in Eq. (3.102) and the other integrals in Eqs. (3.100), (3.101), (3.103) are as follows

72

3 Evaluation of Definite Integrals II: Applications to Various Types of Integrals

1 PV

dx 0

√ √



log x π2 1 √

= 2 + 1 − Li2 2−1 − log2 3 2 x− 2−1 (3.110)

1 dx

√

log x √

= Li2 2−1 x− 2+1

(3.111)

dx

√

log x √

= Li2 − 2−1 x+ 2+1

(3.112)

dx

√

log x √

= Li2 − 2+1 x+ 2−1

(3.113)

0

1 0

1 0

Accordingly, 1 8PV

dx

x4

0

   √



 x2 log x 1 1 √ Li2 2 − 1 − Li2 − 2−1 − 1− √ = 1+ √ 2 − 6x + 1 2 2  2 √ √ √



 π 1 2 + 1 − Li2 2 − 1 − Li2 − 2+1 − log2 3 2

(3.114) We could stop here if we wanted; the dilogarithms represent a sufficiently closed form. But as was mentioned previously, the great thing about dilogarithms is that they satisfy a Brobdingnagian set of identities. While we are still learning about them, the hard part about using dilogarithms is finding the right identity that will simplify an expression. So we will just present some that will help us; keep in mind that there may be others we could use as well. The following identity is valid for z < 0:   1 π2 1 =− − log2 (z) Li2 (z) + Li2 z 6 2

(3.115)

The following identity is valid for z ∈ (0, 1): 

    1−z 1−z π2 − Li2 − = Li2 (−z) − Li2 (z) + − log z log Li2 1+z 4 1+z (3.116) √

In Eq. (3.115), set z = − 2 − 1 ; the result is 1−z 1+z



√ √ √





π2 1 Li2 − 2 − 1 + Li2 − 2+1 =− 2+1 − log2 6 2

(3.117)

3.2 Integration Over the Unit Circle

In Eq. (3.116), set z = z). Accordingly,

√

73

2 − 1. For this value of z, it turns out that z = (1 − z)/(1 +

√ √



π 2

1 2 − 1 − Li2 − 2−1 = 2+1 (3.118) − log2 8 2 √ √



We actually need Li2 2 − 1 + Li2 − 2 + 1 in Eq. (3.114). Fortunately, we can get this from adding Eqs. (3.117) and (3.118); the result is Li2

Li2

√

√

√ √





π2 − log2 2 − 1 + Li2 − 2+1 =− 2+1 24

(3.119)

And with this, we have removed the dilogarithms from the final result. This is not necessarily the final goal, as dilogarithms are straightforward to compute. Nevertheless, expressing this purely in terms of elementary functions may leave us with the feeling that there is a better way to evaluate the original integral. And perhaps there is, but that’s not the point. The point is, we have demonstrated that we can extend the contour integration methods developed in this chapter to new classes of integrals; the journey is more important than the destination. Let’s put this all together. Substituting Eqs. (3.117) and (3.119) into Eq. (3.98), we get π dx 0

  1 x cos x = 1 − √ 1 + sin2 x 2  √

π2  1 2 log 2+1 − 2 8   √

π2  1 1 log2 + 1+ √ 2+1 − 2 8 2

(3.120)

Recall there was an additional term beside the Cauchy principal value integral in Eq. (3.98). We finally have arrived at our destination: π dx 0

√

π2 x cos x 2 = log 2 + 1 − 4 1 + sin2 x

(3.121)

It is amazing how simple the result is given how much mathematics has been involved. Is there a simpler way? Probably. But again, the point of this exercise is to apply complex integration methods to integrals to which such methods have not been traditionally applied. Moreover, interestingly, by adding the two integrals discussed here we get a nontrivial result:

74

3 Evaluation of Definite Integrals II: Applications to Various Types of Integrals

π dx 0

√

x (sin x + cos x) (2 − sin x cos x) 2 = log 2 + 1 2 + cos2 x sin2 x

(3.122)

The reader is challenged to evaluate this integral using purely real methods. (Note: like the two component integrals, one can seem to make progress using integration by parts but you then run into integrating an arctangent of some rational combination of trig functions.) Essentially singular We have mentioned essential singularities; these are those singularities that blow up faster than any power of z. For example, many trig or exponential functions of 1/z will have an essential singularity at z = 0. We occasionally are tasked with evaluating residues at essential singularities, and these pose some problems for those relying on formulae for residues of low-order poles. But still, how to evaluate such residues? The answer, as in all cases of nonsimple poles, is to analyze Laurent series and pluck out coefficients of 1/z. It really is straightforward conceptually; the real problems may lie in computation, which frequently involves summing infinite series. Let’s look at an integral over the unit circle as an illustrative example. We seek the value of the following integral. 2π dθ e

a cos θ

2π cos (sin θ ) = Re

0

d θ ea cos θ+i sin θ

(3.123)

0

To write as a complex integral, make the same substitution as above and, after a little manipulation, get  −i

|z|=1

  1 1 dz −1 exp (a + 1)z + (a − 1)z z 2 2

(3.124)

There is an essential singularity at z = 0; the way to find the residue there is to expand the exponential into a Taylor/Laurent series and find the coefficient of z 0 (because there is a factor of z −1 inside the integral already:  −i

|z|=1

∞ k dz  1  (a + 1)z + (a − 1)z −1 k z 2 k!

(3.125)

k=0

Using the binomial theorem, we write the sum as ∞ k    1  k (a + 1)m (a − 1)k−m z 2m−k 2k k! m=0 m k=0

(3.126)

3.3 Residue Reduction Contours

75

We get the coefficient of z 0 in the sum, for each k, when m = k/2. This only works for even values of k. The result is that the coefficient of z 0 is, when |a| > 1: k ∞  2

 a +1 = I0 a2 − 1 2k 2 2 (k!)

(3.127)

k=0

The integral is the real part of, by the residue theorem, i2π times the residue of the essential singularity at z = 0. Therefore 2π

d θ ea cos θ cos (sin θ ) = 2π I0



a2 − 1

(3.128)

0

where I0 is the modified Bessel function of the first kind of order zero. It should be clear what the result is when |a| ≤ 1.

3.3 Residue Reduction Contours Many times we are faced with evaluating contour integrals in which the integrand has many poles inside the contour. The value of the integral would then be expressed as a sum over the residues of each pole. Frequently, this can be unwieldy. The purpose of this section, then, is to explore ways to reduce the number of residues needed to evaluate an integral. Mostly, this involves using a different contour than one might normally use. And many times, the change in contour also requires a change in integrand. Someone once said that evaluating definite integrals using the Residue Theorem is an art. Based on a first course in Complex Analysis, this is mostly true. The intention here, however, is to introduce what we need based on some logic and deduction. Wedge The first example here is one that is more or less standard and perhaps which with the reader is familiar. The problem is to evaluate the following integral: ∞ dx 0

xm 1 + xn

(3.129)

with m, n ∈ Z, n − m ≥ 2, and m ≥ 0. Let’s evaluate this integral using the usual technique for integrals over the positive real axis: consider the following integral:  dz C

zm log z 1 + zn

(3.130)

76

3 Evaluation of Definite Integrals II: Applications to Various Types of Integrals

where C is the usual keyhole contour in Fig. (3.1). I leave the demonstration that the integrals over the small and large arcs vanish as R → ∞ and  → 0 to the reader. The contour integral is accordingly equal to  C

zm dz log z = −i2π 1 + zn

∞ dx 0

xm 1 + xn

(3.131)

By the Residue Theorem, the contour integral is equal to i2π times the sum of the residues at the poles inside C. Note that the the interior of C includes a large circle with the origin and positive real axis excluded as a branch cut. Note that arg z ∈ [0, 2π ) because of the way we defined the branch cut. C contains all of the n poles of the integrand; the reader may verify that the poles are at zk = exp (i(2k + 1)π/n). The contour integral is then i2π

n−1  k=0

 Res z=zk

zm log z 1 + zn



  n−1 i2π  i2π(m+1−n)( 2k+1 ) iπ(2k + 1) n e n n k=0 ! n−1 " n−1   2π 2 iπ m+1 m+1 m+1 i2π k i2π k n n ke + e = 2 e n 2 n

=

k=0

(3.132)

k=0

It will be an exercise for the reader to demonstrate that, upon setting the expression in Eq. (3.131) equal to Eq. (3.132), the integral is found to be ∞ 0

  m+1 xm π π dx = csc 1 + xn n n

(3.133)

Was that really as bad as it may have been implied? If the reader has as much expertise with geometric series as one may assume in profiting from the subject matter of this book, then the above sums should not pose that much of a challenge. That said, even for someone practiced in this area, the evaluation of the above sums can be a messy affair. And such a messy affair—providing such a relatively simple result—makes one wonder of there is a more efficient way at arriving at that result. If you have been exposed to the Residue Theorem already, then you likely know where this is going. Rather than blindly applying the Residue Theorem as we did above, we find a contour that satisfies the following: 1. The contour encloses as few poles as possible, preferably one. 2. The integral we wish to evaluate results from evaluation of each portion of the contour.

3.3 Residue Reduction Contours

77

Fig. 3.4 Wedge contour enclosing a single pole at z = exp (iπ/7)

In this case, the reader may be familiar with a wedge contour that encloses only a single pole. Figure 3.4 illustrates such a wedge for the case n = 7. Let’s denote the wedge contour by W and evaluate the contour integral over W . Note that, due to the nature of the wedge, the factor of log z is not needed in the integrand. This contour is relatively simple and has only three components:  W

zm dz = 1 + zn

R 0

xm dx + iRm+1 1 + xn 0

+ ei2π(m+1)/n R

2π/n 

dθ 0

ei(m+1)θ 1 + Rn einθ (3.134)

tm dt 1 + tn

  Note that, on the third piece of W —the sloped line—we set z = exp i2π m+1 , as n m+1 2π n is the wedge angle. Also note how fortuitous it is that the denominator of the integrand in the third integral matches that in the first integral. Actually, “fortuitous” is not the right word—the denominators match by design. Let’s estimate the second integral using the ML-lemma: # # 2π/n # #  i(m+1)θ # m+1 # m+1 e # ≤ 2π R #iR d θ # 1 + Rn einθ ## n Rn − 1 #

(3.135)

0

The above expression vanishes as R → ∞ because n − m ≥ 2. The remaining terms of the contour integral may be combined in this limit to produce  W

  zm dz = 1 − ei2π(m+1)/n 1 + zn

∞ dx 0

xm 1 + xn

(3.136)

78

3 Evaluation of Definite Integrals II: Applications to Various Types of Integrals

The contour integral is also equal to i2π times the residue at the pole at z = eiπ/n ; accordingly we may write ∞

  1 − ei2π(m+1)/n

dx 0

xm 1 = −i2π eiπ(m+1)/n n 1+x n

(3.137)

From this, the result in Eq. (3.133) is confirmed. We can now see how the use of a different contour (and integrand) simplified the computation of a definite integral. We will show a few more examples of this. Box, Or here’s one that looks like a sinh Let’s try to evaluate the following definite integral. ∞ dx −∞

sinh ax sinh bx cosh cx

(3.138)

where |a| + |b| < |c|. (Why?) Let’s first try our go-to semicircular contour of radius R in the upper half plane. Clearly there will be an infinite set of poles along the imaginary axis. But before we trouble ourselves with that, let’s write out the usual contour integral. (The astute reader, and the non-astute reader probably can see that this is not going to end well.)  C

sinh az sinh bz = dz cosh cz

R dx −R

sinh ax sinh bx cosh cx

π + iR

    iθ sinh bReiθ iθ sinh aRe   dθ e cosh cReiθ

(3.139)

0

We can estimate the second integral as R → ∞ using the ML-Inequality. # # #   #  # π # sinh aReiθ  sinh bReiθ  # iθ iθ ## # sinh bRe sinh aRe # # #iR d θ eiθ #     # # # ≤ π RMaxθ ## # cosh cReiθ cosh cReiθ # # 0 (3.140) And therein lies the problem. The quantity in Eq. (3.140) will grow without bound as R → ∞ for θ = π/2. This does not mean the whole technique is impossible, but it will involve a delicate cancellation of this oscillatory behavior with infinite amplitude with the infinite sum of residues, which makes one shudder with terror. There is obviously a better way, but how to find it? If we take a look at how the integrand behaves, it is clear that that the integrand converges along the real axis and diverges along the imaginary axis. To reflect this behavior, let’s imagine the contour is a rectangle with sides parallel to the real axis that increase in length without bound,

3.3 Residue Reduction Contours

79

and parallel to the imaginary axis that are fixed in length. In this way, we obtain the original definite integral without having to deal with the awful behavior at sufficient distance from the origin along the imaginary axis. A good part of this is we can define the height of the rectangle parallel to the imaginary axis such that the rectangle encloses only a single pole of the integrand. We will find, however, that a change in contour may necessitate a change in integrand to get the desired definite integral. So let’s consider the following contour integral over C, where C is a rectangle with vertices at the points −R, R, R + iπ/c, −R + iπ/c. Note this rectangle encloses the pole at z = iπ/(2c) and no others. 

sinh az sinh bz dz = cosh cz

C

R dx −R

sinh ax sinh bx cosh cx

π/c sinh a(R + iy) sinh b(R + iy) + i dy cosh c(R + iy) 0

(3.141)

R −

dx

−R

sinh a(x + iπ/c) sinh b(x + iπ/c) cosh(cx + iπ/c)

π/c sinh a(−R + iy) sinh b(−R + iy) − i dy cosh c(−R + iy) 0

The reader should be able to see that the second and fourth integrals, over the vertical sides of C, vanish in the limit as R → ∞ because |a| + |b| < |c|. The contour integral in this limit is then 

sinh az sinh bz = dz cosh cz

C

∞ dx

sinh ax sinh bx cosh cx

dx

sinh a(x + iπ/c) sinh b(x + iπ/c) cosh (cx + iπ/c)

−∞ ∞

− −∞

   ∞ πa

πb sinh ax sinh bx = 1 + cos cos dx 2c 2c cosh cx − sin

πa

2c

 sin

πb 2c

−∞

 ∞ dx −∞

cosh ax cosh bx cosh cx

(3.142)

80

3 Evaluation of Definite Integrals II: Applications to Various Types of Integrals

Well, who saw that coming? The contour integral is a linear combination of the integral we seek and an integral we never asked for. The usual reaction to a complication like this is to figure out another way to evaluate the integral. Nevertheless, it turns out that the extra integral can be evaluated just like we outlined for the integral we intend to evaluate. That is, we consider the contour integral 

cosh az cosh bz dz = cosh cz

C

∞ dx −∞

cosh ax cosh bx cosh cx

∞ − −∞

dx

cosh a(x + iπ/c) cosh b(x + iπ/c) cosh(cx + iπ/c)

 ∞ πb sinh ax sinh bx = − sin sin dx 2c 2c cosh cx −∞    πa

πb + 1 + cos cos 2c 2c ∞  cosh ax cosh bx dx cosh cx πa



(3.143)

−∞

Hopefully the reader can see where this is going. We need the residues of the respective integrands at the poles. By the Residue Theorem,  dz C

 C

  sinh az sinh bz sinh az sinh bz = i2π Res z=iπ/(2c) cosh cz cosh cz   πa

πb −2π sin sin = c 2c 2c

  cosh az cosh bz cosh az cosh bz = i2π Res dz z=iπ/(2c) cosh cz cosh cz   πa

πb 2π cos cos = c 2c 2c

(3.144)

(3.145)

We will evaluate the desired integral by solving a 2 × 2 linear system. Before we write down the system, let’s define

3.3 Residue Reduction Contours

81

∞ I1 :=

dx

sinh ax sinh bx cosh cx

(3.146)

dx

cosh ax cosh bx cosh cx

(3.147)

−∞ ∞

I2 := −∞

The linear system to be solved is ⎛ ⎛

⎞ ⎞         − sin π2ca sin π2cb − sin π2ca sin π2cb 1 + cos π2ca cos π2cb 2π I 1

⎠ ⎠ ⎝ ⎝ =       I2 c 1 + cos π2ca cos π2cb − sin π2ca sin π2cb cos π2ca cos π2cb

(3.148) The result of solving this system is ∞ −∞

     π π(a + b) π(a − b) sinh ax sinh bx = sec − sec . dx cosh cx 2c 2c 2c

(3.149)

The algebra is not trivial and should be verified. But we can simplify the algebra in the first place by considering different contours and integrands. For example, we can also derive the result in Eq. (3.149) by shifting the contour downward so that the center of the rectangle is at the origin rather than at z = iπ/(2c). Let’s briefly walk through this. Now C is a rectangle with vertices at −R − iπ/(2c), R − iπ/(2c), R + iπ/(2c), and −R + iπ/(2c). So nominally, we may consider the following contour integral  dz

sinh az sinh bz cosh cz

(3.150)

C

A little thought renders this integrand problematic. Note that there are poles of this integrand on C. We will deal with such poles in the next chapter. We would rather maintain the pole at the rectangle center, as we did in the example above. So we can make a minor change to the integrand and now consider the contour integral  dz

sinh az sinh bz sinh cz

(3.151)

C

As the reader can see, at least this integrand has a single pole at the origin within C. Moreover, the fact that the integrals parallel to the real axis are symmetrically situated with respect to the real axis will make the algebra somewhat simpler than what we just experienced. This of course begs the question: why did we go with the messy algebra first? Because it is a lot easier to understand the reasoning behind changing the integrand when the context of the straightforward solution has been provided.

82

3 Evaluation of Definite Integrals II: Applications to Various Types of Integrals

Verification that this contour integral provides the solution in Eq. (3.149) is left as an exercise. Chirped! We’ve talked plenty about having to define integrands and contours to evaluate a definite integral, and that sometimes the integrand of the contour integral is not the same as that of the definite integral. Here we will go through a detailed example that should give the reader an idea of how we determine a combination of integrand and contour. The task is to evaluate the following integral: ∞ 0

  cos π x2

dx 2π 1 + 2 cosh √ x 3

(3.152)

(N.B. The title of this subsection refers to the quadratic argument of the cosine in the numerator of the integrand.) Again, no reason to not try the usual semicircular contour in the UHP. Note that we may have a better chance of success by converting the cosine into an exponential.  dz C

eiπx

R

2

1 + 2 cosh



2π √ z 3

=

dx −R

eiπz

2

1 + 2 cosh



2π √ x 3

π d θ eiθ

+ iR 0

e

(3.153)

iπR2 ei2θ

1 + 2 cosh



2π √ Reiθ 3



It is far from clear that the second integral vanishes—or even converges—as R → ∞. For example, in the neighborhood of θ = π/2, the numerator and denominator become oscillatory and as R → ∞, highly oscillatory, with poles along the imaginary axis. To

matters worse,

the poles of the integrand in the contour integral are at make 3k+1 3k+1 z = i √3 and z = i √3 for k = 0, 1, 2, .... As a result, the contour integral is equal to  dz C

eiπz

2

1 + 2 cosh



2π √ z 3

= 2e

−i π3

∞ 

√

e−i3πk e−i3 2

3πk

cos

√

3π k

(3.154)

k=0

Maybe we could evaluate the sum in closed form. But we still have to contend with the integral over the arc of radius R. This is enough to justify finding another way to solve the problem. So in the spirit of the previous subsection, let’s contain a few poles within a rectangular contour. And as we discovered there, rectangles symmetric about the origin may make things easier. (This of course is not a guarantee but seems like at least a rule of thumb for a first try.)

3.3 Residue Reduction Contours

83

But of course, we have no idea what the width of the rectangle should be nor the integrand of the contour integral. So let’s define the contour as a rectangle R with vertices −R − ib, R − ib, R + ib, R − ib, where we hope to determine the value of b. So without any other data, let’s see what the following contour integral tells us.  dz C

eiπz

2

1 + 2 cosh



2π √ z 3



(3.155)

which, as R increases without bound, becomes (as the integrals over the sides parallel to the imaginary axis vanish) ∞ −∞

⎛

⎞ iπ(x−ib)2 iπ(x+ib)2 e e

−

⎠ dx ⎝ 2π 2π √ 1 + 2 cosh √ (x − ib) 1 + 2 cosh (x + ib) 3 3

(3.156)

Now,           2π 2π 2π 2π 2π cosh √ (x ± ib) = cosh √ x cos √ b ∓ i sinh √ x sin √ b 3 3 3 3 3 (3.157) √ part of the cosh; in that case It appears b = 3/2 would get rid of the imaginary√ the rectangle is centered at the origin and has width 3. Accordingly, the contour integral as expressed in Eq. (3.156) becomes  dz C

eiπz

2

1 + 2 cosh



2π √ z 3

= −2e−i3π/4

∞

−∞

√

3π x

dx 2π −1 + 2 cosh √ x 3 2

eiπx sinh

(3.158)

The integral on the RHS is unfortunately a far cry from the one we actually want to evaluate in Eq. (3.152). But that’s OK: this is precisely the sort of progress we need to try again. For example, note that the resulting definite integral has a factor of

√ 3π x in the numerator of the integrand and a difference instead of a sum in sinh the denominator. Along these lines, maybe the following contour integral will work:  C

2

eiπz √





dz 2π sinh 3π z 2 cosh √ z −1 3

(3.159)

In evaluating this contour integral, we note that ! ! √ ""

√ √ 3 3π x ± i 3π x sinh = ∓i cosh 2

(3.160)

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3 Evaluation of Definite Integrals II: Applications to Various Types of Integrals

The cosh would look like trouble except the reader should note the change in sign because sin (3π/2) = −1. Building on the previous observations, we find that ∞ 2 2 eiπz eiπx −i3π/4 √



= i2e

dz dx 2π 2π √ √ sinh 3π z 2 cosh z − 1 2 cosh x +1 −∞ C 3 3 (3.161) And there we have it—the contour integral of Eq. (3.159) is equal to some factor times the definite integral we seek. Before we go ahead and finish off the calculation via the Residue Theorem, let’s briefly review how we got here. 

1. First, we tried the most obvious way to attack the given integral in the complex plane by using a semicircular contour in the UHP with an unchanged integrand. The result was a messy sum over the infinite set of residues and an integral over the semicircle that oscillates wildly as a function of R. Not ideal to say the least. 2. Next, we determined a rectangular contour that would help, still using the original integrand. There we found the resulting definite integral wasn’t what we sought but the integrand might be used to guide us to an integrand that would return the desired definite integral. 3. Finally, taking a cue from the previous integrand, we defined a contour integral with a new integrand over the rectangle, and this indeed got us what we wanted. Now that we have established a contour integral that returns the original definite integral, we may now apply the Residue Theorem to the contour integral: the contour integral in Eq. (3.161) is equal to i2π times the sum of the residues of the poles inside C. Of course, we need to find those poles and residues. Fortunately it is not terribly √ √ difficult in this case. The poles within C are at z = 0, z = ±i 3/3, and z = ±i 3/6. The residues are as follows: ⎡ ⎤ iπz 2 e 1 √



⎦ = √ Res ⎣ (3.162) z=0 2π 3π sinh 3π z 2 cosh √ z − 1 3 ⎡ ⎤ 2 iπz e e−iπ/3 ⎣ √



⎦ = √ Res (3.163) √ 2π z=±i 3/3 sinh 2 3π 3π z 2 cosh √ z −1 3 ⎡ ⎤ iπz 2 e e−iπ/12 ⎣ ⎦





(3.164) Res = − √ √ 2π 2π z=±i 3/6 sinh 3π z 2 cosh √ z −1 3

Accordingly,

i2e

−i3π/4

∞

−∞

2

eiπx

= i2π dx 2π 2 cosh √ x + 1 3



1 e−iπ/3 e−iπ/12 + √ − √ π 3π 3π

 (3.165)

3.4 Definite Integrals Evaluated Between Branch Points

85

At this point, the reader can work out the arithmetic for themself. The value of the original integral in Eq. (3.152) is then simply the real part of the result of that arithmetic:   √ √ ∞ cos π x2 2+ 2− 6

= (3.166) 2π 8 √ 1 + 2 cosh x 0 3 We may as well show the integral we get for free by taking the imaginary part of the arithmetic in Eq. (3.165): ∞ 0

  √ √ √ sin π x2 2+ 6−2 3

= 2π 8 1 + 2 cosh √ x 3

(3.167)

It is the intent of this book to demonstrate more systematic ways to use complex integration to evaluate difficult integrals, and it is hoped that the reader can apply the thinking provided in this section to other difficult integrals. How difficult? It turns out that the results in Eqs. (3.166) and (3.167) still cannot be reproduced by various computer algebra systems as of the time of this writing.

3.4 Definite Integrals Evaluated Between Branch Points Frequently we are faced with integrals whose upper and lower limits are branch points of an integrand. One example that comes to mind is an evaluation of a Beta Function for non-integer values. If this topic is treated at all in a first course, one is told to use a “dog bone” contour that avoids a branch cut between the branch point integration limits. Then, one is told to consider the Residue at Infinity and given a formula for such a residue, which contributes to the value of the definite integral as any other residue. The reality is that the branch cuts of the integrands of contour integrals used in evaluating integrals between branch points extend beyond the integration limits. If we limit ourselves to those limits and an additional residue, we cheat ourselves of an ability to evaluate ever more integrals using complex integration. In this Section, we will look at a few simple examples. We will see even more interesting examples when we study asymptotic techniques. A dogged approach to branch point integrals Let’s begin, as usual, by considering an integral that is commonly encountered: 1 0

dx x3/2 (1 − x)−1/2

(3.168)

86

3 Evaluation of Definite Integrals II: Applications to Various Types of Integrals

Fig. 3.5 Dog-bone contour about branch points at z = 0 and z = 1

Most readers may recognize this as a beta function integral 

5 1 B , 2 2

 =

5 1

2 2

(3)

(3.169)

The usual way one is taught to evaluate this integral using complex integration methods is to set up a dog bone contour extending over the branch cut between the branch points at x = 0 and x = 1. We consider the following integral 

dz z 3/2 (z − 1)−1/2

(3.170)

C

where C represents a “dog-bone” contour as in Fig. 3.5. The reader may have seen this sort of contour before and may have been bothered by the fact that this contour, rather than exclude the branch points at z = 0 and z = 1, it instead seems to go out of its way to include them. The cure for this is to stress that the contour C is oriented outward rather than inward. Accordingly, if the contour is oriented outward, we need to consider the residue at infinity. For example, if we are evaluating the contour integral about C of a function analytic within C, f (z), then the residue at infinity is given by  Res f (z) = − Res

z=∞

z=0

1 f z2

  1 z

(3.171)

The astute student should be able to see that the relation defined in Eq. (3.171) is based on the substitution z = 1/w, dz = −dw/w2 . We evaluate the integral in Eq. (3.168) by evaluating the following contour integral clockwise—recall, this is because the contour in oriented outward rather than inward.  dz z 3/2 (z − 1)−1/2 (3.172) C

We chose z − 1 to induce the different arguments over each side of the branch cut. The reader should understand that each factor in the integral has its own branch cut. For example, z 3/2 has a branch cut over (−∞, 0), while (z − 1)−1/2 has a branch cut

3.4 Definite Integrals Evaluated Between Branch Points

87

over (−∞, 1). In light of these branch cuts, if it seems a bit strange that C intersects both branch cuts in the neighborhood to the left of z = 0, it is because it is strange. The contour C is merely an abbreviation for the contour that is really intended. We can ignore this curious detail because in this case, the integrals over the arcs around the branch points vanish as their radius R → 0. Moreover, it turns out that, in this case—and virtually all cases shown in first treatments—the difference in arguments between either side of the branch cut to the left of the integration interval is a multiple of 2π . This practically renders the integrand as analytic over that portion of the branch cut. In the end, the crossing of the branch cut to the left of the integration interval is not as strange as it appears in this case. So with that discussion behind us, let’s evaluate the contour integral. Note that above the real axis in the integration interval, arg z = 0 and arg (z − 1) = π and below the real axis arg z = 0 and arg (z − 1) = −π . (Note that z 3/2 has no branch cut in the integration interval so its argument does not change going from one side of the real axis to the other there.) Accordingly,  dz z

3/2

(z − 1)

−1/2

1−  R

=

dx x

3/2 −iπ/2

e

−1/2

(1 − x)

1−  R

R

C

dx x3/2 eiπ/2 (1 − x)−1/2

− R

1−  R

dx x3/2 (1 − x)−1/2

= −i2

(3.173)

R

The contour integral is also equal to i2π times the residue at infinity: 

 dz z

3/2

(z − 1)

−1/2

C

 −1/2  1 −3/2 1 −1 = −i2π Res 2 z z=0 z z   1 −1/2 = −i2π Res 3 (1 − z) z=0 z 3 = −i2π 8

(3.174)

Accordingly, letting R → 0: 1

dx x3/2 (1 − x)−1/2 =

3π 8

(3.175)

0

This is all well and good and the reader may find such problems worked out like we did above in many textbooks. But let’s work the same problem without any abbreviations. Let’s compute the same contour integral but with C as follows (Fig. 3.6).

88

3 Evaluation of Definite Integrals II: Applications to Various Types of Integrals

Fig. 3.6 Unabbreviated dog-bone for branch points at z = 0 and z = 1

We may simply allow R → 0 and the contour integral is  dz z

3/2

(z − 1)

−1/2

0 =e

iπ

dx ei3π/2 x3/2 e−iπ/2 (x + 1)−1/2

R

C

1 +

dx x3/2 e−iπ/2 (1 − x)−1/2

0

1 −

dx x3/2 eiπ/2 (1 − x)−1/2

0

+e

−iπ

R

dx e−i3π/2 x3/2 eiπ/2 (x + 1)−1/2

0

π + iR

 −1/2 dx eiθ R3/2 ei3θ/2 Reiθ − 1

(3.176)

−π

The reader should see that the integrals about either side of the branch cut to the left of the integration interval cancel in this case. The reason is that the exponents in the integral, i.e., the arguments of the beta function sum to an integer. Accordingly,  C

dz z 3/2 (z − 1)−1/2 = −i2

1 0

dx x3/2 (1 − x)−1/2 + iR2

π −π

  1 −1/2 d θ ei2θ 1 − iθ (3.177) Re

3.4 Definite Integrals Evaluated Between Branch Points

89

The integral over θ provides the residue at infinity. To see this, assume R > 1 and we may Taylor expand with respect to R. The reader should note that the only integrated term that is nonzero is that which cancels out the factor of ei2θ in the integrand, i.e., the 1/R2 term in the expansion:        1 3 1 1 −1/2 1 1 1 1 (3.178) − − 1 − iθ =1+ + + O iθ 2 i2θ Re 2 Re 2! 2 2 R e R3 All we care about is the coefficient of  dz z

3/2

(z − 1)

−1/2

1 , R2

1 = −i2

which is 38 . Accordingly,

dx x3/2 (1 − x)−1/2 + i

3π 4

(3.179)

0

C

which provides the result seen in Eq. (3.175). That said, it is interesting to note that the above evaluation does not require that we take R → ∞; rather, only that R > 1. Generally speaking, if the difference in argument of the integrand about either side of the branch cut is a multiple of 2π , the integrals out to the large circle cancel and we can use the dog-bone and residue at infinity. At this point, the reader should expect that this not always the case. Easier than falling off a log So the natural next step to take is to consider p + q that is not an integer in Eq. (3.168). We are ready to develop the general case based on the contour in Fig. (3.6). That said, we do not yet possess all of the tools we will need for evaluation of some of the more general cases. For that, we will need some very interesting methods of approximating the integrals along the branch cuts which we will develop later. But do not be disappointed! We will work out an example in full right here that is not reliant on the more advanced techniques we will explore in the chapters to come. This example is fairly close to the beta function example we have been studying. Without further ado, let’s evaluate the following integral using the complex integration techniques described here: 1 0

log x dx √ x(1 − x)

(3.180)

Truthfully, this integral is simply a derivative of a beta function. But that’s not the point. Rather, we are testing out a new idea on a relatively simple case so we can focus on the idea rather than excessively messy algebra. So we begin by defining the usual contour integral:  log z (3.181) dz √ z(z − 1) C

90

3 Evaluation of Definite Integrals II: Applications to Various Types of Integrals

where C is the contour in Fig. (3.6). Let’s write this out; because the singularities have exponent > −1, it is safe to ignore the contributions about the small arcs and let their radius R → 0. Accordingly, 

log z = eiπ dz √ z(z − 1)

C

0 + 1

0

log x + iπ + dx √ √ eiπ/2 xeiπ/2 x + 1

R

log x + e−iπ dx √ √ −iπ/2 xe 1−x π

+ iR −π

1 dx √ 0

log x √ xeiπ/2 1 − x

R dx

e−iπ/2

0

log x − iπ √ −iπ/2 √ xe x+1

log R + iθ d θ eiθ √ √ Reiθ Reiθ − 1 R

= −i2π 0

π + i log R −π

dx − i2 √ x(x + 1)

1 0

log x dx √ x(1 − x)

    π 1 −1/2 1 −1/2 d θ 1 − iθ − d θ θ 1 − iθ Re Re −π

(3.182) In this case, we will need to allow R to take on increasingly large values. By allowing R to increase without bound, the fourth integral on the RHS of Eq. (3.182) vanishes. The second integral corresponds to the integral whose value we seek. This leaves the first and third integrals. In this case, the first integral is actual simpler than the original integral as we have done away with the log in the numerator. The first integral is in fact fairly easily evaluated: R 0

√

dx =2 √ x(x + 1)

R 0

√ √ du R+ R+1 = 2 log √ 1 + u2

(3.183)

As R increases, the nonvanishing behavior of the integral is log R + 2 log 2. It is clear that in this limit, the third integral has nonvanishing behavior log R. Putting this together, we see that the contour integral is equal to, as R → ∞,  C

log z = −i2π (log R + 2 log 2) − i2 dz √ z(z − 1)

1 0

log x + i2π log R dx √ x(1 − x)

(3.184) We note that, of course, the divergent log R terms cancel. By Cauchy’s Theorem, the contour integral is zero; accordingly,

3.5 Sums

91

1 dx √ 0

log x = −2π log 2 x(1 − x)

(3.185)

To review, the integral over the portion of the contour usually ignored in conventional treatments is that which provides the value we seek. Moreover, the divergent behavior of the integral over this portion is canceled by the divergent behavior of the integral about the circle of radius R. In the more complicated examples we will see later, the contributions will come from both of these integrals.

3.5 Sums Well, we did advertise evaluating integrals and sums, so it is time to uphold that end of the bargain. Some first treatments will see the Residue Theorem be applied to infinite sums—infinite over all integers, not just the nonnegative ones. We will not deviate from that requirement here; there are places that treat such sums using residues. Rather, we will be expanding, after a review and some general musing, the class of sums that may be evaluated using the Residue Theorem. This is made possible, as has been virtually everything else discussed in this chapter, by reconsidering the basics. Poles at the integers The problem generally is to evaluate the following: ∞ 

f (n)

n=−∞

where f is meromorphic in C and has poles that are not real integers and has behavior for | Re z| → ∞ such that the sum converges. The idea is to express the sum in terms of a contour integral which may be evaluated using Cauchy’s Theorem or, more specifically, the Residue Theorem. So nominally, we may consider a contour integral of the following form:  dz H (z)f (z)

(3.186)

C

where H is a meromorphic function with poles at the real integers. Accordingly, C is a contour that encompasses all of the real line in addition to the poles of f . Additionally, H is defined such that the contour integral vanishes as it encompasses increasingly many poles. A well-known H (z) is π cot (π z). In terms of residues this works because π cos (π n) =1 (3.187) Res [π cot (π z)] = z=n π cos (π n)

92

3 Evaluation of Definite Integrals II: Applications to Various Types of Integrals

Of course, the other requirement is that the contour integral vanishes as the contour encompasses all of the poles of the integrand. A contour that works here is the square with vertices N ± 21 (1 ± i). The contour integral is then N + 21



 π

dz f (z) cot (π z) = π

dx f

       1 1 cot π x − i N + x−i N + 2 2

−N − 21

C

N + 21



+ iπ

dy f

     1 1 N + + iy cot π N + + iy 2 2

−N − 21 N + 21



−π

dx f

       1 1 x+i N + cot π x + i N + 2 2

−N − 21 N + 21



− iπ

dy f

     1 1 −N − + iy cot π −N − + iy (3.188) 2 2

−N − 21

We consider the value of the contour integral as N → ∞. To estimate the value of the contour integral in this limit, we apply the ML-inequality: # # # #   # # # dz f (z) cot(π z)# < 8 N + 1 # # 2 # # C #   # #    # ## # # 1 # #cot π N + 1 (1 + i) # #f N + (1 + i) # # # # 2 2

(3.189)

And here is another major reason we chose H (z) = π cot (π z): the magnitude of the cotangent approaches unity as N → ∞: # # # eπ (N + 21 )(1+i) + e−π (N + 21 )(1+i) # # # # π N + 1 (1+i) # −π (N + 21 )(1+i) # #e ( 2) −e # # 1 # # < #1 + 2e−π (N + 2 )(1+i) #

#    # # # #cot π N + 1 (1 + i) # = # # 2

≤ 1 + 2e−π (N + 2 ) 1

(3.190)

which clearly approaches unity as N → ∞. Accordingly, if for some α > 0, #   #   # # 1 −(1+α) 1 # #f N+ (1 + i) # ≤ N + # 2 2

(3.191)

3.5 Sums

93

as N → ∞, then the magnitude of the contour integral is bounded by # # # #  −α # # # dz f (z) cot (π z)# < 8 N + 1 # # 2 # #

(3.192)

C

in this limit, and accordingly the contour integral vanishes in this limit. Of course, the contour integral is also equal to the sum of the residues at its poles. As the contour integral is zero, then the following must hold: i2π

∞ 

f (n) + i2π

n=−∞

M 

  π Res f (z) cot (π z) = 0

m=1

z=zm

(3.193)

where M is the number of poles {zm } of f in the complex plane away from the real integers. This brings us to the following well-known result: ∞ 

f (n) = −π

n=−∞

M  m=1

  Res f (z) cot (π z)

z=zm

(3.194)

The above framework may be used for alternating sums: ∞ 

(−1)n f (n)

(3.195)

n=−∞

It should not surprise the reader that H (z) = π csc (π z). The requirement for the behavior of f as N → ∞ is that α > −1; we leave the proof of this for the exercises. We will work through a standard example for review. Then we will go through examples that do not quite fit this framework but still benefit from a complex integration treatment. An example made from whole coth We will be concerned with the following sum: ∞  n=1

∞ 1 1 1  1 = − 2 n2 + a2 2 n=−∞ n2 + a2 2a

(3.196)

−1  If we let f (z) = z 2 + a2 , we can apply the relation in Eq. (3.194) to evaluate the infinite sum on the RHS:

94

3 Evaluation of Definite Integrals II: Applications to Various Types of Integrals

   cot (π z) 1 = −π Res 2 z=k n2 + a2 z + a2 n=−∞ k∈±ia   cot (iπ a) cot (−iπ a) + = −π i2a −i2a   −i coth (π a) i coth (π a) = −π + i2a −i2a π = coth (π a) a ∞ 

Accordingly,

∞  n=1

Similarly,

n2

1 π 1 = coth (π a) − 2 2 +a 2a 2a

∞  (−1)n+1 n=1

n2

+

a2

=

1 π csch (π a) − 2 2a 2a

(3.197)

(3.198)

(3.199)

To set up a discussion we will have later, we can look at the behavior of these sums as a → 0 and a → ∞. We note that 2

1 + z2 + ... cosh z 1 z coth z = = + + ... = z3 sinh z z 3 z + 6 + ... 1 z 1 1 = = − + ... csch z = 3 z sinh z z 6 z + 6 + ...

(3.200)

Using this, we may deduce ∞  1 π2 = n2 6 n=1 ∞  (−1)n+1 n=1

n2

=

π2 12

(3.201) (3.202)

Moreover, we may also infer the behavior of the sums as a → ∞: ∞  n=1 ∞  n=1

1 1 π − = + ... n2 + a2 2a 2a2

(3.203)

(−1)n+1 1 = 2 + ... 2 2 n +a 2a

(3.204)

3.5 Sums

95

We will have a chance to revisit behaviors like these later when we discuss asymptotic expansions of integrals and sums. Poles up, down, left, and right The problem is to evaluate the following sum: ∞  n=1

(−1)n+1 sinh (π n)

n3

While we can apply the formula in Eq.(3.194), let’s take a step back and formulate a contour integral that will produce the sum we want. So let’s start with the contour integral we used to derive the formula in Eq.(3.194):  dz C

π z 3 sinh (π z) sin (π z)

(3.205)

We have simple poles at z = ±1, ±2, ... and z = ±i, ±i2, as well as a fifth-order pole at z = 0. By the residue theorem, the contour integral is equal to i2π times the sum of the residues at the poles inside C. Let’s compute the residues. Let n ∈ Z: 

 π (−1)n Res 3 = 3 z=n z sinh (π z) sin (π z) n sinh (π n)   (−1)n π = 3 Res 3 z=in z sinh (π z) sin (π z) n sinh (π n)

(3.206) (3.207)

So we have four-fold symmetry of the residues; that is, the residues are equal at z = ±n, ±in. Accordingly,  dz C

  N  π (−1)n+1 π = −4 + Res z 3 sinh (π z) sin (π z) n3 sinh (π n) z=0 z 3 sinh (π z) sin (π z) n=1

(3.208) We have to compute that residue at z = 0, a fifth-order pole. Again, trying to derive a simple formula for such a high-order pole is pointless. Rather we will perform a Laurent expansion about the pole: π 1 1 =



z 3 sinh (π z) sin (π z) π z 5 1 + π 2 z 2 + π 4 z 4 + ... 1 − π 2 z 2 + π 4 z 4 + ... 6 120 6 120 ⎞ ⎛ " ! "2 ! π2 2 π2 2 1 ⎝ π4 4 π4 4 = 1− z + z + ... + z + z + ... + ...⎠ 6 120 6 120 π z5 ⎛ ⎞ ! " ! "2 π2 2 π4 4 π2 2 π4 4 × ⎝1 + z − z + ... + z − z + ... + ...⎠ 6 120 6 120 " ! 4 1 π 4 z + ... = 1+ 90 π z5

(3.209)

96

3 Evaluation of Definite Integrals II: Applications to Various Types of Integrals

The reader may fill in the gaps of the above derivation as an exercise. Accordingly, 

 π π3 = Res 3 z=0 z sinh (π z) sin (π z) 90

(3.210)

so the contour integral is  dz C

N  π3 (−1)n+1 π = −4 + 3 3 z sinh (π z) sin (π z) n sinh (π n) 360 n=1

(3.211)

We need to show that the contour integral vanishes (or is equal to a constant, but here it is zero). The reader may show in the exercises that, when C is the square of side 2N + 1 centered at the origin, the magnitude of the contour integral is bounded as N → ∞ as follows: # # # # −(2N +1) π2 # # π # dz #≤ e (3.212) # z 3 sinh (π z) sin (π z) ## (2N + 1)2 # C

And we finally obtain the result we sought: ∞  n=1

π3 (−1)n+1 = n3 sinh (π n) 360

(3.213)

While we could have derived this result using the formula in Eq. (3.194), referring back to the original contour integral will prepare us for evaluation of sums using residues that are not shown in a first treatment. An irrational approach to summation We want to evaluate the following sum: ∞  n=1

n3

1 √ sin ( 2π n)

(3.214)

√ This is an interesting sum. The 2 term in the sine guarantees that the sine is never zero. In the meantime, let’s write down a contour integral that will produce the value of the above sum. So as before we write the following contour integral:  dz C

π cot (π z) √ sin ( 2π z)

z3

(3.215)

3.5 Sums

97

Well, if we just think about it √ before scribbling residues, we not only have poles at z = n ∈ Z but also at z = n/ 2. This will produce another sum—a pretty scarylooking sum—that we would have to figure out before we compute the sum of interest. So, to our utter surprise, the formulation that led to a Eq. (3.194) is not much use here. What can we do? We could just give up and try to find some other way to compute the sum. But the reader can figure out if that were the case, we would not be discussing this problem. Obviously, the key here is finding a contour integral that will produce the sum of interest alone, along with a constant term. Because we have no solid ideas of what this contour integral should look like, let’s look somewhere close by. For example, the reason the usual cotangent factor we use in the contour integral fails us here is because it produces yet another sum we will have to deal with, and frankly we have no way of dealing with that sum yet. We could use the cosecant in an attempt to avoid that nasty second sum, but the reader should be able to anticipate the problem: we will end up with the alternating sum which in this case is no help. In a situation like this in which we are our path to a solution seems blocked at every turn, we are forced to be a little creative. For example, one way to make progress is to introduce a parameter a whose value is to be determined. For example, let’s try the following.  π cot (π z) (3.216) dz 3 z sin (aπ z) C

Unfortunately, this contour integral has the same problem as that in Eq. (3.215) in that the residues at z = n/a produce a whole other sum, even nastier than the original. Nevertheless, we can learn from this. For example, the residues of the integrand at z = n/a contain a factor of (−1)n . What this tells us is that the integrand we seek should produce another factor of (−1)n to cancel this out. So let’s try the following contour integral,  π (3.217) dz 3 z sin (aπ z) sin (π z) C

which by the Residue Theorem satisfies the following equation: 1 i2π

 dz C

∞ ∞   (−1)n (−1)n π 2   = 2 + 2a 3 3 3 z sin (aπ z) sin (π z) n sin (aπ n) n sin πn a n=1 n=1

+ Res

z=0 z 3

π sin (aπ z) sin (π z)

(3.218)

How should we determine the value of a? While it is easy enough to eliminate values of a that won’t work, we need to figure out a value that does. The easiest way to go about this is, instead of guessing what value(s) or a gets us the sum we want, we should instead look for any value of a that creates the right conditions to allow us to move forward.

98

3 Evaluation of Definite Integrals II: Applications to Various Types of Integrals

The stroke that will move us forward is shockingly mundane: observe that (−1)n = cos (π n). This allows us to play around to get us something plausible without any more significantly creative effort: (−1)n sin (aπ n) = sin (aπ n) cos (π n) = sin (aπ n) cos (π n) ± cos (aπ n) sin (π n) = sin ((a ± 1) π n)

(3.219)

Comparing this result to the analogous term in the original sum gives us a first crack at a: √ (3.220) sin ((a ± 1) π n) = sin ( 2π n) √ which implies that a = 2 ± 1. Note that the ±1 eliminates the factor of (−1)n in the sum we never wanted in the first place, i.e., the residue of the csc (π z) term in the contour integral. This √ of course is by design but that doesn’t make it uninteresting. Let’s take a = 2 − 1. We still have yet to deal with the second sum on the RHS of Eq. (3.218). What happens when we plug in this value of a? ∞

2  √ (−1)n (−1)n   √



a 2 − 1 = πn n3 sin a 3 2 + 1 πn n=−∞ n=−∞ π n sin 2

∞ 

=

∞

2  √ 2−1 n=1

n3

1 √ sin ( 2π n)

(3.221)

It’s a miracle! The two sums may be combined into the one sum we sought. We may now write out the contour integral we needed the whole time: 

π √



sin 2 − 1 π z sin(π z) C   ∞

2  √ 1 2−1 =2 1+ √ 3 n sin( 2π n) n=1 π √



+ Res z=0 3 z sin 2 − 1 π z sin(π z)

1 i2π

dz

z3

(3.222)

We leave it as an exercise to show that Res z=0

z 3 sin

√

13π 3 √



= 2−1 90 2 − 1 π z sin (π z) π

And we can finally produce an exact result for the sum of interest:

(3.223)

3.5 Sums

99 ∞  n=1

1 13π 3 =− √ √ n3 sin ( 2π n) 360 2

(3.224)

This is an amazing result, and we derived it using a theme common to all of the boundary-pushing results we obtained throughout this chapter: by going back to the basics rather than relying on formulae derived in first treatments. In the chapters to follow, we will continue to take similar hard looks at topics the reader may have encountered in advanced undergraduate or graduate math or physics courses.

Problems 3.1 (a) Show that the following recurrence relation is satisfied:  m   m+1 k

k=0

(i2π )m−k Ik = −

N 

Rn logm+1 zn

n=1

where

∞ Ik =

dx f (x) logk x 0

and Rn = Res f (z) z=zn

with f : C → C has N isolated simple poles in the complex plane away from the positive real axis. (b) As defined in Eq. (3.33) and based on the recurrence relationship in (a), find the universal polynomials P3 , P4 , and P5 . (c) Evaluate the following integrals: ∞ (i)

dx

log2 x x2 + 2x cos (πβ) + 1

β ∈ [−1, 1]

dx

log4 x x2 + 2x cos (πβ) + 1

β ∈ [−1, 1]

0

∞ (ii) 0

Can you find a way to provide an independent check on your answer to (i)? 3.2 Show that the second integral in Eq. (3.41) vanishes as R → ∞. Hint: break up into real and imaginary parts.

100

3 Evaluation of Definite Integrals II: Applications to Various Types of Integrals

3.3 (a) Evaluate

∞

dt log (1 + xt)t −p−1

0

where p ∈ (0, 1) and x > 0. Carefully consider the contour to be drawn and the branch cuts from each term in the integrand. (b) Evaluate ∞ log (1 + x3 ) dt (1 + x)2 0

Carefully consider the contour to be drawn and the branch cuts from each term in the integrand. 3.4 (a) Evaluate

π dx 0

x sin x 1 + a2 − 2a cos x

for the cases (i) 0 < a < 1 and (ii) a ∈ C. (b) Evaluate

π dx 0

x3 sin x 1 + a2 − 2a cos x

for the case a > 1. 3.5 Evaluate

π dx

x2 p − q cos x

dx

xm log x xn + 1

0

where p > q. 3.6 (a) Derive Eq. (3.96). (b) Derive Eq. (3.109). 3.7 Evaluate

∞ 0

where n > m + 1. Can you work with higher powers of log?

3.5 Sums

101

3.8 Laplace’s equation may be used in any variety of physical situations; for example steady-state temperature over a particular domain. The solution to the problem of finding the steady-state temperature φ(x, y) over a strip of width 2a as follows: ∂ 2φ ∂ 2φ + = 0 ∀x > 0 and |y| < a ∂x2 ∂y2 such that φ(x, ±a) = 1 ∀x > 0 φ(0, y) = 0 ∀|y| < a The solution to this equation may be found using transform techniques: φ(x, y) =

2 π

∞ dk

sin kx cosh ky k cosh ka

0

Evaluate the integral on the RHS by setting up a contour and integrand to produce a closed form for φ. 3.9 (a) Evaluate

∞ dx 0

coth2 x − 1 (coth x + x)2 + π 2 /4

The challenge is to find a proper integrand; the contour should be clear. The pole(s) result from the integrand; you should prove that the pole(s) you found are the only ones within the contour. This last task may be accomplished via Rouche’s Theorem. (b) Evaluate ∞ cosh π x cos π x2 dx sinh2 π x 0

3.10 Evaluate (a)

1

dx x−2/3 (1 − x)−1/3 log x

0

(b)

1 0

dx x−1/3 (1 − x)−2/3 log x

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3 Evaluation of Definite Integrals II: Applications to Various Types of Integrals

(c)

1

dx x−3/4 (1 − x)−1/4 log x

0

(d)

1

dx x−1/4 (1 − x)−3/4 log x

0

3.11 Evaluate (a)

∞  coth π n

n3

n=1

(b)

∞  n=1

n5

sin

(c) ∞ cot  n=1

1 √

2π n

√

2π n n3

3.12 Set up a contour integral for a sum in which the summand has a branch point singularity. For example, describe in as much detail as possible how one would go about evaluating ∞  f (n) √ n2 + a2 n=1

Chapter 4

Cauchy Principal Value

Abstract In the previous section, we defined the Cauchy Principal Value of an integral; these usually result when there is a small detour in a contour to avoid a singularity otherwise enclosed by the contour. In this chapter, we will explore the concept in cases where there are singularities on a contour. First, we will explore how to deal with removable singularities; we will see that, despite an integrand being finite at a removable singularity, we must treat these with caution and still detour around them. We will extend the definition of a Cauchy Principal value to contour integrals in general and define residues for poles on contours. After going through some examples for residue reduction contours discussed in the previous chapter, we will apply this concept to Hilbert transforms, in which we find the imaginary part of a function analytic in the upper half-plane, over the real line in terms of the real part, and viceversa; this involves evaluating Cauchy Principal Values of an integrand with a pole on the real line. We will apply the concept of Hilbert transforms to analytic signals in physics, which are analytic functions in the upper-half plane used to represent real and imaginary parts of complex physical quantities, e.g., index of refraction. We will then briefly discuss the Hilbert transform in the context of transforms generally and introduce the Dirac delta function as the link between transforms and their inverses.

4.1 Removable Singularities Removable singularities occur in integrands in which a zero in a denominator that would normally produce a pole (or in some cases, a branch point) is “canceled” out by the zero in the numerator of equal or higher order. That is, consider an integrand that is the ratio of two functions p and q, each of which have the same zero z 0 . If, in the neighborhood of z 0 , p(z) = p0 (z − z 0 )m +  p and p(z) = q0 (z − z 0 )n + q , where  p,q are smaller than their respective leading terms in the limit as z → z 0 , then the singularity at z 0 is removable is m ≥ n. Note that neither m nor n is required to be an integer.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 R. Gordon, Complex Integration, Undergraduate Lecture Notes in Physics, https://doi.org/10.1007/978-3-031-24228-1_4

103

104

4 Cauchy Principal Value

An easy example blown sky-high to illustrate what we’re doing here To illustrate the shpiel above, let’s begin by evaluating the following integral: ∞

log x x2 − 1

dx 0

(4.1)

It’s fairly simple to evaluate the integral using a mapping from x → 1/u and Taylor 2 expanding the denominator to see that the integral is equal to π4 . But that’s not the point; rather, we want to explore the possibilities of using complex integration to evaluate more difficult problems. The more salient thing to observe is that the singularity at x = 1 is canceled by the x log there. Accordingly, the singularity at x = 1 is removable, in that lim x→1 xlog 2 −1 = 1 . Because the singularity is removable, it would seem we can use a simple keyhole 2 contour as in Fig. 3.1. In that regard, we consider the following contour integral: 

log2 z z2 − 1

dz C

(4.2)

which according to Fig. 3.1 is split into the following portions:  γ1

 γ2

 γ3

 γ4

log2 z = dz 2 z −1

R

log2 x x2 − 1

dx R

log2 z = iR dz 2 z −1 log2 z =− dz 2 z −1

2π dθ eiθ 0

R dx R

log2 z = i R dz 2 z −1

(log R + iθ )2 R 2 ei2θ − 1

(log x + i2π )2 x2 − 1

0 dφ eiφ 2π

(4.3)

(4.4)

(4.5)

(log  R + iφ)2  2R ei2φ − 1

(4.6)

...and before we even try to estimate the integrals over γ1 and γ3 , we find trouble upon combining the integrals over γ1 and γ3 :  γ1 +γ3

log2 z = −i4π z2 − 1

R R

log x + 4π 2 dx 2 x −1

R R

dx x2 − 1

(4.7)

4.1 Removable Singularities

105

The second integral on the RHS blows up when R ≥ 1 and 0 <  R < 1. But...wait! Can’t we just call the integral a Cauchy principal value and make the integral finite? Well, yes and no. No, because the Cauchy principal value has a specific meaning: it is the result of a detour around a pole. Here, we have not presented any detour because there isn’t really a pole; rather the “pole” is a removable singularity. But as we see from Eq. (4.7), even if the singularity is removable, we still have to treat that singularity as we would any pole and provide a detour around it. So it looks like we will need to use the contour in Fig. 3.2 to get the Cauchy principal value integral we need, even though the singularity is removable. So let’s go ahead and list the eight contour segments.  γ1

 γ2

 γ3

 γ4

 γ5

log2 z =− dz 2 z −1

log2 z =− dz 2 z −1

R dx

log2 z = z2 − 1

 dz

log2 z = z2 − 1

2π

R

(log x + i2π )2 x2 − 1

0 dφ eiφ 2π

1−  R

dx R

0

(log  R + iφ)2  2R ei2φ − 1

log2 x x2 − 1

  log2 1 +  R eiφ dφ e  2 1 +  R eiφ − 1 iφ

π

R dx 1+ R

(log x + i2π )2 x2 − 1

   2 log 1 +  R eiφ + i2π dφ e  2 1 +  R eiφ − 1 dx

log2 z = i R dz 2 z −1

(log R + iθ )2 R 2 ei2θ − 1

iφ

1−  R

log2 z = i R dz 2 z −1

γ6

γ8

0

log2 z dz 2 = i R z −1

dz

γ7

dθ eiθ

1+ R π





2π

log2 z = iR dz 2 z −1

log2 x x2 − 1

(4.8)

(4.9)

(4.10)

(4.11)

(4.12)

(4.13)

(4.14)

(4.15)

Before we continue, a brief word. Yes, it sure seems like a lot of work to verify that a definite integral is equal to zero when real techniques provide that result in a fraction of a second in one’s head. But we are not verifying the value of that integral. Rather, we are using this very simple case to demonstrate how we can deal with

106

4 Cauchy Principal Value

tricky situations that include removable singularities. This will be to our benefit in more complex cases. So let’s begin combining some of these integrals. Remember, we want to recover some Cauchy principal value integrals to prevent things from blowing up on us. For example, if we combine γ6 and γ8 and take the limit as  R → 0 we get:  γ6 +γ8

⎛ 1− ⎞  R R 2 2 x x log2 z log log ⎠ dz 2 dx 2 dx 2 = lim ⎝ +  R →0 z −1 x −1 x −1 R

R = PV

dx 0

1+ R

log2 x x2 − 1

(4.16)

Similarly, combining the integrals over γ2 and γ4 results in another Cauchy principal value in the limit as  R → 0:  γ2 +γ4

log2 z = −P V dz 2 z −1

R dx 0

(log x + i2π )2 x2 − 1

(4.17)

And accordingly the sum of the Cauchy principal value integrals puts us where we wanted to be in the original (and incorrect) approach:  γ2 +γ4 +γ6 +γ8

log2 z = PV dz 2 z −1

R dx 0

log2 x − (log x + i2π )2 x2 − 1 R

= −i4π P V 0

log x + 4π 2 P V dx 2 x −1

R 0

dx (4.18) x2 − 1

Note that, because of the removable singularity, we can remove the P V from the first integral on the RHS but not the second. This is what we wanted:  γ2 +γ4 +γ6 +γ8

log2 z = −i4π dz 2 z −1

R 0

log x + 4π 2 P V dx 2 x −1

R

dx −1

x2 0

(4.19)

And because we generated Cauchy principal value integrals, we will need to deal with the detours around the poles. We add the contributions of integrating over γ3 and γ5 , keeping in mind that the integral over γ3 has an extra i2π term resulting from being on the side of the branch cut where 1 = ei2π . Accordingly, as  R → 0:

4.1 Removable Singularities

 γ3

107

log2 z dz 2 = i R z −1 =i

1 2

π

   2 log 1 +  R eiφ + i2π dφ e  2 1 +  R eiφ − 1 iφ

2π π

dφ (−4π 2 ) 2π

= i2π 3

(4.20)

From the calculation above, should be clear that, as  R → 0:  dz γ7

log2 z =0 z2 − 1

(4.21)

Finally, we take R → ∞ and it should also be clear that, in this limit,  dz

log2 z =0 z2 − 1

(4.22)

dz

log2 z =0 z2 − 1

(4.23)

γ1



γ5

So the contour integral may be written in the limits  R → 0 and R → ∞ as  C

log2 z = i2π 3 − i4π dz 2 z −1

∞ 0

log x + 4π 2 P V dx 2 x −1

∞ 0

dx x2 − 1

(4.24)

OK, so we’re done, right? Apply Cauchy’s Theorem by setting the contour integral 2 to zero and abracadabra, we have verified the contour integral to be equal to... π2 ...? 2 Wait, wasn’t this supposed to be π4 Did we get things wrong again? No, we forgot the additional pole at z = −1, which is within the contour C. So instead, the contour integral is equal to i2π times the residue at z = −1. This of course, should be easy for us as the pole at z = −1 is simple. The residue there is then (iπ )2 /(2(−1)) and accordingly  dz C

π2 log2 z = i2π = iπ 3 z2 − 1 2

(4.25)

108

4 Cauchy Principal Value

And combining Eqs. (4.24) and (4.25), and equating real and imaginary parts, we get ∞ dx 0

∞ PV

π2 log x = x2 − 1 4

(4.26)

dx =0 −1

(4.27)

x2 0

So, yes, we will need to treat removable singularities as we do any singularities and provide detours around a pole. Of course, here we really only needed to detour around one side of the branch cut but the general point is clear. Note also that we frequently get a Cauchy principal value integral for free when working out the original integral. No complaints there! Squares Are Not Necessarily Dull Let’s now move on to something a little more interesting; we will evaluate ∞ dx  0

log2 x

2 x2 − 1

(4.28)

The integrand is the square of the previous integrand. But the evaluation of this integral will lead us into new areas that might have seemed as signs that we should quit. Spoiler: we won’t quit. Another point is that the real evaluation of this integral is far from trivial so we are on more solid ground to use complex integration in the evaluation. We begin by writing down the contour integral we will need to evaluate the integral we want:  log3 z dz  (4.29) 2 z2 − 1 C

Armed with the knowledge we learned about removable singularities in the previous example, we begin by writing down the integrals over the eight contour segments from Fig. 3.2.  γ1

log3 z dz  2 = i R z2 − 1

 dz  γ2

log3 z 2 = − z2 − 1

2π 0

(log R + iθ )3 dθ eiθ  2 R 2 ei2θ − 1

R

1+ R

dx

(log x + i2π )3  2 x2 − 1

(4.30)

(4.31)

4.1 Removable Singularities

 γ3

 γ4

 γ5

log3 z dz  2 = i R z2 − 1 log3 z dz  2 = − z2 − 1

dz  γ6

γ7

 γ8

dφ e

2π

(log  R + iφ)3 dφ eiφ  2  2R ei2φ − 1

(4.34)

dx  R

0

log3 x

2 x2 − 1

(4.35)

  log3 1 +  R eiφ dφ e  2 2 1 +  R eiφ − 1 iφ

π

R dx  1+ R

(4.32)

(4.33)

R

0

  3 log 1 +  R eiφ + i2π

 2 2 1 +  R eiφ − 1

(log x + i2π )3  2 x2 − 1

dx

log3 z dz  2 = i R z2 − 1

iφ

2π

1−  R

log3 z 2 = z2 − 1

log3 z dz  2 = z2 − 1



π

1−  R

log3 z dz  2 = i R z2 − 1





109

(4.36)

log3 x

2 x2 − 1

(4.37)

So more for reasons of having a really cool narrative, let’s first observe that the integral over γ7 vanishes as  R → 0. The integral over γ3 , however, is interesting. Expanding the numerator and denominator of the integrand, we see that two of the terms vanish while the other two do not. What remains is  dz  γ3

log3 z 2 = i R z2 − 1

π dφ eiφ 2π

  −12π 2 log 1 +  R eiφ − i8π 3  2 2 R eiφ +  2R ei2φ

and as  R → 0 ⎡



3

dz  γ3

2

log z 2π ⎣ 2 2 = −i3π (−π ) + 2 R z −1

π

2π

= i3π 3 + 2π 4 − i

dφ e−iφ +

π

⎤ dφ (− R )⎦

2π

3

4π R

(4.38)

where we have ignored terms that vanish as  R → 0. Obviously this diverges as  R → 0. We will need to find an equal and opposite divergence somewhere. This will involve an adventure. As we did before, let’s combine all of the integrals over the real line and get a Cauchy principal value integral out of it. We consider the limit as  R → 0

110

4 Cauchy Principal Value

 γ2 +γ4 +γ6 +γ8

⎡ 1− ⎤  R R 2 2 3 log3 z ⎦ d x −i6π log x+ 12π  log x + i8π dz  + 2 = ⎣ 2 z2 − 1 x2 − 1 0

1+ R

R = −i6π

log2 x

R

2 + 12π P V −1 0 ⎡ 1− ⎤  R R ⎦  dx  + i8π 3 ⎣ + 2 x2 − 1 dx 

0

x2

2

dx  0

log x

2 x2 − 1 (4.39)

1+ R

Let’s pause here and take stock. We now have three terms: the original integral that is finite because of the removable singularity, a Cauchy principal value integral, and...is that third term a Cauchy principal value too? Well, the idea is that a c.p.v. is finite, the answer is...probably not. Generally speaking, the pole should be simple in a c.p.v., and the third term has a double pole. A bit of thought, though, gets us to realize that this third term should include the divergent term we need to cancel the divergent term from the detour. That’s OK, but we are going to see that we will need to maintain a fine level of precision to make everything work. So let’s consider the following integral. 1−  R

 0

arcsin (1− R )

dx x2

2 = −1

dθ sec3 θ 0

arcsin (1− R ) 1 sec θ tan θ + log (sec θ + tan θ ) 0 = 2   1 2 − R 1 − R + log = 2 R (2 −  R ) 4 R

(4.40)

And the other one. R  1+ R

dx

arcsin  (1/(1+ R )

2 = x2 − 1

  dθ sec3 θ − sec θ

arcsin (1/R)

arcsin (1/(1+ ) 1 sec θ tan θ − log (sec θ + tan θ ) arcsin (1/R) R = 2     1 1 2 + R 1 + R +h − log (4.41) = 2 R (2 +  R ) 4 R R where h(x) → 0 as x → 0. That is, the antiderivative vanishes as R → ∞ and it is not productive to write out its exact value. It is important that the above steps be understood. It should be noted that we have not yet applied any expansions of, e.g., the arcsine, during the computation; this allows a more precise accounting when we evaluate the integrals in terms of trig functions. This level of precision—

4.1 Removable Singularities

111

only expanding for small  R —is typical when we are canceling divergences in the presence of Cauchy principal value integrals. We may combine the expressions in Eqs. (4.40) and (4.41) to produce ⎡ 1− ⎤    R ∞ 1 − R 1 2 − R ⎣ ⎦  dx  = + + log 2 2 R (2 −  R ) 4 R x2 − 1 0

1+ R

1 + R 2 R (2 +  R )   2 + R 1 − log 4 R   2 − R 1 2 −  2R 1 = + log  R 4 −  2R 4 2 + R +

(4.42)

What’s interesting about the expression in Eq. (4.42) is that there is no constant term; rather there is just the divergent term and then terms that vanish as  R → 0. Accordingly, as R → ∞:  γ2 +γ4 +γ6 +γ8

log3 z dz  2 = −i6π z2 − 1

R dx  0

log2 x

2 x2 − 1

R + 12π P V 2

dx  0

+i

log x

2 x2 − 1

4π 3 R

(4.43)

And there it is! The divergent term in Eq. (4.43) cancels out the divergent term in Eq. (4.38), as we predicted. It should be stressed again that an advantage of working problems out that require canceling divergent terms is the built-in check on the answer: if the divergent terms do not cancel then there is a mistake somewhere. With the integrals over γ1 , γ5 , and γ7 vanishing as  R → ∞ and R → ∞, we may write the contour integral as  C

log3 z dz  2 = −i6π z2 − 1

∞ dx  0

log2 x

2 x2 − 1

∞ + 12π P V 2

dx  0

log x

4 3 2 + 2π + i3π x2 − 1

(4.44)

112

4 Cauchy Principal Value

We’re not done because, like the previous example, we need to consider the pole at z = eiπ . In this case, however, that is a double pole so we are probably better off expanding the integrand into a Laurent series about z = eiπ . Let z = eiπ + ζ ; then   log3 eiπ + ζ log3 z Res    = Res  ζ =0 (−1 + ζ )2 − 1 2 z=eiπ z 2 − 1 2 (iπ + log (1 − ζ ))3  2 ζ =0 4ζ 2 1 − ζ2   −iπ 3 + 3π 2 ζ + · · · (1 + ζ + · · · ) = Res ζ =0 4ζ 2 2 3 π 3π −i = 4 4 = Res

(4.45)

Now we can apply the Residue Theorem: ∞ − i6π

dx  0

log2 x

2 x2 − 1

∞ + 12π P V 2

dx  0

log x

4 3 2 + 2π + i3π x2 − 1

3π 3 π4 +i = 2 2

(4.46)

And, finally, all of this analysis gets us to our destination: ∞ dx 

π2 2 = 4 x2 − 1

(4.47)

dx 

π2 2 = − 8 x2 − 1

(4.48)

0

∞ PV 0

log2 x

log x

That was lovely! That said, we would not be completely honest with ourselves without acknowledging that there may be an easier way of attacking this problem with contour integration.1 That easier way is addressed in the problems. So why dd we do it the hard way? Three reasons: (1) to illustrate the canceling of divergences that is so useful and (2) to allow us to generalize this problem in cases where the reduction contour won’t work so well.

1

Thanks to @RandomVariable on math.stackexchange for pointing this out.

4.1 Removable Singularities

113

Fig. 4.1 Contour with branch points at z = 0 and z = −1 and removable singularity at z = 1

An example that combines removable singularities with a co-located pole and branch point Now let’s combine the quirkiness of removable singularities with the fun of having two branch cuts and a co-located pole and branch point. This example may be worth the price of this book alone! The integral we will evaluate is ∞ dx 0

log x log (1 + x) x2 − 1

(4.49)

So as promised, a removable singularity at x = 1, branch points at x = 0 and x = eiπ , as well as a co-located pole at x = eiπ . This will be an experience for sure, and along the way we will demonstrate our facility we built over the past examples. Let’s consider the following contour integral:  dz C

log2 z log (1 + z) z2 − 1

(4.50)

wherein C is shown in Fig. 4.1. So there are 12 contour segments. The integrals along four of the segments vanish as  R → 0 and R → ∞, but we will write them initially so to not skip too many steps. That said, once enough experience has been gained it is not necessary to list those vanishing integrals. Without further ado...  γ1

log2 z log (1 + z) = dz z2 − 1

 dz γ2

1−  R

dx R

log2 z log (1 + z) = i R z2 − 1

log2 x log (1 + x) x2 − 1

0 dφ eiφ π

    log2 1 +  R eiφ log 2 +  R eiφ  2 1 +  R eiφ − 1

(4.51)

(4.52)

114

 γ3

 γ4

 γ5

4 Cauchy Principal Value

log2 z log (1 + z) = dz z2 − 1

R dx 1+ R

log2 z log (1 + z) = iR dz z2 − 1

dz γ6

  (log R + iθ )2 log 1 + Reiθ dθ e R 2 ei2θ − 1

π

iθ

0

1+  R

log2 z log (1 + z) = eiπ dz z2 − 1



log2 x log (1 + x) x2 − 1

dx R

(log x + iπ )2 (log (x − 1) + iπ ) x2 − 1

−π

log2 z log (1 + z) = i R z2 − 1

dφ eiφ π

(4.53)

(4.54)

(4.55)

   2 log 1 −  R eiφ + iπ (log  R + iφ)  2 −1 +  R eiφ − 1 (4.56)

 dz γ7

 γ8

log2 z log (1 + z) = iR dz z2 − 1

 dz γ9

 γ10

log2 z log (1 + z) = e−iπ z2 − 1

log2 z log (1 + z) = z2 − 1

R dx 1+ R

2π

(log x + iπ )2 (log (x − 1) − iπ ) x2 − 1

  (log R + iθ )2 log 1 + Reiθ dθ e R 2 ei2θ − 1 iθ

π

1+  R

dx R

log2 z log (1 + z) = i R dz z2 − 1

π

(log x + i2π )2 log (1 + x) x2 − 1

(4.57)

(4.58)

(4.59)

   2   log 1 +  R eiφ + i2π log 2 +  R eiφ dφ e  2 1 +  R eiφ − 1 iφ

2π

(4.60)  dz γ11

 γ12

log z log (1 + z) = z2 − 1 2

 R dx 1− R

log2 z log (1 + z) = i R dz z2 − 1

(log x + i2π ) log (1 + x) x2 − 1 2

0 dφ e 2π

  + iφ)2 log 1 +  R eiφ  2 1 +  R eiφ − 1

iφ (log  R

(4.61)

(4.62)

Well! It seems we have a lot to do here...but again, we recognize groups of integrals and can treat each group together. For example, we should be able to see by now that the integrals over γ2 , γ4 , γ8 , and γ12 vanish as  R → 0 and R → ∞. That leaves eight integrals. Then we should see that we can combine γ1 , γ3 , γ9 , and γ11 into a single Cauchy principal value integral:

4.1 Removable Singularities

 γ1 +γ3 +γ9 +γ11

115

log2 z log (1 + z) = PV dz z2 − 1

R R

 2  log x − (log x + i2π )2 log (1 + x) dx x2 − 1

R = −i4π

log x log (1 + x) x2 − 1

dx R

R + 4π P V 2

dx R

log (1 + x) x2 − 1

(4.63)

Four down, four to go. We may combine γ5 and γ7 and note that the log (x − 1) terms cancel:  γ5 +γ7

log2 z log (1 + z) dz = i2π z2 − 1

R dx

(log x + iπ )2 x2 − 1

dx

log2 x x2 − 1

dx

log x x2 − 1

1+ R

R = i2π 1+ R

R − 4π

2 1+ R

R − i2π

3

dx −1

x2 1+ R

(4.64)

Let’s evaluate from most to least simple. The third integral on the right is easy; we consider the limit as R → ∞: ∞ 1+ R

  1 2 + R dx = log x2 − 1 2 R

(4.65)

For the second integral from the right we may also take  R → 0 and see that the integral is one-half of the value of the integral we considered at the beginning of this section: ∞ ∞ 1 π2 log x log x = = (4.66) dx 2 dx 2 x −1 2 x −1 8 1

0

116

4 Cauchy Principal Value

And the for the first integral, we will find that contour integration methods do not help much as ∞ log2 x dx 2 =0 (4.67) x −1 0

Rather, we map from [1, ∞) → [0, 1] and Taylor expand the denominator: ∞ 1

log2 x dx 2 =− x −1 =− =2

1 dx 0 ∞  

log2 x 1 − x2

1

d x x 2k log2 x

k=0 0 ∞  k=0

1 (2k + 1)3

7 = ζ (3) 4

(4.68)

where ζ (3) is the Riemann zeta function ζ (s) evaluated at s = 3. Putting these all together as  R → 0:  dz γ5 +γ7

7π log2 z log (1 + z) π4 3 = i ζ (3) − iπ + iπ 3 log  R log 2 − z2 − 1 2 2

(4.69)

So we have a divergence as  R → 0, and again, we will need to find a canceling divergence. Ten down, two to go.   Let’s evaluate the integral over γ6 . Note that when we expand the log 1 −  R eiφ term, the contributions to the integral for all terms of order  R and higher vanish. Accordingly, as  R → 0:  dz γ6

log2 z log (1 + z) = −iπ 3 log  R z2 − 1

(4.70)

A-ha! We have canceled out the divergent term in Eq. (4.69). 11 down, 1 to go. Finally, let’s evaluate the integral over γ10 . This one is structurally the same as the integral over γ6 , in which the first log term will only provide vanishing contributions as  R → 0. Accordingly, in this limit:  dz γ10

log2 z log (1 + z) = i2π 3 log 2 z2 − 1

(4.71)

4.2 Poles on Contours

117

Whew. Now we can write an expression for the contour integral as  R → 0 and R → ∞: 

log2 z log (1 + z) = −i4π z2 − 1

dz C

∞ dx 0

log x log (1 + x) x2 − 1

∞ + 4π 2 P V

dx 0

+i

log (1 + x) x2 − 1

π4 7π ζ (3) + iπ 3 log 2 − 2 2

(4.72)

The reward for dealing with such a complicated contour is that the expression for the contour integral excludes any pole and accordingly, the contour integral is zero. Accordingly, we can write down the result, and as usual, get a Cauchy principal value integral for free. ∞ dx 0

7 π2 log x log (1 + x) = ζ (3) + log 2 x2 − 1 8 4

∞ PV

dx 0

π2 log (1 + x) = x2 − 1 8

(4.73)

(4.74)

Making it through these three examples should be enough to understand the caution needed to deal properly with removable singularities. Now let’s move onto singularities that are not so removable.

4.2 Poles on Contours Sometimes we have to deal with poles on contours in places other than the real line. These are more insidious because they may arise in complex-integration evaluations of real integrals when we may not expect it. The point of this section is to learn when to expect them through a few examples. We will then realize what we have been seeing all along: when we take a detour around a pole on a contour and integrate around the pole over a fraction of a full circle, the residue theorem scales by that fraction. Let’s begin with a relatively simple example. A Dimple in a Box Let’s evaluate the following integral: ∞ dx 0

x 1 + ex

(4.75)

118

4 Cauchy Principal Value

The knee-jerk reaction is to consider the following contour integral:  dz C

z log z 1 + ez

(4.76)

where C is a keyhole contour. Of course, we have seen similar things when we developed residue reduction contours and we did so with good reason: the contour integral in Eq. (4.76) only brings us misery in the form of a divergent series that is only tempered by a divergent integral along the large circular arc. Clearly we want something like a rectangle as we used previously. The difference here, however, is that there is no symmetry about the imaginary axis and the integral is only over [0, ∞). So we may define a box with vertices at z ∈ {0, R, R + i2π, i2π }. No problem...except it is clear that we have a pole at z = iπ , which lies on the box! Accordingly, we will have to detour around the pole, but in this case the detour is a semicircular dimple. This is illustrated in Fig. 4.2, where the radius of the dimple is  R as usual. It is tempting to write down the contour integral as  dz C

z 1 + ez

But what will happen is, the factor of z in the numerator of the integrand will cancel when the integrals along γ1 and γ3 (Fig. 4.2) are combined. Of course, we know what to do in such scenarios by now: increase the exponent of z in the denominator by one, so we work with the contour integral  dz C

Fig. 4.2 Rectangular contour with a pole on edge at z = iπ

z2 1 + ez

(4.77)

4.2 Poles on Contours

119

To this end, let’s write out the integrals over the various contour segments:  γ1

 γ2

z2 dz = 1 + ez

dz γ3

γ4

 γ5

0

z2 = 1 + ez

dx R

z2 =i 1 + ez

(4.79)

(x + i2π )2 1 + ex

(4.80)

0

0

(4.78)

(R + i y)2 1 + e R ei y

dy

π+  R

dy 2π

z2 dz = i R 1 + ez

dz

x2 1 + ex

2π

z2 dz =i 1 + ez

 γ6

dx

z2 dz =i 1 + ez





R

(i y)2 1 + ei y

(4.81)

2  iπ +  R eiφ dφ 1 − e R eiφ

−π/2 

π/2

0 dy π− R

(i y)2 1 + ei y

(4.82)

(4.83)

It should be clear how this will proceed, but before we do, a comment. In the previous section in which we dealt with removable singularities on the real axis, we had contributions to the integral from a semicircle about the removable singularity; that, however, was a result of the integral along that segment coinciding with a side of a branch cut. But here, there is no branch cut; rather, the semicircle appears from the asymmetry of the integrand. Note that in the section in which we introduced residue reduction contours, we dealt with symmetric integrands and had no need to close the rectangle at Im z = 0. With that said, we can combine the integrals along γ1 and γ3 , and as R → ∞:  γ1 +γ3

z2 dz = 1 + ez

∞ dx 0

x 2 − (x + i2π )2 1 + ex ∞

= −i4π 0

x dx + 4π 2 1 + ex

∞ 0

dx 1 + ex

(4.84)

In the limit as R → ∞, the integral along γ2 rapidly approaches zero. We can combine the integrals along γ4 and γ6 and get a Cauchy principal value integral:

120

4 Cauchy Principal Value

 γ4 +γ6

z2 dz = i PV 1 + ez

2π dy

y2 1 + ei y

dy

y2 (1 + cos y − i sin y) (1 + cos y)2 + sin2 y

0

2π = i PV 0

 = i PV

2π

0

=i

1 2

2π

y 2 (1 + cos y) + PV dy 2 + 2 cos y

2π dy y 2 +

1 PV 2

0

2π dy y 2 tan

dy 0

y 2 sin y 2 + 2 cos y

y 2

0

4π 3 1 =i + PV 3 2

2π dy y 2 tan

y

(4.85)

2

0

In the limit as  R → 0, the integral over the semicircle γ5 becomes  γ5

z2 dz = i R 1 + ez

−π/2 

dφ π/2

(iπ )2 − R eiφ

= −i(−π 2 )(−π ) = −iπ 3

(4.86)

We now have the expression for the contour integral:  dz C

z2 = −i4π 1 + ez

∞ dx 0

x π3 +i + 4π 2 x 1+e 3

∞ 0

dx 1 + PV 1 + ex 2

2π dy y 2 tan

y 2

0

(4.87) Equating imaginary parts and by Cauchy’s theorem, we have ∞ dx 0

x π2 = x 1+e 12

(4.88)

There is another way to look at the derivation of this result in Eq. (4.88). The integral over γ5 would appear to be just another residue calculation, except for the fact that the integral is over a semicircle rather than a full circle. Just for laughs, let’s compute the residue of the integrand at the pole z = iπ :

4.2 Poles on Contours

121

Res

z=iπ

z2 = π2 1 + ez

(4.89)

This is interesting, because if set the contour integral sans the integral over γ5 , equal iπ (rather than i2π ) times this residue, we reproduce the result in Eq. (4.88). Sort of like a half-residue theorem! We will see such half-residues appear in several places. As usual, let’s look at the other (real) part and we get ∞ 0

dx 1 = − 2 PV 1 + ex 8π

2π dy y 2 tan

y 2

(4.90)

0

In the exercises, we will relate the integral on the RHS to a sort-of well-known integral and show that the RHS is equal to log 2. Poles on vertices Because of the Cauchy principal value integral, we can extend the shortcut that is the Residue Theorem to semicircles. In the previous section, we had a pole on an edge of a contour and that resulted in a semicircular detour. What happens when a pole lies on a contour vertex? Well, we’re about to find out. Let’s consider the following integral: ∞ dx

sin x −αx e sinh x

(4.91)

0

Nominally, we require that Re α > 0, but for reasons that will be apparent soon, we will restrict α to positive integer values, i.e., α = n ∈ N. We consider the following contour integral:  dz

sin z −nz e sinh z

(4.92)

C

where C is the contour illustrated in Fig. 4.3. Wait! How on earth did we come to this contour, with a quarter-dimple on the upper-left vertex? Nominally, the function sinh z has zeroes on the imaginary axis at integers times π . Nominally we might draw a rectangle with vertices at −R, R, R + iπ, −R + iπ , which would leave us with a half-dimple. (Note, drawing the rectangle between the poles will result in unnecessary complication.) But the exponential destroys that symmetry, so we are stuck with the rectangle at 0, R, R + iπ, iπ . The pole at z = 0 is mitigated by the zero of sin z there, i.e., the singularity at z = 0 is removable and, as we learned in the previous section, we need not deal with it because the removable singularity is not along a branch cut. So we will have a rectangle with just a quarter-dimple at z = iπ .

122

4 Cauchy Principal Value

Fig. 4.3 Rectangular contour with a pole on vertex at z = iπ

But let’s take this a step further. We learned in the previous example that we can replace a semicircular dimple with a half-residue at the pole. In the same vein, there’s no reason we can’t simply replace the quarter-dimple with a quarter-residue! It will at least simplify things a little. So we ignore γ5 in Fig. 4.3 and write out the integrals over the other contours. 

sin z −nz e = sinh z

γ1

R

sin x −nx e sinh x

dx

(4.93)

0



sin z −nz e =i sinh z

γ2

π

sin (R + i y) −n(R+i y) e sinh (R + i y)

dy 0



sin z −nz e = −P V sinh z

γ3

R dx 0



sin z −nz e = −i P V sinh z

γ4

π dy 0

sin (x + iπ ) −n(x+iπ) e sinh (x + iπ ) sin (i y) −iny e sinh (i y)

(4.94)

(4.95)

(4.96)

It should be noted that the “P V ” in integrals over γ3 and γ4 are to be meant loosely because the pole is at an endpoint of the integration interval. It should be clear that as R → ∞, the integral over γ2 goes to zero. We note that sinh (x + iπ ) = − sinh x

(4.97)

sin (x + iπ ) = cosh π sinh x + i sinh π cos x

(4.98)

so that the integral over γ3 becomes  γ3

sin z −nz e = (−1)n cosh π sinh z

∞ dx

sin x −nx e sinh x

0

∞ + i(−1) sinh π P V n

0

cos x −nx e sinh x

(4.99)

4.2 Poles on Contours

123

We were able to take the “P V ” out of the real part because the singularity became removable. The integral over γ4 is a little tricky because it is easy to lose track of the signs. If we do everything right, we get for that integral  γ4

sin z −nz e =− sinh z

π

sinh y sin ny − i P V dy sin y

0

π dy

sinh y cos ny sin y

(4.100)

0

Again, we were able to to take the “P V ” out of the real part because the singularity became removable. We can write down an expression for the contour integral, but let’s delay that for a second and instead, let’s apply the fractional Residue Theorem for lack of a better name: remember the pole at z = iπ is on a vertex of the contour C and accordingly, as C is a rectangle the arc around it sweeps out an angle π/2. Accordingly, by the fractional Residue Theorem:  dz

sin z −nz sin z −nz π e e = i Res sinh z 2 z=iπ sinh z

C

= (−1)n

π sinh π 2

(4.101)

The contour integral is real. So let’s just pluck out the real parts of the integrals along the contour segments above and we get the following equation:   1 + (−1)n cosh π

∞

sin x −nx e dx − sinh x

0

π dy

π sinh y sin ny = (−1)n sinh π sin y 2

0

(4.102) There are a couple of observations here worth mentioning. First, it’s kind of nice that all of the integrals we plucked out have integrands finite at z = iπ . Second, it seems less than nice that the integral we seek is in terms of another integral that looks just as difficult. On that second point, it will turn out not to be the case and the integral we need to evaluate will be quite easy and will lead us to a result that would seem immensely difficult to obtain without complex analysis. We just need to remember a very cool trig identity that can be proven easily using, e.g., induction: ⎧ m−1  ⎪ ⎪ ⎪ 2 cos (2k + 1)y n = 2 m ⎪ ⎨

sin ny k=0 = m−1  ⎪ sin y ⎪ ⎪ ⎪ cos 2ky ⎩1 + 2 k=0

where m ≥ 1.

(4.103) n = 2m −1

124

4 Cauchy Principal Value

Note that if n were instead a more general positive real number we would not be able to get a nifty expression for the integral and you’d likely be better off using real-type analysis to express the integral in terms of some horror like a complex digamma. But we are all about using complex analysis here and integral values of n make using our methods viable. To simplify the discussion going forward, we will only discuss the cases where n is even, i.e., n = 2m. (Don’t worry, the other case where n is odd will be in the exercises.) For n = 2m, the integral we want is as follows: ∞ 0

m−1  π 2 sin x −2 mx π e dx = tanh + dy sinh y cos (2k + 1)y sinh x 2 2 1 + cosh π k=0 π

0

(4.104) The integrals on the RHS are all quite easy: π

π dy sinh y cos (2k + 1)y = Re

0

dy sinh (1 + i(2k + 1)) y 0

cosh (1 + i(2k + 1)) π − 1 1 + i(2k + 1) 1 + cosh π =− 1 + (2k + 1)2

= Re

(4.105)

And we finally have an expression for the integral: ∞ 0

π  1 sin x −2 mx π e = tanh − 2 + 2k + 1 sinh x 2 2 2k k=0 m−1

dx

(4.106)

This is a lovely expression for which a CAS has yet to do for arbitrary integer values of m. To appreciate this amazing expression, let’s churn out a few values: ∞ dx

π sin x −2x π e = tanh − 1 sinh x 2 2

(4.107)

dx

π 6 sin x −4x π e = tanh − sinh x 2 2 5

(4.108)

dx

π 83 sin x −6x π e = tanh − sinh x 2 2 65

(4.109)

0

∞ 0

∞ 0

and we can just keep on going because

4.3 The Hilbert Transform

∞

125

sin x −2(m+1)x dx = e sinh x

0

∞ dx 0

sin x −2 mx 1 − e sinh x 2 m2 + 2 m + 1

(4.110)

Another interesting aspect of this result is what happens as m → ∞. Obviously, the integral goes to zero, but the result on the RHS implies that ∞  k=0

π π 1 = tanh 2k 2 + 2k + 1 2 2

(4.111)

The proof of this will be an exercise. The hope here is that these methods become less intimidating to use and that there are problems more easily treated in the complex plane, many more than are usually presented in a first course in Complex Analysis. At this point, we will shift gears somewhat. While the next section is centered on Cauchy Principal Value, it will present a gateway to the next chapter.

4.3 The Hilbert Transform We are going to shift gears somewhat and talk a little about Physics, and specifically, electromagnetism (although this discussion applies to many areas of Physics). The reason for this is to motivate our discussion about the next chapter on Transforms, an important topic in its own right but one we will need in the final chapter on Asymptotics which will allow us to extend some of our integration techniques in frankly stunning ways. When we consider Maxwell’s Equations in a linear and isotropic medium, we consider the relationship between the electric field E and the displacement field D as a function of frequency ω, D(ω) = f (ω)E(ω). The quantity f (ω) is known as the permittivity of the medium in which the fields are measured. Generally, f (ω) is complex. One of the main concerns with such relations in frequency space is that the fields obey causality in the time domain. The concept of causality in the time domain is related to analyticity in the frequency domain in that, by Titchmarsh’s Theorem causality is equivalent to the permittivity f being analytic in the upper half plane. Because f is analytic in the upper-half plane, let’s consider the following contour integral:  f (z) (4.112) dz z−ω C

where C is the contour illustrated in Fig. 4.4. If ω were complex, this would look like Cauchy’s Integral Formula. But as ω ∈ R, we use a semicircular detour. By the fractional Residue Theorem, the contour integral

126

4 Cauchy Principal Value

Fig. 4.4 Contour for illustrating Hilbert transforms

is equal to iπ f (ω). Accordingly, R PV −R

f (ω ) dω +iR ω − ω 

π

  f Reiθ = iπ f (ω) dθ e Reiθ − ω iθ

0

(4.113)

We consider what happens as R → ∞; we would like the integral over θ to vanish in this limit. By the ML-Inequality, we conclude that, for the integral to vanish, f must satisfy the following condition:    lim Max  f Reiθ  = 0

R→∞ θ∈[0,π]

(4.114)

Assuming f satisfies Eq. (4.114), we may write down the following equation satisfied by f : ∞ f (ω ) PV dω = iπ f (ω) (4.115) ω − ω −∞

The significance of Eq. (4.115) is seen when we consider the real and imaginary parts of f , i.e., f (ω) = u(ω) + iv(ω), then we may find a relationship between these real and imaginary parts. To make things a little more concrete, the complex permittivity is equal to n 2 (ω) − 1, where n(ω) is the complex index of refraction of the medium, where the imaginary part of the index represents a degree of attenuation of an amplitude of an electric field propagating into the medium. When we take the real and imaginary parts of Eq. (4.115), we have the following equations. 1 u(ω) = P V π

∞

dω

−∞

1 v(ω) = − P V π

∞ −∞

v(ω ) ω − ω

dω

u(ω ) ω − ω

(4.116)

(4.117)

4.3 The Hilbert Transform

127

Equations (4.116) and (4.117) are collectively known as Hilbert Transforms (or, in more commonly in Physics, Kramers-Krönig relations). They allow us to determine the imaginary part of a complex function analytic in the upper half plane on the real line in terms the imaginary part of such a function on the real line, and vice-versa. We refer to such a complex function as an analytic signal of the quantity represented by the real or imaginary part along the real line. Note that we may assume that the permittivity in the time domain is real so that in the frequency domain it satisfies f ∗ (ω) = f (−ω)

(4.118)

i.e., u(−ω) = u(ω) and v(−ω) = −v(ω). We will see why this is in the next chapter. A concrete example of Hilbert transform pairs We can compute an imaginary part of a function analytic in an upper half plane, alone the real line, given the real part along the real line using the Hilbert transform. As we are discussing Physics, it may be a little weird that we have negative values of the frequency. While in the abstract we may provide some interpretation to such negative values, it is better in many cases to just deal with positive frequencies. This is easy to do and involves a trick we have employed before: map the interval from (−∞, ∞) to [0, ∞). So, for example, ∞ u(ω ) 1 dω  v(ω) = − P V π ω −ω −∞ ⎛ ⎞ 0 ∞   1⎝ u(ω ) u(ω ) ⎠ =− + P V dω  dω  PV π ω −ω ω −ω −∞ 0 ⎛ ⎞ ∞ ∞   u(−ω ) u(ω ) ⎠ 1 + P V dω  = − ⎝−P V dω  π ω +ω ω −ω 0 0 ⎛ ⎞ ∞ ∞    1 u(ω ) u(ω ) ⎠ + P V dω  = − ⎝−P V dω  π ω +ω ω −ω 0

=−

2ω PV π

∞ 0

dω

0

u(ω ) ω2 − ω2

(4.119)

So we have an expression for the imaginary part of a permittivity over a positive frequency spectrum in terms of its imaginary part. We can similarly write down the vice-versa: ∞ ω v(ω ) 2 (4.120) u(ω) = P V dω 2 π ω − ω2 0

128

4 Cauchy Principal Value

So with that out of the way, we can move on a provide an example of a real computation. For example, let’s compute the imaginary part of a permitivitty, i.e., the imaginary part of the square of the index of refraction minus 1, given its real part. In this example, we will work with, for a > 0,   2 u(ω) = Re n 2 (ω) − 1 = e−aω

(4.121)

By Eq. (4.119), the imaginary part of the permittivity is 2ω PV v(ω) = − π

∞

e−aω 2 ω − ω2 2

dω

0

(4.122)

We evaluate the integral, how else?, using a contour in the complex plane. In this case, the Cauchy Principal Value integral sort of forces our hand; otherwise, this would be a fairly straightforward integral using real methods. Since we are in slightly new territory, the contour C will simply be presented as follows. We evaluate  C

e−az z 2 − ω2 2

dz

(4.123)

where C is the following contour illustrated in Fig. 4.5 where the wedge angle is π/4. As has become our practice, we will combine the integrals over γ1 + γ3 into a single Cauchy PV integral and use the fractional Residue theorem. Accordingly,  γ1 +γ3

e−az dz 2 = PV z − ω2 2



0

e−az z 2 − ω2

dz

e−az = −eiπ/4 2 z − ω2

2



2

e−aω ω2 − ω2 2

dω

π/4 = iR dθ eiθ

dz γ4

γ5

R

0

R

e−a R e R 2 ei2θ − ω2

(4.125)

e−iat it 2 − ω2

(4.126)

2 i2θ

2

dt 0

(4.124)

As R → ∞, the integral over γ4 will vanish. The integral over γ1 + γ3 is the Cauchy PV integral we want. We will need to evaluate the integral over γ5 , which is best done at this point using Feynman-like tricks. To begin, we express the integral over γ5 in a way that makes our next move 2 obvious. That is, we will being a factor of eaω into the integral as follows:

4.3 The Hilbert Transform

129

Fig. 4.5 Contour for Hilbert transform computation

∞ − eiπ/4 0

e−iat 2 = −eiπ/4 e−aω it 2 − ω2 2

dt

∞

e−a (it −ω ) it 2 − ω2 2

dt 0

2

(4.127)

We define a function F as ∞ F(a) = 0

e−a (it −ω ) dt it 2 − ω2 2

2

(4.128)

We note that since ω > 0, the denominator is never zero. Accordingly, we may differentiate F and get 

F (a) = −e

aω2

∞

dt e−iat

0

1 2 = e−iπ/4 eaω 2



2

π a

(4.129)

The proof that the integral converges and is equal to the value in Eq. (4.129) is left to the exercises. So now we must integrate over a to get back F(a): √ a π −iπ/4 2  F(a) = F(0) + e da  a −1/2 eω a 2 0 √

a √ −iπ/4 2 2 = F(0) + πe du eω u 0

= F(0) + e−iπ/4

√

π ω

√ ω  a

dv ev

2

0

√  π −iπ/4 = F(0) + erfi aω e 2ω

(4.130)

130

4 Cauchy Principal Value

where erfi is analogous to the error function and is defined as x

2 erfi (x) = √ π

dv ev

2

(4.131)

0

We must also compute ∞ F(0) =

1 it 2 − ω2

dt 0

∞ = −i

1 t 2 + iω2

dt 0

1 = 2π 2

Res

z=ei3π/4 ω

= −eiπ/4

π 2ω

z2

1 + iω2 (4.132)

So the integral over γ5 is then  γ5

√  e−az π −aω2 π −aω2 e e dz 2 =i − erfi aω z − ω2 2ω 2ω 2

(4.133)

The contour integral is then, as R → ∞  C

e−az dz 2 = PV z − ω2 2

∞ 0

√  e−aω π −aω2 π −aω2 e e dω 2 +i − erfi aω ω − ω2 2ω 2ω 2



(4.134) By the fractional Residue Theorem, the contour integral is also equal to  C

e−az e−az = iπ Res z=ω z 2 − ω2 z 2 − ω2 2

dz

2

e−aω = iπ 2ω

2

(4.135)

Putting this all together, the imaginary pieces cancel (whew!) and we finally get, for the imaginary part of the permittivity, v(ω) = −e−aω erfi 2

√  aω

(4.136)

which is something called a Dawson’s Function. The analytic signal representing the complex permittivity analytic in the upper half plane, one the real line, is

4.3 The Hilbert Transform

131

√  2  f (ω) = e−aω 1 − ierfi aω

(4.137)

What makes the Hilbert transforms, transforms Let’s return to our Hilbert transform pairs: 1 u(ω) = P V π

∞

dω

−∞

1 v(ω) = − P V π

∞

v(ω ) ω − ω

dω

−∞

u(ω ) ω − ω

We can look at Eq. (4.117) as a forward transform, producing an imaginary part of an analytic signal on the real axis in terms of the real part. We can also look at Eq. (4.116) as an inverse transform, providing the real part back to us. In that spirit, let’s look at substituting Eq. (4.117) into the integral in Eq. (4.116). ⎛ ⎞ ∞ dω ⎝ 1 u(ω") ⎠ dω  − PV ω − ω π ω − ω −∞ −∞ ⎛ ⎞ ∞  ∞  dω 1 ⎠ (4.138) = PV dω u(ω ) ⎝− 2 P V π (ω − ω )(ω − ω )

1 u(ω) = P V π

∞

−∞

−∞

Here we assume that the reversal of integration can be carried out. For these Cauchy principal value integrals it is not trivial to show this, and we will have to resort to a trick to keep this together, a trick with which you may be utterly familiar. For now, let’s just play along. The effect of the Cauchy PV integral in the parentheses in the RHS in Eq. (4.138) is to sift values of ω from all values of ω to produce the LHS. This is a common phenomenon in transforms in general and in this case, we have ∞ u(ω) = P V

dω u(ω )δ(ω − ω)

(4.139)

−∞

where 1 δ(ω − ω) = − 2 P V π 

=

1 PV π2

∞ −∞

(ω

∞ −∞

ω (ω

dω − ω )(ω − ω ) dω − (ω − ω))

(4.140)

132

4 Cauchy Principal Value

In other words, 1 δ(x) = 2 P V π

∞ −∞

dω ω (ω − x)

(4.141)

Before we continue, let’s review a few facts about the delta function we may remember from Physics:  0 x = 0 δ(x) = (4.142) ∞ x =0 ∞ d x δ(x) = 1

(4.143)

−∞

Does the definition in Eq. (4.141) match that in Eqs. (4.142) and (4.143)? Looking directly at Eq. (4.141), it it should be clear that the condition in Eq. (4.142) is satisfied; a quick partial fractions expansion should see to that. Verifying whether Eq. (4.143) is verified is not so trivial and is in fact interesting enough to demonstrate here. ∞

 d x δ(x) =

−∞

d x δ(x) ∀ > 0 − 

= −

=

⎛ 1 dx ⎝ 2 PV π

1 PV π2

1 = 2 PV π 1 = 2 PV π 2 = 2 PV π 2 = 2 PV π

∞ −∞ ∞

−∞ ∞

−∞ ∞

0

1 0

dω ω

∞ −∞ 

−

⎞ dω ⎠ ω (ω − x)

dx ω − x

   ω +   1   dω  log   ω ω − 

  ω + 1 1   dω log  ω ω − 1

  ω + 1 1   dω log  ω ω − 1     ∞ 1+ω 2 ω+1 1 1 + 2 P V dω log dω log ω 1−ω π ω ω−1 1

4.3 The Hilbert Transform

133

4 = 2 PV π

∞ 1

  1 ω+1 dω log ω ω−1    u= ω+1 ω−1

8 = 2 π =

8 π2

∞ du

log u u2 − 1

du

log u 1 − u2

1

1 0

And we know what to do with this! This is an integral with a removable singularity, which for this case it is easiest to recognize as half the result we derived earlier in 2 the chapter. And the result is—surprise surprise!— π8 . And Eq. (4.143) is verified. By verifying that the expression in Eq. (4.141) is indeed a delta function we have established the pair in Eqs. (4.117) and (4.116) is indeed a transform pair, able to produce a real part of an analytic signal along the real axis from an imaginary part and vice versa. This is an important part of our story because much of the work ahead is related to other transforms which will not only be great applications of the complex integration we have developed over the past three chapters but will also be the basis for other techniques we will develop here and will provide many new arrows in our quivers. Before we transition, let’s think about the energy in a signal. Typically, if the real part of an analytic signal u(ω) represents an electric field amplitude, then under some conditions, u(ω)2 is proportional to an energy density. Accordingly, let’s evaluate an integral over the energy density. ∞

∞ dω u(ω) = P V 2

PV −∞

−∞ ∞

= PV −∞

⎛ 1 dω ⎝ P V π

∞

⎞⎛ ⎞ ∞  v(ω ) ⎠ ⎝ 1  v(ω ) ⎠ dω  dω  PV ω −ω π ω −ω

−∞ ∞

dω v(ω )P V

−∞



−∞

dω v(ω )

1 PV π2

∞ −∞

dω (ω − ω )(ω − ω )

This is assuming we can make the change in order of integration, and here we assume we can. An important thing to note is that when we square an integral, we write the product out twice using different dummy variables over which to integrate; in this case we used ω and ω . It is easier to use simple distinctions in situations like this. (Remember, notation is how we understand mathematics and is accordingly of great practical importance to our work.) So, that said we may realize thet the rightmost integral above is, by Eq. (4.141), the beloved delta function, and applying the sifting property we get a really cool result.

134

4 Cauchy Principal Value

∞

∞ dω u(ω) = P V

dω v(ω)2

2

PV −∞

(4.144)

−∞

We will see relations like this in the next chapter, where the integral of a square of an amplitude is equal to the integral of the square of its transform. We will refer to such relations here as Parseval’s equalities.

Problems 4.1 Evaluate the following integrals. ∞ (a)

dx

log x n∈N n>2 xn − 1

dx

log2 x n∈N n>2 xn − 1

0

∞ (b) 0

∞ (c)

dx 0

log2 x n∈N n>2 (x n − 1)2

4.2 (a) For the integral in Eq. (4.28), find a residue reduction contour that will get us to the same answer. (b) For the integrals in Prob. 1.1, for what values of n will a residue reduction contour work? 4.3 Evaluate the following integrals. ∞ (a)

dx 0

log x log (1 + x) n∈N n>2 x 2n − 1 ∞

(b)

dx 0

log x log (a + x) a>0 (x − 1)(x + a) ∞

(c)

dx

log x log2 (1 + x) x2 − 1

dx

log2 x log (1 + x)  2 x2 − 1

0

∞ (d) 0

4.3 The Hilbert Transform

135

4.4 (a) Prove Eq. (4.111) using complex integration. (b) Evaluate the following integral. ∞

sin x −(2 m+1)x e sinh x

dx 0

where m ∈ N. Conjecture a value of a sum in the limit as m → ∞ and show, using complex integration, that the conjecture is true. 4.5 Evaluate the following integral. ∞ dx

  sin π x 2 cosh (π x) sinh2 (π x)

0

by defining a proper integrand and contour for a contour integral. As a hint, keep in mind the contours we have studied in this chapter. 4.6 Evaluate the following integral for k ∈ R and |a| < π . ∞ dx −∞

sinh (ax) ikx e sinh (π x)

by defining a proper integrand and contour for a contour integral. As a hint, keep in mind the contours we have studied in this chapter. 4.7 Show that the integral used in the derivation of Eq. (4.129), ∞

dt e−iat

2

0

is convergent. (Hint: construct a contour integral over a residue reduction contour.) 4.8 (a) Verify that the analytic signal in Eq. (4.137) satisfies the condition in Eq. (4.114). (b) Verify that the real and imaginary parts of the analytic signal in Eq. (4.137) satisfy Parseval’s equality, i.e., Eq. (4.144). 4.9 (a) Determine the analytic signal f (ω) when its real part on the real line is given by 1 u(ω) = 1 + ω2 Also verify that the real and imaginary parts of f (ω) satisfy Parseval’s equality, i.e., Eq. (4.144).

136

4 Cauchy Principal Value

(b) Determine the analytic signal f (ω) when its real part on the real line is given by u(ω) =

sin ω ω

We will have the tools to verify Parseval’s equality in the next chapter. (c) Determine the analytic signal f (ω) when its real part on the real line is given by  u(ω) =

1 |ω| ≤ 1 0 |ω| > 1

Also verify that the real and imaginary parts of f (ω) satisfy Parseval’s equality.

Chapter 5

Integral Transforms

Abstract Our study of Hilbert transforms was part of a natural progression of our deep dive into complex integration and the Cauchy principal value integrals that frequently result from poles on contours. We are now at a stage where we can study other useful integral transforms and use what we learn to attack other integrals and sums. In this chapter, we will explore Fourier, Laplace, and Mellin transforms. We will be taking a unified approach to these transforms and will demonstrate that these three integral transforms are outgrowths of one another, even though they have vastly different properties and accordingly different benefits for us. The Fourier transform has properties much like the Hilbert transforms but has a usefulness to us that is transcendent. The Laplace transform also brings its own set of techniques and we will see that the computation of its inverse is a perfect fit for this book. Finally, the Mellin transform will provide one of the most shocking and satisfying uses for our goal of expanding the class of integrals that may be attacked using complex integration. In this chapter we will be studying the common integral transforms of Mathematical Physics: Fourier, Laplace, and Mellin. Our discussion will be at once unifying but also a deep dive into some aspects of them that illustrate the techniques we have developed in the previous chapters. Remember, the point of the book is to have some fun by going places people generally don’t go. To motivate the discussion, let’s sort-of-kind-of review what we did with Hilbert transforms, but at a higher level. When we spoke of an integral transform pair, we defined relations like the following for functions u(x ), x ∈ I and U (x), x ∈ J :  U (x) = PV dx K(x , x)u(x ) (5.1) I

where K is a transform kernel. For example, in the Hilbert transform, I = J = R and K(x, x ) = π(x1 −x) . The inverse transform is then 



u(x ) = PV

dx L(x, x )U (x)

(5.2)

J

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 R. Gordon, Complex Integration, Undergraduate Lecture Notes in Physics, https://doi.org/10.1007/978-3-031-24228-1_5

137

138

5 Integral Transforms

where L is the inverse transform kernel. For example, in the Hilbert transform, L(x, x ) = − π(x1 −x) . By combining the transform and inverse transform, we get a delta-function relation: ⎡ ⎤     ⎣    ⎦ u(x ) = PV dx L(x, x ) PV dx K(x , x)u(x ) J

 = PV



dx u(x )PV

I

 = PV

I

dx K(x , x)L(x, x )

J 





dx u(x )δ(x − x )

(5.3)

I

and we may associate the delta function with the following integral: 

δ(x − x ) = PV

dx K(x , x)L(x, x )

(5.4)

J

By performing similar operations with the forward transform, we have another expression for the delta function: δ(x − x ) = PV



dx L(x , x )K(x , x)

(5.5)

I

Keep in mind that we are glossing over the fact of whether it makes any sense to reverse the order of integration, etc., but then again delta functions are not exactly rigorous-math-friendly concepts anyway. The important thing is, for our purposes, they work well enough. We can also get some insight into Parseval-type equalities by considering, for example,  PV I

2  dx u(x ) = PV 

 I

dx L(x, x )U (x) PV

J



= PV

I



dx L(x , x )U (x )

J



dx PV

 dx U (x) PV

J

dx U (x )PV

J

dx L(x, x )L(x , x )

(5.6)

5.1 The Fourier Transform

139

Equation (5.6) is pretty telling. We expected a delta function as in Eq. (5.5) so that we could define a Parseval equality between the square of functions and their transforms, but we don;t really have that. Rather, we have a condition that should be satisfied for a Parseval equality to exist for is that, in the language of operators, L is the Hermitian conjugate of K. As we will see, this is not always satisfied and accordingly not all of the transforms we study will have a corresponding Parseval equality.

5.1 The Fourier Transform Most people who are reading this book have likely come across the Fourier transform in some shape or form. A facile way of going about introducing it is to simply use the notation above and let K(x, k) = Aeikx and L(k, x) = Be−ikx , where AB = 1/(2π ). We would take I = J = R and we can just take off from there. The truth is more subtle than this. The Fourier transform is not simply an integral over the real line of a function times eikx . This subtlety will provide the Fourier transform with the ability to help us tackle whole new classes of integrals, whether via a Parseval equality, a convolution relation, or considerations of the analyticity of the Fourier transform in regions of the complex plane. The goal here is two-fold: (1) To be able to attack new classes of integrals and sums, and (2) to segue into other transforms where we will develop whole new arrows for our quivers. Plancherel’s Theorem and Jordan’s Lemma Let’s begin by just writing down The Formula for a Fourier transform of a function f :R→R ∞ F(k) = dx f (x)eikx (5.7) −∞

We define the inverse transform of F : R → R as 1 f (x) = 2π

∞

dk F(k)e−ikx

(5.8)

−∞

There are of course any number of conventions that are used in Mathematics and Electrical Engineering, but we will use this terribly asymmetric and inconvenient Physics convention because, well, it is how we make sense of the world as physicists. Before we plow ahead, we should address the subtlety of Eqs. (5.7) and (5.8). What do we mean by the infinite integration limits? What sort of functions can f and F be? Generally speaking, we can write such an integral as a limit:

140

5 Integral Transforms

∞

M dx f (x)e

=

ikx

−∞

lim

M ,N →∞ −N

dx f (x)eikx

(5.9)

The problem with this definition is that the function f must be absolutely convergent. This unnecessarily restricts the sort of functions we can work with to evaluate cool integrals. Rather, we require the Fourier transform to approach its infinite integration limits at the same rate; this forms the basis of Plancherel’s theorem. To this effect, define M FM (k) =

dx f (x)eikx

(5.10)

−M

If we can establish the existence of the inverse Fourier transform F(k) as follows: 1 f (x) = lim 2π M →∞

M

dk F(k)e−ikx

(5.11)

dk |FM (k) − F(k)|2 = 0

(5.12)

−M

then

M lim

M →∞ −M

That is, F is defined in the limit of a mean-square, rather than pointwise, convergence. Consequently, f belongs to the class L2 (−∞, ∞) of functions squareintegrable over the real line, i.e. ∞ dx |f (x)|2 < ∞

(5.13)

−∞

To illustrate this, let’s work out an example. Consider the contour integral  dz C

eikz z

(5.14)

where for k > 0, C is the contour illustrated in Fig. 5.1. The contour integral in Eq. (5.14) is equal to, by the fractional residue theorem, R PV −R

eikx dx +i x

π 0

d θ eikR cos θ e−kR sin θ = iπ

(5.15)

5.1 The Fourier Transform

141

Fig. 5.1 Contour for illustrating a Fourier transform when k > 0

As always, we consider the limit as R → ∞. Here, we are faced with an integral that may or may not appear familiar; we usually encounter these in first courses in Complex Analysis in the discussion of Jordan’s Lemma. We may estimate the magnitude of the integral as R → ∞ as follows.   π  π     i d θ eikR cos θ e−kR sin θ  ≤ d θ e−kR sin θ     0

0

π/2 = 2 d θ e−kR sin θ 0

π/2 ≤ 2 d θ e−2kRθ/π 0

∞ ≤2

d θ e−2kRθ/π

0

π = kR

(5.16)

We note that in the transition between the second and third lines above, we made use of the nifty inequality sin θ ≥ 2θ ∀θ ∈ [0, π/2]. (This is almost trivially easy to π see in a plot.) We can make a couple of observations about Eq. (5.16). 1. We spoke of the Fourier transform applying to the class of square-integrable rather than absolutely convergent (a much more restricted class) integrals, and here is a prime example. The absolute value of the integrand here is not integrable over the real line, yet its square certainly is. 2. It is a bit weird to see the integral of an integrand exponentially vanishing with R ending up vanishing relatively weakly (i.e., as 1/R) with R. This should instill in us a sense of caution as we make estimate of integrals over large arcs of integrands we will encounter when computing Fourier transforms.

142

5 Integral Transforms

Fig. 5.2 Contour for illustrating the Fourier transform when k < 0

Keeping this in mind, let us write down our finding for k > 0 as R → ∞: ∞ dx

PV −∞

eikx = iπ x

(5.17)

This is a lovely result but by no means complete. Recall that we subjected ourselves to the restriction k > 0. Note that if this were not the case, the integral over the large arc in the contour integral would not converge and the result is accordingly meaningless for k < 0. What do we do in this case? This should not be much of a mystery: we instead close the contour below the real axis, as shown in Fig. 5.2. The difference here is in the orientation of the contour C: if C is to be oriented positive, as is required for the Residue theorem, then we must traverse the real axis from right to left. This will induce a sign change in the result; that is, for k < 0: ∞ dx

PV −∞

eikx = −iπ x

(5.18)

Accordingly, we can combine this into a single, much more useful result for all values of k. ∞ eikx PV = iπ sgn(k) (5.19) dx x −∞

where sgn is the signum function, which is just ±1 according to the sign of its argument. There is some really neat stuff coming ahead that will allow us to evaluate tricky integrals and provide new context to some of the stuff we did in the last chapter back here. But first, let’s go back and discuss Jordan’s lemma but maybe in a different way than is usually taught in a first, more formal treatment. As stated throughout this book (and likely, the back cover, the Preface, any ads Springer Nature in all its wisdom may put out in attempts to sell this book,...), we are far less interested in making sweeping, formal statements that are the domain of professional mathematicians. This book is instead aimed at those who enjoy messing

5.1 The Fourier Transform

143

around with integrals and sums and like to learn things through examples. And in the spirit of this book, let’s explore the practical difference between the integrals we encounter in taking Fourier transforms (aka, Fourier integrals) and the integrals with which we have been monkeying up until now. OK, Jordan’s lemma. Let’s consider the general Fourier integral. Consider a function f meromorphic in the complex plane. As we did in the previous example, we will consider the following contour integral for k > 0  dz f (z)eikz

(5.20)

C

where C is a semicircle in the upper half-plane of radius R. This contour integral is equal to R π

ikx dx f (x)e + iR d θ eiθ f Reiθ eikR cos θ e−kR sin θ (5.21) −R

0

As usual, we consider the limit as R → ∞. The first term obviously becomes the Fourier transform of f when k > 0. As we did in the previous example, we bound the magnitude of second term in this limit. To do this, we assume that f is bounded over the large arc as follows. 

 Max f Reiθ  = M (R)

(5.22)

θ∈[0,π]

Then by the ML-inequality,    π  π   

 iR d θ eiθ f Reiθ eikR cos θ e−kR sin θ  ≤ R d θ f Reiθ  e−kR sin θ     0

0

π ≤ RM (R)

d θ e−kR sin θ

0

π/2 = 2RM (R) d θ e−kR sin θ 0

π/2 ≤ 2RM (R) d θ e−2kRθ/π 0

∞ ≤ 2RM (R)

d θ e−2kRθ/π

0

π M (R) = k

(5.23)

144

5 Integral Transforms

Fig. 5.3 Contour for evaluating integral in Eq. (5.24)

This is the essence of Jordan’s lemma. In a typical integral as we saw in previous chapters, the requirement for convergence of an improper integral of F over the real line is that limR→∞ RM (R) = 0, which implies that M (R) must decrease faster than 1/R. But for a Fourier integral, the condition for convergence is merely limR→∞ M (R) = 0. This allows for the Fourier transformation of a wide range of functions, even beyond those indicated by Plancherel’s Theorem. Note that we can restrict the analysis to k > 0 without loss of generality. An example of the sort of Fourier transform allowed via Jordan’s lemma, let’s take the Fourier transform of f (x) = |x|−1/2 . ∞ dx |x|

−1/2 ikx

e

∞ = 2 Re

−∞

dx eikx

2

(5.24)

0

It isn’t completely clear that we made things better by making the substitution x → x2 . But as anyone who has worked out the solution to Prob. 4.7 knows, to evaluate the integral on the RHS we consider the following contour integral for k > 0.  2 dz eikz (5.25) C

in which C is the wedge contour shown in Fig. 5.3 which wedge has an angle of π/4. The contour integral segments are  dz e

ikz 2

γ1

R =

dx eikx

(5.26)

0

 dz e

ikz 2

π/4 2 i2θ = iR d θ eiθ eikR e

γ2

(5.27)

0

 dz e γ3

2

ikz 2

0 =e

iπ/4 R

dt e−kt

2

(5.28)

5.1 The Fourier Transform

145

We can show that the magnitude of the integral over γ2 vanishes as R → ∞:     π/4  2 ikz  dz e  ≤ R d θ e−kR2 sin 2θ     γ2

0

1 = R 2

π/2 2 d θ e−kR sin θ 0

π/2 1 2 ≤ R d θ e−2kR θ/π 2 π ≤ 4kR

0

(5.29)

Then by Cauchy’s Theorem, ∞ dx e

ikx2

∞ =e

iπ/4

0

0

1 = eiπ/4 2

dt e−kt



2

π k

(5.30)

Accordingly, the Fourier transform of f (x) = |x|−1/2 is by Eq. (5.24), ∞ −∞

√ π dx |x|−1/2 eikx = √ k −1/2 2 2

(5.31)

for k > 0. The case k < 0 has a similar result and its evaluation is left to the exercises. The final result for all values of k = 0 is ∞ dx |x| −∞

−1/2 ikx

e

√ π = √ |k|−1/2 2 2

(5.32)

The result in Eq. (5.32) is really interesting: it says that the Fourier transform of an inverse square root function is a constant times an inverse square root function. If we look at the Fourier transform as an operator in a Hilbert space L2+a (−∞, ∞) (i.e., square-integrable functions) for any a > 0, then the function |x|−1/2 is an eigenfunction of the Fourier transform operator in that Hilbert space.1 1

The notion of Hilbert space is not discussed in this book, but is not as esoteric as it appears. For details about differential and integral operators in Hilbert space and their eigenfunctions, please see S. Holland, “Applied Analysis by the Hilbert Space Method,” Dover (2007).

146

5 Integral Transforms

To review, Plancherel’s Theorem defines the inverse of the Fourier transform operator, while Jordan’s lemma expands the space of functions that may be Fourier transformed. Convolution: It’s All Your Faltung Frequently the concept of convolution invades our dreams and reveals itself to be incredibly useful. For example, consider two functions f and g with respective Maclurin series expansions about z = 0: ∞

f (z) =

fn z n

n=0 ∞

g(z) =

gn z n

n=0

Then we can also express the function h(z) = f (z)g(z) as a Maclurin series as well: h(z) = f (z)g(z)

∞  ∞  = fm z m gn z n m=0

=

n=0

∞ ∞

fm gn z m+n

m=0 n=0

=

∞

hk z k

(5.33)

k=0

where hk =

k

f gk−

(5.34)

=0

That is, a product of a pair of functions, each with a Maclurin series expansion having respective series coefficients, has a Maclurin series expansion which coefficients is a convolution of the respective series coefficients. Similarly, we can define a convolution between two functions f and g, each ∈ L2 (−∞, ∞). Analogous to Eq. (5.34), we may state the Convolution Theorem for Fourier transforms:

5.1 The Fourier Transform

147

Theorem 5.1 If f and g ∈ L2 (−∞, ∞) have Fourier transforms ∞ dx f (x)eikx

F(k) = −∞ ∞

G(k) =

dx g(x)eikx −∞

Then the convolution of f and g, f ∗ g, satisfies the following. ∞ −∞

1 dx f (x )g(x − x ) = 2π 



∞



dk F(k)G(k)e−ikx

(5.35)

−∞

We will leave the proof of this to the exercises. A nice example comes from the concept of line-broadening in lasers. Frequently, line-broadening is the result of a convolution between two beam shapes. For example, in homogeneous broadening, the profile is a Lorentzian, defined as fa (x) =

1 a 2 π a + x2

(5.36)

Its Fourier transform is, for k > 0, Fa (k) =

a π

∞ dx −∞

= i2a Res =e

eikx a2 + x2 eikz + z2

z=ia a 2 −ak

(5.37)

where a > 0. When k < 0, the result is the same except we close below the real line and use the pole at z = −ia. The result is, for all values of k, Fa (k) = e−a|k|

(5.38)

Then if we perform homogeneous broadening for two separate, e.g., collision processes with widths a and b, then the aggregate beam shape is

148

5 Integral Transforms

ab fa ∗ fb = 2 π

∞ −∞

dx



a2 + x2 b2 + (x − x )2

1 2ab 



Res 2 2 π z=ia a + z b2 + (x − z)2  1



+ Res 2 z=x+ib a + z 2 b2 + (x − z)2 

2ab 1 1

+

=i π i2a b2 + (x − ia)2 i2b a2 + (x + ib)2 =i

b 1 1 a



+ 2 2 2 2 2 π x − a − b − i2ax π x + a − b2 + i2bx 1 a+b = π x2 + (a + b)2 =

(5.39)

While the Residue Theorem simplifies the evaluation of the convolution integral, the algebra leading to Eq. (5.39) is far from trivial and it is left as a challenge to work it out by hand. Now compare the mess above leading to Eq. (5.39) to its equivalent in frequency space via the Convolution Theorem: 1 fa ∗ f b = 2π 1 = 2π

∞

dk e−a|k| e−b|k| e−ikx

−∞

0

ak bk −ikx

dk e e e −∞

⎛

∞

1 + 2π ⎞

1 Re ⎝ dk e−(a+b+ix)k ⎠ π 0   1 1 = Re π a + b + ix 1 a+b = 2 π x + (a + b)2

∞

dk e−ak e−bk e−ikx

0

=

(5.40)

That was a lot easier, but it is satisfying to see the Convolution Theorem verified like that. This example also illustrates an amazing bit of mathematics supporting a rather intuitive result in Physics: that the convolution of Lorentzians of a first width and a second width is a Lorentzian of a width equal to the sum of the widths. The Lorentzian is an example of a stable distribution: a Gaussian is another. (Although interestingly, the convolution of a Lorentzian and a Gaussian, known as a Voigt distribution, is not similarly stable.)

5.1 The Fourier Transform

149

Fig. 5.4 Illustrating the computation of the convolution integral in Eq. (5.43) for T = 1

The above example with the Lorentzians illustrated analysis involving functions that stretched out to infinity. The Convolution Theorem is equally valid with functions of finite support, i.e., that are zero outside of an interval. It will turn out that the ease of computing convolution integrals in such cases are more in the eye of the beholder. Another example will illustrate why. Let’s take the convolution of a rectangular impulse f with a triangular impulse g. Note here we will use time/temporal frequency (t, ω).  f (t) =  g(t) =

T t ∈ [0, T ] 0 t ∈ {t t < 0} ∪ {t t > T }

(5.41)

t t ∈ [0, 2 T ] 0 t ∈ {t t < 0} ∪ {t t > 2 T }

(5.42)

We will compute f ∗ g both directly and using the Convolution Theorem. First, directly. ∞







∞

dt f (t )g(t − t ) =

f ∗g= −∞

dt  f (t − t  )g(t  )

(5.43)

−∞

These problems can be tricky and it always helps to draw a picture, as in Fig. 5.4. We can then write down the following.  T t  ∈ [t − T , t]  f (t − t ) = (5.44) 0 t  ∈ {t t  < t − T } ∪ {t  t  > t}

150

5 Integral Transforms

The challenge is to determine where the convolution changes functional form. Note that the convolution f ∗ g must be continuous so this serves as a check on the results. For example, when t < 0, it is clear that f ∗ g(t) = 0. The first transition occurs at t = 0, when the right edge of the rectangle intersects the triangle. (As can be seen, drawing a good picture in these problems is crucial.) Between t = 0 and t = T , the convolution is for t ∈ [0, T ] t f ∗ g(t) =

dt  T · t  =

1 2 Tt 2

(5.45)

0

The next transition occurs at t = T , when the left corner of the rectangle intersects the triangle. For t ∈ [T , 2T ], the convolution is t f ∗ g(t) =

1 dt  T · t  = T 2 t − T 3 2

(5.46)

t−T

The next transition occurs at t = 2T , when the rectangle is sticking out of the triangle to the right. For t ∈ [2T , 3T ], the convolution is 2 T f ∗ g(t) =

dt  T · t  =

3 1 2 1 T 4 T − (t − T )2 = T 3 + T 2 t − Tt 2 2 2 2

(5.47)

t−T

The convolution is clearly zero for t ≥ 3T . Note that the width of the convolution is the sum of the widths of the two pulses convolved. The final result is, for all t ∈ R: ⎧ ⎪ 0 ⎪ ⎪ ⎪ 1 2 ⎪ ⎪ ⎨2t f ∗ g(t) = T 2 t − 21 T 3 ⎪ ⎪ ⎪ 3 T 3 + T 2 t − 1 Tt 2 ⎪ 2 2 ⎪ ⎪ ⎩0

t 3T . It looks almost the same as the case we just did, expect now the contour integral is in the opposite direction. The effect is that the integration limits of the integral we evaluate are different: 

∞ dω

PV −∞

1 2T −i 2 ω3 ω 

2π = −iR

dφ e π

iφ



ei(3 T −t)ω

1 2T − i 2 i2φ 3 i3φ R e R e



ei(3 T −t)R e

iφ

(5.59)

In essence, the end result of repeating the analysis we just did for the case t < 3T is that the sign of the second term changes. That’s it. So then end result for all t ∈ R is, once and for all:

154

5 Integral Transforms

T PV i2π

∞

 dω

−∞

1 2T −i 2 ω3 ω



ei(3 T −t)ω =

2 T (T − t) 1 + (3 T 3 + 2 T 2 t − Tt 2 ) sgn (3 T − t) 2π R 4

(5.60)

Similarly, we may write down the results for the other pieces, the derivations of which are exercises and are quite similar to what we saw above. T PV i2π



∞ dω −∞

1 2T −i 2 3 ω ω



ei(2 T −t)ω = −

2 Tt 1 3 4 T − Tt 2 sgn (2 T − t) + 2π R 4 (5.61)

T PV i2π

∞ dω −∞

1 i(T −t)ω 2 T (T − t) 1 e = − T (t − T )2 sgn (T − t) ω3 2π R 4 (5.62)

T PV i2π

∞ dω −∞

1 i(T −t)ω 2 Tt 1 e =− − Tt 2 sgn (−t) ω3 2π R 4

(5.63)

Putting this altogether, we get our convolution:



1 3 1 3 3 T + 2 Tt − Tt 2 sgn (3 T − t) − 4 T − Tt 2 sgn (2 T − t) 4 4

1 2 1 + T − 2 T 2 t + Tt 2 sgn (T − t) + Tt 2 sgn (t) (5.64) 4 4

f ∗ g(t) =

Happily, albeit unsurprisingly, the divergent terms canceled. Do the remaining terms match what we got with Eq. (5.48)? Yes they do, and (again, unsurprisingly) it is an exercise to demonstrate this. We have demonstrated both the correctness and the usefulness of the Convolution Theorem with the above examples. For continuous functions, we have seen that we can convert potentially messy convolution computations into much simpler finite integrals, even when the convolution computations are simplified via the Residue Theorem. For signals of finite support, we can see that the Convolution Theorem may replace the fiddling with geometry into a powerful algebraic method, albeit one in which we may need to manipulate divergences. But it is these divergences which provide checks on the correctness of the result, something not provided in a conventional convolution computation. Before we move on to the next section, it is worth mentioning that the Hilbert transform is a convolution. If we recall the result in Eq. (5.19), we can make a few observations.

5.1 The Fourier Transform

155

1 u(ω) = PV π

∞

d ω

−∞

v(ω ) ω − ω

∞

1 dt V (t)iπ sgn (t)eiωt π −∞ ⎤ ⎡∞  ∞ = i ⎣ dt V (t)eiωt − dt V (−t)e−iωt ⎦ =

⎡ = i⎣

0

0

∞

∞ dt V (t)eiωt −

0

⎤ dt V (t)e−iωt ⎦

0

∞ = −2 Im

dt V (t)eiωt

(5.65)

0

where 1 V (t) = 2π

∞

d ω v(ω)e−iωt

(5.66)

−∞

is the temporal signal corresponding to the imaginary part of the analytic signal. Because V (−t) = V (t), it follows - as asserted in the previous chapter - that the real part of the analytic signal depends only on the temporal signal (F.T.of the imaginary part of the analytic signal) for t > 0. That is, the Hilbert transform is the result of a causal relationship in the analytic signal, and now we can appreciate why. Parseval’s Equality It should be plain that the inverse Fourier transform kernel is the conjugate of the Fourier transform kernel (to within a constant) so that the Fourier transform satisfies a Parseval’s equality. If f ∈ L2 (−∞, ∞) and F is the Fourier transform of f , then ∞

1 dx |f (x)| = 2π

∞ dk |F(k)|2

2

−∞

(5.67)

−∞

Parseval’s equality allows us to evaluate some fairly insane integrals, so long as we can evaluate some FTs. We can start to have a little fun by imply reviewing some basic Fourier transforms. First, we can easily derive the FT of f (x) = sin x/x by using Eq. (5.19) (which is turning out to be a really, really useful result):

156

5 Integral Transforms

∞ ∞

⎤ ⎡ ∞ ∞ sin x ikx 1 ⎣ eix ikx e−ix ikx ⎦ dx e = PV dx e − PV dx e x i2 x x ∞ ∞ ⎡ ⎤ ∞ ∞ 1 ⎣ ei(k+1)x ei(k−1)x ⎦ = − PV dx PV dx i2 x x ∞

∞

 π sgn (k + 1) − sgn (k − 1) = 2  π k ∈ [−1, 1] = 0 {k k < −1} ∪ {k k > 1}

(5.68)

So because of Parseval, we can now say that ∞ −∞

sin2 x 1 dx 2 = x 2π

1 dk π 2 −1

=π

(5.69)

And why not f (x) = sin2 x/x2 while we are at it: ∞ dx ∞

∞

sin2 x ikx 1 e = − PV 2 x 4

dx ∞

∞ − 2PV

dx ∞

e2ix ikx e + PV x2

∞ dx ∞

e−i2x ikx e x2

eikx  x2

(5.70)

Now we need to evaluate the contour integral  dz C

eikz z2

(5.71)

where C is shown in Fig. 5.1 for k > 0 and Fig. 5.2 for k < 0. We can then deduce that for k > 0, ∞ PV −∞

eikx dx 2 = −iR x =−

i R

0

iφ

d φ eiφ π

0

eikR e R2 ei2φ

  d φ e−iφ 1 + ikR eiφ + . . .

π

2 = − πk R

(5.72)

5.1 The Fourier Transform

157

Again, for k < 0 the effect is to simply change the integration limits over φ ∈ [π, 2π ]. So we then have the following: ∞ dx

PV −∞

eikx 2 = − π |k| 2 x R

(5.73)

where, as expected, we carry the divergences and expect them to cancel when we tally the net integral. The result is as follows. ∞ dx ∞

sin2 x ikx π e = (|k + 2| + |k − 2| − 2|k|) x2 4 !  k ∈ [−2, 2] π 1 − |k| 2 = 0 {k k < −2} ∪ {k k > 2}

(5.74)

So we can now easily show that ∞ −∞

sin4 x 1 dx 4 = x 2π

2

  |k| 2 dk π 1 − 2 2

−2

2π = 3

(5.75)

While the above integrals evaluated using Parseval may be nifty - and likely seen before by many here - we can do so much more. For example, we may generalize Parseval for functions f and g ∈ L2 (−∞, ∞) with respective Fourier transforms F and G: ∞ −∞

1 dx f (x)g(x) = 2π

∞ dk F(k)G(k)

(5.76)

−∞

This may be derived by substituting f → f + eiθ g, F → F + eiθ G in Eq. (5.67). Upon that substitution, we find ⎡ e−iθ ⎣

∞ dx f (x)g(x) −

−∞

⎡

+e ⎣

∞

iθ

−∞

1 2π

∞

⎤ dk F(k)G(k)⎦

−∞

1 dx f (x)g(x) − 2π

∞ −∞

⎤ dk F(k)G(k)⎦ = 0

(5.77)

158

5 Integral Transforms

As Eq. (5.77) is valid for all values of θ , then each term in brackets is separately zero. We can now evaluate a whole slew of new integrals by computing some Fourier transforms. So let’s start simple: ∞ −∞

1

sin3 x 1 dx 3 = x 2π

−1

  |k| dk π 2 1 − 2

  k dk 1 − 2

1 =π 0

3π = 4

(5.78)

We can perform some odd integrals where Parseval may be applied after a little manipulation: ∞ −∞

arctan ax arctan bx dx = x2

∞

1 dx

−∞

0

1

a du 1 + a 2 x 2 u2 1

= ab

du 0

1 = 2π

∞ dv −∞

0

1

du u

0

1 =π

1

dv v

0

du u

0

1

dv v

0

1 = π ab

dv 0

b 1 + b2 x2 v2

dx



1 + a2 u2 x2 1 + b2 v2 x2

∞

dk π e−|k|/(au) π e−|k|/(bv)

−∞

∞

dk e−( au + bv )k 1

1

0

1 du

0

1

dv 0

1 au + bv

  b = π a du log 1 + au 0   a! b + π b log 1 + = π a log 1 + a b 1

(5.79)

Note that it appears this example could be done with just the plain old Residue Theorem, and that can be an exercise. But the algebra is a bit more involved than that we see using Parseval. Nevertheless, it is always a good idea to be aware of all

5.1 The Fourier Transform

159

of the arrows in our quivers when attacking any integration or summation problem: if we hit a roadblock using Parseval, it is nice to know that the Residue theorem may apply, for example. Here is a rather complicated example to illustrate that, like integration by parts, it takes a little experience to know how to split things up. Consider the following integral. ∞ √

sin x erfi x e−sx (5.80) J (s) = dx x 0

where, again, 2 erfi (y) = √ π

y dt et

2

(5.81)

0

The hope here is that by now we recognize that a factor of sin x/x inside of an integral simply becomes finite support in the frequency domain. Accordingly, we want to see if we can compute the FT of the rest of the integrand, i.e., (NB (x) is the Heaviside step function, zero when x < 0 and one when x > 0.) ∞ −∞

√

2 dx erfi x e−sx (x)eikx = √ π 2 = √ π 1 =

∞

⎡ 1 ⎤  √ 2 dx x ⎣ dt ext ⎦ e−sx eikx

0

0

1

∞ dt

0

dx

√

xe−(s−t

2

−ik )x

0

−3/2 dt s − ik − t 2

0

=

1 √ (s − ik) s − 1 − ik

(5.82)

All things considered, this is a delightfully simple form! And now the integral we seek is simply the integral of this simple function over [−1, 1]: 1 π J (s) = 2π

1

dk √ (s − ik) s − 1 − ik

−1 √ s−1+i 

= −i √ s−1−i

dv 1 + v2

(5.83)

160

5 Integral Transforms

The latter step was achieved by the substitution v2 = s − 1 − ik. The integration limits are complex but because the integrand is analytic along any path of integration that avoids v = ±i, we can simply integrate using the Fundamental Theorem of Calculus. We will leave the resulting algebra (which admittedly is a little messy) to the exercises and produce the final result. ⎞ ⎛ ! $ # # ∞ (1 + (s − 1)2 + 1 + 2 (s − 1)2 + 1 − 2(s − 1) √ −sx sin x 1 ⎟ ⎜ erfi x e dx = log ⎝ ⎠ ! $ # # x 2 2 + 1 − 2 (s − 1)2 + 1 − 2(s − 1) (s − 1) (1 + 0

(5.84) Parsevel’s Equality gives us a hugely useful way to evaluate integrals. Note that, when the functions involved are periodic, we have Fourier series rather than Fourier integrals. In this case, Parseval’s equality may be used to evaluate sums in terms of simple integrals. Some such sums will be in the exercises. Analyticity of the Fourier Transform in the Complex Plane2 We have played around with Fourier transforms for k ∈ R, but happens when k ∈ C? That is, does the Fourier transform still have its very nifty convergence properties per Plancherel’s Theorem, and if so, how much more useful can it be for us? The answer is, sometimes, and extremely useful. In fact, so useful it will help us shepherd in the super-useful Laplace and Mellin transforms. So let us now consider a Fourier transform of a function f : C → C, where z = x + iy. When f was a real function, we considered the behavior of f over the real line. In contrast, let us assume that f is analytic in a strip S = {z y ∈ (y− , y+ )}, parallel to the real axis, where y− < 0 and y+ > 0. Our analysis is guided by the following theorem. Theorem 5.2 Let A, B, C, D ∈ R. If the function f analytic in the strip S = {z y ∈ (y− , y+ )} satisfies the asymptotic relation lim |f (z)|e−τ− x = A

x→∞

lim |f (z)|e−τ+ x = B

x→−∞

where τ− < 0 and τ+ > 0, then the Fourier transform of f is a function F : C → C such that F(k), k = σ + iτ is analytic in the strip Sˆ = {k τ ∈ (τ− , τ+ )}, and within ˆ the strip S,

2

The main gist of the following two subsections may be found in passages from P. Morse & H. Feshbach, Methods of Theoretical Physics, Feshbach Publishing LLC (2005), Sect. 4.8, pp. 459– 461 and 466–467. The author is grateful to the Sylvia Feshbach Trust and Mrs. Andrea Feshbach in particular for permission to use the passages herein.

5.1 The Fourier Transform

161

lim |F(k)|ey+ σ = C

σ →∞

lim |F(k)|ey− σ = D

σ →−∞

To see why Theorem 5.2 is true, consider the behavior of F along the real axis: ∞ F(σ + iτ ) =

dx f (x)eiσ x e−τ x

(5.85)

−∞

Uniform convergence at the upper limit of integration requires that eτ− x e−τ x decays exponentially so that it must be that τ− < τ ; same at the lower limit of integration requires that eτ +−x e−τ x decays exponentially so that is must be that τ+ > τ . Now consider F(k) in the strip Sˆ and the convergence of the integral representing F(k). Because f is analytic in the strip S, we may take a path of integration along a line parallel to the real axis within S. In this case, for some y ∈ (y− , y+ ), F(σ + iτ ) = e

−σ y

∞

dx f (x + iy)ei(σ x−τ y) e−τ x

(5.86)

−∞

so that |F(k)| exponentially decays as σ → ∞. For y near y+ , convergence occurs only when σ > 0; for y near y− , convergence occurs only when σ < 0. Accordingly, Theorem 5.2 is true. The above assumes that f ∈ L2 (−∞, ∞) for all y ∈ S. To see how this might change when y = 0, consider the function f (z) = sin z/z, where z = x + iy. Then ∞ dx |f (x + iy)|2 = π −∞

sinh 2y 2y

(5.87)

(The derivation of this is of course an exercise.). While f ∈ L2 (−∞, ∞) for all finite y, the integral grows exponentially with y so that f is “less” L2 with increasing y. Let’s assume there is an f that is not L2 (−∞, ∞), but there exists a τ0 ∈ R such that a function g(x) = f (x)e−τ0 x is ∈ L2 (−∞, ∞). Then we may apply Plancherel’s Theorem to g, and g may be written in terms of its inverse Fourier transform G. g(x) = f (x)e

−τ0 x

1 = 2π

∞ −∞

dk G(k)e−ikx

(5.88)

162

5 Integral Transforms

so that f (x) =

1 2π

1 = 2π

∞

dk G(k)e−(k+iτ0 )x

−∞ ∞+iτ  0

d κ G(κ − iτ0 )e−iκx

(5.89)

−∞+iτ0

And because 1 G(k) = 2π

∞

(5.90)

−∞

then G(κ − iτ0 ) =

dx f (x)e−τ0 x eikx

1 2π

∞ dx f (x)eiκx = F(κ)

(5.91)

−∞

So we if we think about this a little, we can conclude that, despite f not being in L2 (−∞, ∞) on the real axis, on the line τ = τ0 on which it is in L2 (−∞, ∞) , we can write the following inverse Fourier relation. 1 f (x) = 2π

∞+iτ  0

dk f (k)e−ikx

(5.92)

−∞+iτ0

Let’s pause and take a little stock before moving on. A function f needs to be square integrable, i.e., in L2 (−∞, ∞) on the real axis for it to have an inverse FT by Plancherel’s Theorem. But while some functions f may not be square integrable on the real axis, its transform be square integrable on a line parallel to the real axis in the complex plane, e.g., along a line τ = τ0 when k = σ + iτ . In that case, the inverse transform is defined along that line Im k = τ0 and is determined through Eq. (5.92). The above discussion assumed that there was a single value of τ0 along which the inverse transform existed. This may not always be the case. For example, it may be the case that f is square integrable over x ∈ (0, ∞) at τ = τ0 , but f is square integrable over x ∈ (−∞, 0) at τ = τ1 . Along this line, define the following functions for τ0 > 0 and τ1 < 0.

5.1 The Fourier Transform

163

 f+ (x) =  f− (x) =

0 x0

(5.93)

f (x)e−τ1 x x < 0 0 x>0

(5.94)

We take τ0 as the minimum value possible to make f+ square integrable, and τ1 as the maximum value possible to make f− square integrable. The corresponding Fourier transforms are as follows for k = σ + iτ . ∞ F+ (k) =

dx f+ (x)eikx −∞ ∞

=

f (x)e−τ x eiσ x

(5.95)

0

for all τ > τ0 , and ∞ F− (k) =

dx f− (x)eikx −∞

0 =

f (x)e−τ x eiσ x

(5.96)

−∞

for all τ < τ1 . It then follows that 1 f+ (x) = 2π 1 f− (x) = 2π

∞+iτ  0

dk F+ (k)e−ikx

(5.97)

dk F− (k)e−ikx

(5.98)

−∞+iτ0 ∞+iτ  1

−∞+iτ1

Adding the above inverse transforms, we may conclude 1 f (x) = 2π

∞+iτ  0

dk F+ (k)e −∞+iτ0

−ikx

1 + 2π

∞+iτ  1

dk F− (k)e−ikx

(5.99)

−∞+iτ1

We note that F+ is analytic in the upper half-plane above k = iτ0 and F− is analytic in the lower half-plane below k = iτ1 . We stress that this is a special case in which

164

5 Integral Transforms

the function f is square integrable along two separate lines in the complex plane, according to whether x is positive or negative. This special case, however, will be very useful to us in deriving the amazing Poisson sum formula. The Amazing Poisson Sum Formula We have used the Residue Theorem to evaluate sums of meromorphic functions to great and in some case dramatic effect. But given the above (admittedly dry) discussion, we will now have an opportunity to evaluate sums of (piecewise) squareintegrable functions. Let’s consider the problem of evaluating the following sum. P=

∞

f (αn)

n=−∞

Let f = f+ + f− as defined above, i.e., having inverse transforms 1 f+ (x) = 2π 1 f− (x) = 2π

∞+iτ  0

dk F+ (k)e−ikx

−∞+iτ0 ∞+iτ  1

dk F− (k)e−ikx

−∞+iτ1

The Fourier transform F+ is analytic for τ > τ0 , τ0 > τ0 , as shown in Fig. 5.6. The Fourier transform F− is analytic for τ < τ1 , τ1 < τ1 , as shown in Fig. 5.6.

Fig. 5.6 Regions of analyticity for Poisson sum

5.1 The Fourier Transform

165

It is then convenient to break the sum P into two pieces: P+ =

∞

f+ (αn)

(5.100)

n=0

P− =

−1

f− (αn)

(5.101)

n=−∞

where  0 ∞ ∞+iτ 1 P+ = dk F+ (k)e−ikαn 2π n=0

(5.102)

−∞+iτ0

Because F+ is analytic in the upper-half plane above the line of integration, F+ is uniformly convergent at the integration limits and the summation and integration operations may be exchanged. Accordingly, 1 P+ = 2π 1 = 2π

∞+iτ  0

dk F+ (k)

∞

e−ikαn

n=0

−∞+iτ0 ∞+iτ  0

dk −∞+iτ0

F+ (k) 1 − e−iαk

(5.103)

Again, because F+ is analytic in the upper half-plane above the line of integration, we can consider a contour integral 1 2π

 dz C

F+ (z) 1 − e−iαz

(5.104)

We further may set τ0 < 0 so that the contour C encloses poles of the integrand along the real axis, i.e., at z = 2π m/α for m ∈ Z. Accordingly, ∞

F+ (z) 1 − e−iαz m=−∞   ∞ 2π m 1 = F+ α m=−∞ α

P+ = i

Res

z=2πm/α

(5.105)

166

5 Integral Transforms

For F− we close below the line of integration but we end up with a similar result: ∞

F− (z) z=2πm/α 1 − e−iαz m=−∞   ∞ 2π m 1 = F− α m=−∞ α

P− = i

Res

(5.106)

The amazing Poisson sum formula results from simply adding P+ and P− :   ∞ 1 2π m f (αn) = F α m=−∞ α n=−∞ ∞

(5.107)

Let’s work through some examples worthy of the name “amazing.” Bizarre behavior of sums of powers of sincs, explained! Let’s begin by asking for the value of the following sum: S1 =

∞ n=−∞

f1 (n) =

∞ sin n n n=−∞

(5.108)

To evaluate S1 using the Poisson sum formula, we need to evaluate the Fourier transform F1 of f1 , which as we know is π when k ∈ [−1, 1] and zero otherwise. The sum on the RHS of Eq. (5.107) is over the Fourier transform of argument 2π n. Note that when n = 0, F1 (2π n) = 0. Therefore, S1 = F1 (0) = π . Boom. What about ∞ ∞ sin2 n S2 = f2 (n) = (5.109) n2 n=−∞ n=−∞ To evaluate S2 using the Poisson sum formula, we need to evaluate the Fourier transform F2 of f2 , which as we know is π 1 − 21 |k| when k ∈ [−2, 2] and zero otherwise. The sum on the RHS of Eq. (5.107) is over the Fourier transform of argument 2π n. Note that when n = 0, F2 (2π n) = 0. Therefore, S2 = F2 (0) = π . Whoa. It should be clear that the Fourier transform of fp (x) = sinp x/x is nonzero only within k ∈ [−p, p]. Accordingly, by the Poisson sum formula, the sum Sp = Fp (0) for p ∈ {1, 2, 3, 4, 5, 6}, as all of these integers are less than 2π . Interestingly, for these values of p, we have ∞ ∞ sinp n sinp x = dx Sp = np xp n=−∞ −∞

(5.110)

5.1 The Fourier Transform

167

The infinite sum is equal to the infinite integral for these values of p. So from our previous examples we have S3 = 3π/4 and S4 = 2π/3. To compute S5 and S6 , we will need the Fourier transform of f3 (x), which will be derived in the exercises as

∞ F3 (k) =

dx −∞

sin3 x ikx e x3

⎧ ⎪ ⎪ ⎪0 ⎪ π ⎪ ⎪ + k)2 ⎨ 8 (3

= π4 3 − k 2 ⎪ ⎪ ⎪ π (3 − k)2 ⎪ 8 ⎪ ⎪ ⎩0

k < −3 −3 < k < −1 −1 < k < 1 1 2π so that F7 (2π ) = 0. Then again, it is easy to see that F7 (4π ) = 0, so that the Poisson sum formula provides us with the still simple relation S7 = F7 (0) + 2F7 (2π ), as Fp is even for all integer p. Accordingly, we may no longer say that the infinite sum of sincs is equal to the infinite integral over sincs when the power is 7 or greater. What does it look like? We will have to construct the Fourier transform F7 (k) of f7 (x); by now it should be clear how to go about this, even if the bookkeeping will get a little messy. Again, we leave the details for the problems: ∞ F7 (k) =

dx −∞

sin7 x ikx π  (k + 7)6 sgn (k + 7) − 7(k + 5)6 sgn (k + 5) e = x7 92120 + 21(k + 3)6 sgn (k + 3) − 35(k + 1)6 sgn (k + 1) + 35(k − 1)6 sgn (k − 1) − 21(k − 3)6 sgn (k − 3)  + 7(k − 5)6 sgn (k − 5) − (k − 7)6 sgn (k − 7) (5.114)

In this case it is easier to express the transform in terms of the signums rather than rearrange as we just seek a few values from it rather than the entire shape of the resulting curve. Given this expression, we may simply write down (OK, after some nontrivial manipulation) the exact expression for S7 : S7 =

∞ π sin7 n 5887π + (7 − 2π )6 = 7 n 11520 23040 n=−∞

(5.115)

Just for the record, we derived all of the above expressions by hand, only using a CAS to verify the derived results. We will stop here for now and leave additional discussion to the exercises. Taking stock, we see that the Poisson Sum formula allows us to express complicated sums in terms of another sum just having one or a few terms in certain cases, i.e., when the summand has a Fourier transform of finite support. In these case, the algebraic nature of the formula explains some bizarre behavior for sums of powers of sincs. We now proceed to a wholly different application of the Poisson sum formula, but the theme is similar: we replace a nasty-looking sum with one only having very few terms.

5.1 The Fourier Transform

169

Computing the error function of complex argument to arbitrary precision Recall that the error function of real argument x is defined as follows. 2 erf x = √ π

x

dt e−t

2

(5.116)

0

In discussions of what a closed-form is, it has been proposed that a closed form is an expression the value of which requires fewer computing resources to generate than an integral or summation representation. In that respect, the error function is a closed form because of the various simple expressions that may be used to compute its value to arbitrary precision over any interval on the real line. What about values in the complex plane? In the complex plane, we may define 2 erf z = √ π

z

d ζ e−ζ

2

(5.117)

0

Because e−ζ is analytic everywhere, the above integral is independent of the path  used between ζ = 0 and ζ = z. Consider, then, the special case where  is the path that runs from 0 to x along the real axis, then from x to z = x + iy parallel to the imaginary axis. Seen this way, the error function of a complex number is equal to 2

2 2 erf (x + iy) = erf x + i √ e−x π

y 0

2 2 du e cos 2xu + √ e−x π u2

y

2

du eu sin 2xu 0

(5.118) Keep in mind that we consider erf x a given closed form, and we are seeking erf z for z = x + iy. So to compute the complex erf, we need to evaluate these integrals in some form. What we will find is...we can’t, exactly. But we can get as close as we’d like, and to do that we will be using the Poisson sum formula.3 First, we will need the Poisson sum formula in a shifted form: ∞

f (n + x) =

n=−∞

∞

ei2πmx F(2π m)

(5.119)

m=−∞

We now consider f (x) = e−a x , a > 0; its Fourier transform is F(k) = Let u = ax and we get 2 2

√ π −k 2 /(4a2 ) e . a 3

This stems from an idea taken from H. E. Salzer, “Formulas for calculating the error function of a complex variable,” Math. Comp. 5, 67–68 (1951).

170

5 Integral Transforms ∞

e

−(u+na)2

n=−∞

' √ & ∞ π u! −m2 π 2 /a2 = e cos 2π m 1+2 a a m=1

(5.120)

The key observation here is that we can choose any value of a we wish and this equation holds true. In fact, we can choose a value of a such that the sum on the RHS may be ignored. Let’s call this sum (u):  ∞ ∞  ! u 2 2 2 2 2 2   e−n π /a cos 2π n  ≤ e−n π /a (5.121) |(u)| = 2    a n=1 n=1 We then rearrange Eq. (5.120): a e [1 + (u)] = √ π

&

u2

1+2

∞

' e

−n2 a2

cosh 2nau

(5.122)

n=1

So what we essentially did was use the Poisson sum formula applied to Gaussians 2 in order to replace the factor eu in the integrals in Eq. (5.118). Admittedly, this was far from trivial but in hindsight, it is clear that this factor was troublesome and would need replacement if we were to make any progress. 2 2 It is noted that, when a = 1/2, |(u)| ≤ e−4π + e−16π + · · · ≈ 5 × 10−17 . Thus, while the result we attain will not be exact, it will achieve double precision for any value of z in the complex plane and will be, for all intents and purposes, a formula for the complex error function. To get the value of an integral, we multiply both sides of Eq. (5.122) and integrate to get 2 2 √ e−x π

y 0

2 2 du eu sin 2xu ≈ √ π

y 0

a du √ π

& 1+2

∞

' 2 2 e−a n cosh 2nau sin 2xu

n=1

1 − cos 2xy = 2a 2π x  ∞ 4a −a2 n2 e du cosh 2anu sin 2xu π y

+

n=1

0

1 − cos 2xy = 2a 2π x ∞ 2a −a2 n2 x(1 − cos 2xy cosh 2any) + an sin 2xy sinh 2any + e π x 2 + a 2 n2 n=1

(5.123)

5.2 Laplace Transforms

171

Similarly,

2 √ π

y

2

du eu cos 2xu ≈ 2a

sin 2xy 2π x

0

+

∞ 2a −a2 n2 x sin 2xy cosh 2any + an cos 2xy sinh 2any e π n=1 x2 + a2 n2

(5.124) Putting this all together and plugging in a = 1/2, we get the following result: erf(x + iy) = erf x e−x [(1 − cos 2xy) + i sin 2xy] 2π x 2 ∞ 2 e−k /4 2 + e−x [fk (x, y) + igk (x, y)] + (x, y) π k 2 + 4x2 2

+

(5.125)

k=1

where fk (x, y) = 2x(1 − cos 2xy cosh ky) + k sin 2xy sinh ky

(5.126)

gk (x, y) = 2x sin 2xy cosh ky + k cos 2xy sinh ky

(5.127)

and |(x, y)| ≤ 5 × 10−17 . We can of course adjust the maximum relative error throughout by adjusting the value of a, but here is a formula that provides the error function everywhere in the complex plane out to double precision. Not exact for analytical purposes but exact for numerical purposes. We note here that, as in the since sum formulae in the previous section, we only used one term in the sum from the Poisson sum formula here. The rest of the terms were used to define the relative error  in approximating the integrals in Eq. (5.120). We also note here that we only considered one possible integration path in deriving the exact result in Eq. (5.120). Another integration path is considered in the exercises.

5.2 Laplace Transforms We define a Laplace transform of a function f : [0, ∞) → R as ∞ F(s) = 0

dt f (t)e−st

(5.128)

172

5 Integral Transforms

F is the boundary value on the real line of a meromorphic function in the complex plane. Note that the Laplace transform is an expression of causality and is a special case of a Fourier transform. To see this, define the following function.  f+ (t) =

f (t) t > 0 0 t 0 was the minimum value above the real axis for which the Fourier transform was square integrable. Since we are only concerned with t > 0, we will not need to consider F− or τ1 . With the Laplace transform, we are rotating the axis by π/2. We will discuss the role of τ0 in more detail when we compute inverse Laplace transforms. The expression in Eq. (5.134) seems a bit odd and it is worth examining how the Laplace transform and its inverse cancel each other out, i.e., how we form a delta function. 1 f (t) = i2π =

1 i2π ∞

=

τ0+i∞

ds F(s)est τ0 −i∞ τ0+i∞

⎡

ds ⎣

τ0 −i∞

dt  f (t  )

∞

⎤ dt  f (t  )e−st ⎦ est 

0

1 i2π

0

τ0+i∞



ds es(t−t )

(5.135)

ds est

(5.136)

τ0 −i∞

That is, we may define the delta function as 1 δ(t) = i2π

τ0+i∞

τ0 −i∞

Before we move on, it is worth noting that the inverse Laplace transform kernel is not a Hermitian conjugate of the Laplace transform kernel. Accordingly, we cannot expect a Parseval equality for Laplace transforms. Convolution, or Nobody’s Faltung But Mine Now that we know what a delta function looks like, we can answer the question, what does convolution look like? Or better yet, what is the inverse Laplace transform of the product of two Laplace transforms F(s) and G(s)?

174

5 Integral Transforms

1 i2π

⎡∞ ⎤ τ0+i∞  1  ds F(s)G(s)est = ds ⎣ dt  f (t  )e−st ⎦ i2π 0 τ0 −i∞ ⎤ ⎡∞  ⎣ dt  g(t  )e−st  ⎦ est

τ0+i∞

τ0 −i∞

0

⎡ ∞ ⎤ τ0+i∞  1  = ds ⎣ dt  f (t  ) (t  )e−st ⎦ i2π −∞ τ0 −i∞ ⎤ ⎡ ∞  ⎣ dt  g(t  ) (t  )e−st  ⎦ est −∞ ∞

=





∞



dt f (t ) (t ) −∞

dt  g(t  ) (t  )

−∞

τ0+i∞



ds es(t−t −t



1 i2π

)

τ0 −i∞

∞ =







∞

dt f (t ) (t ) −∞ ∞

=

dt  g(t  ) (t  )δ(t − t  − t  )

−∞

dt  f (t  ) (t  )g(t − t  ) (t − t  )

−∞

t =

dt  f (t  )g(t − t  )

(5.137)

0

The RHS of Eq. (5.137) defines a convolution of two inverse Laplace transforms. This will be of fundamental usefulness to us. Ode to an ODE Laplace transforms are frequently used to solve ordinary differential equations. More to the point, Laplace transforms convert ODEs to algebraic equations. For example, consider the general second-order equation a¨x + b˙x + cx = f (t)

(5.138)

with x(0) = x0 x˙ (0) = x˙ 0 . Recall that the Laplace transform of a time derivative of a function x(t) is

5.2 Laplace Transforms

∞

175

  ˙ −st = x(t)e−st ∞ + s dt x(t)e 0

0

∞

dt x(t)e−st

0

= X (s) − x0

(5.139)

The LT of a time-second derivative is similarly ∞

¨ dt x(t)e

−st

  ˙ −st ∞ + s = x(t)e 0

0

∞

˙ −st dt x(t)e

0

= s2 X (s) − sx0 − x˙ 0

(5.140)

Accordingly, taking Laplace transforms of both sides, we get the algebraic equation:

2 (5.141) as + bs + c X (s) − ax0 s − (a˙x0 + bx0 ) = F(s) Then X (s) =

F(s) + As + B as2 + bs + c

(5.142)

where A = ax0 and B = a˙x0 + bx0 . To find the solution for x(t) we take the inverse Laplace transform: 1 x(t) = i2π

τ0+i∞

ds X (s)est

(5.143)

τ0 −i∞

We may compute this integral by forming a contour integral. Let’s form a contour as illustrated in Fig. 5.7, which is called a Bromwich contour. Here, we closed the contour with a circle of radius R to the left of the integration line. Note that we can write the inverse Laplace transform as

x(t) =

ω0 lim i2π R→∞

√ 2 2  R −τ0 F(s) + As + B st e ds as2 + bs + c √ 2 2

τ0 +i

τ0 −i

(5.144)

R −τ0

Let’s first take homogeoneous the case f (t) = 0 which means F(s) = 0. To compute the inverse Laplace transform, we consider the contour integral 1 i2π

 dz C

Az + B ezt + bz + c

az 2

where for t > 0, C is as in Fig. 5.7. The contour C has two segments γ1 and γ2 :

(5.145)

176

5 Integral Transforms

Fig. 5.7 Bromwich contour for computing inverse Laplace transform, no branch points

 dz γ1

 dz γ2

Az + B ezt = + bz + c

az 2

τ0 −i

Az + B ezt = iR az 2 + bz + c

√ 2 2  R −τ0 As + B est ds 2 as + bs + c √ 2 2

τ0 +i

(5.146)

R −τ0



3π/2−arcsin

τ0 R

AReiθ + B eRt cos θ aR2 ei2θ + bReiθ + c

d θ eiθ π/2+arcsin

τ0 R

(5.147) We can estimate the integral over γ2 by the ML-inequality as R → ∞.       Az + B zt   dz ≤R e  2 az + bz + c   γ2



3π/2−arcsin

π/2+arcsin

τ0 R

  d θ 

τ0 R

AR2 + BR ≤ aR2 − bR + c

3π/2 

d θ eRt cos θ

π/2

AR + BR 2 = aR2 − bR + c 2

≤2

AR + BR aR2 − bR + c 2

  Rt cos θ AReiθ + B e 2 i2θ iθ aR e + bRe + c 



π/2d θ e−Rt sin θ

0

∞

d θ e−2RT θ/π

0

AR + B π = t aR2 − bR + c

(5.148)

5.2 Laplace Transforms

177

which obviously vanishes as R → ∞. Note that we defined the maximum of the integrand at θ = π . The integral over γ1 is i2π times the inverse Laplace transform as R → ∞: 1 i2π

 γ1

τ0+i∞

Az + B 1 dz 2 ezt = az + bz + c i2π

ds τ0 −i∞

F(s) + As + B st e as2 + bs + c

(5.149)

Note that closing the contour to the left is equivalent to indicating t > 0 because on the left, i.e., when θ ∈ [π/2, 3π/2], cos θ < 0 and the integral over γ2 only converges when t > 0. Accordingly, closing to the right is equivalent to indicating t < 0. The contour integral is then equal to the sum of the residues of the poles inside C. And this is where we can make sense of the value of τ0 : we define τ0 such that for t > 0 all of the poles of the integrand lie within C. That way we achieve causality, because for t < 0, where we close the contour C to the right rather than the left, none of the poles will lie within C and the inverse Laplace transform is accordingly zero for t < 0. In other words, we require for causality τ0 > Max Re zk k∈{1..N }

(5.150)

where zk is the kth of N poles of the inverse Laplace transform. With that sorted out, we can easily compute the solution to the ODE as the inverse Laplace transform by summing the residues of the two poles of the integrand. x(t) =

Az+ + B Az− + B ez+ t + 2 ez− t 2 az+ + bz+ + c az− + bz− + c

(5.151)

1# 2 b ± b − 4ac 2a 2a

(5.152)

for t > 0, where z± = −

Let’s also examine the inhomogenoeus case in which x0 = x˙ 0 = 0 and f (t) = 0, i.e., ∃s F(s) = 0. Then for t > 0, 1 f (t) = i2π

τ0+i∞

ds τ0 −i∞

F(s) est as2 + bs + c

(5.153)

We can express the solution directly as an integral over time by the convolution

−1 theorem. Let G(s) = as2 + bs + c . Then by the Residue Theorem, for t > 0,

178

5 Integral Transforms

1 g(t) = i2π =

τ0+i∞

ds τ0 −i∞ z+ t

est as2 + bs + c

e ez− t + 2 2 az+ + bz+ + c az− + bz− + c

(5.154)

By the convolution theorem, we can write down the solution to the ODE for t > 0 as

t x(t) =

dt  g(t − t  )f (t  )

(5.155)

0

The solution to the general equation Eq. (5.138) is the sum of the the homogeoneous solution in Eq. (5.151) and the inhomogeoneous solution in Eq. (5.155). The Heat Equation and Branch Points Let us now move from an ordinary differential equation to a partial differential equation. In the former, we used Laplace transforms to convert the ODE to an algebraic equation, the solution to which we then inverse-Laplace-transformed (what a remarkable verb) to obtain the solution to the ODE. What happens when we apply LTs to a PDE? A very good PDE with which to start is the heat equation. Let’s analyze the temperature distribution u of a one-dimensional semi-infinite rod as a function of position x and time t. The equation we will solve that models this physical situation is as follows. ∂ 2u ∂u =κ 2 (5.156) ∂t ∂x where x > 0 and t > 0. κ > 0 is the heat conductivity of the rod. u(x, 0) = 0 ∀ x > 0

(5.157)

u(0, t) = α(t) ∀ t > 0

(5.158)

We take the Laplace transform with respect to the time t. ∞ 0

∞ ∂u −st  e = u(x, t)e−st 0 + s dt ∂t

∞

dt u(x, t)e−st

0

= −u(x, 0) + sU (x, s) = sU (x, s)

(5.159)

where U (x, s) is the Laplace transform of u(x, t) with respect to time. Note that we applied the initial condition in Eq. (5.157) here.

5.2 Laplace Transforms

179

∞ 0

∂ 2u ∂2 dt 2 e−st = 2 ∂x ∂x

∞

dt u(x, t)e−st

0

∂ U (x, s) ∂x2 2

=

(5.160)

We have replaced the PDE in Eq. (5.156) with an ODE as follows. (We replace the partial derivatives with plain old derivatives because we only have derivatives with respect to x.) d 2 U (x, s) = sU (x, s) (5.161) κ dx2 U (0, s) = A(s) where

∞ A(s) =

(5.162)

dt α(t)e−st

(5.163)

0

is the LT of the temperature distribution at the boundary x = 0. Note that we only have a single boundary condition, and we need another. We will specify the other condition after we form the general solution to Eq. (5.161). That general solution is straightforward to write down. √

U (x, s) = P(s)e−

√ sx/ κ

+ Q(s)e

√ √ sx/ κ

(5.164)

We√note that s ∈ (τ√0 −√i∞, τ0 + i∞). If we require that τ0 > 0, then as x → ∞, √ e− sx/ κ → 0 and e sx/ κ → ∞. Accordingly, if we constrain that the solution U (s, x) remain finite as x → ∞, then this is our second boundary condition. The solution now takes the form √

U (x, s) = P(s)e−

√ sx/ κ

(5.165)

When we apply the condition in Eq. (5.162), we find that P(s) = A(s). The solution to Eq. (5.161) is therefore, after inverse-Laplace-transforming 1 u(x, t) = i2π

τ0+i∞

√

ds A(s)e−

√ sx/ κ st

e

(5.166)

τ0 −i∞

Before we continue, let’s make a couple of simplifications. First, let’s rescale √ the √ x coordinate so we don’t need to carry the factor of κ around, e.g., let x = κy. Second, let’s √take α(t) = α0 , a constant; then A(s) = α0 /s. Accordingly, letting v(y, t) = u κy, t , we now seek to evaluate the following inverse Laplace transform.

180

5 Integral Transforms

Fig. 5.8 Bromwich contour for computing inverse Laplace transform with a branch point

α0 v(y, t) = i2π

τ0+i∞

τ0 −i∞

ds −√sy st e e s

(5.167)

To evaluate the ILT, we form the following contour integral. 

α0 i2π

C

dz −y√z tz e e z

(5.168)

We immediately recognize the presence of a branch point and think...this is going to be fun! OK, maybe not, but sure as heck interesting. As we should know by now, we need to exclude the branch point from the contour; the contour C is shown in Fig. 5.8. By now, we should recognize that the only contour segments that contribute to the contour integral are γ1 , γ3 , γ4 , and γ5 . But because the form of the integrand is new to us, let’s write down each segment and make sure they behave as expected.  γ1

 γ2

dz −y√z tz e e = z

τ0 +i

τ0 −i

√

√ 2 2  R −τ0

√

(5.169)

R2 −τ02

π−arcsin

dz −y z tz e e =i z

ds −√sy st e e s



π/2−arcsin

A R

d θ e−y τ0 R

√

Reiθ/2 tReiθ

e

(5.170)

5.2 Laplace Transforms

181

 γ3

 γ4



dz −y z tz e e =i z

du −i√uy −ut e e u

A R

d φ ey π−arcsin

dz −y z tz e e = z √



dz −y z tz e e =i z

R eiφ/2 tR eiφ

e

(5.172)

du i√uy −ut e e u

(5.173)

R2 −A2

3π/2+arcsin

√

√

A R

√ 2 2 R −A

√

(5.171)

R2 −A2

−π+arcsin

γ5

γ6

dz −y√z tz e e = z √

√





√2 2 R −A

π+arcsin

τ0 R

d θ e−y

√

Reiθ/2 tReiθ

e

(5.174)

A R

Here, R is the radius of the arc in segment γ4 , A is the distance between the segments γ3 and γ5 from the real axis, and R is the radius of the arcs in segments γ2 and γ6 . We are interested in the integrals over the segments in the limits R → ∞, R → 0, and A /R → 0. The √integrals over γ2 and γ6 vanish as R → ∞ because the etz term dominates the e−y z term in the integrand. For example,    π  √  dz −y√z tz   e e  ≤ d θ e−y R cos θ/2 etR cos θ  z   γ2

π/2

π/2 √ = d θ e−y R cos (θ/2+π/4) e−tR sin θ

(5.175)

0

so that both exponential terms vanish as well R → ∞, but the term e−tR sin θ dominates so that the other term is irrelevant. This is true with γ6 as well:   3π/2    √  dz −y√z tz  ≤  e e d θ e−y R cos θ/2 etR cos θ   z   γ6

π

π/2

d θ e−y

= 0

√

R cos (θ/2−3π/4) −tR sin θ

e

(5.176)

182

5 Integral Transforms

so that the first exponential term increases as R → ∞, but is dominated by the decreasing second term so as with γ2 the first term is irrelevant. Thus, γ2 and γ6 vanish as R → ∞. The integral over γ1 becomes the inverse Laplace transform.  γ1

dz −y√z tz e e = z

τ0+i∞

τ0 −i∞

ds −√sy st e e s

(5.177)

The integral over γ4 is easily evaluated as R → 0 and A /R → 0:  γ4

dz −y√z tz e e =i z

−π dφ π

= −i2π

(5.178)

This leaves the integrals over γ3 and γ5 in these limits.  γ3 +γ5

dz −y√z tz e e = z

0 ∞

∞

du −iy√u −tu e e + u

du iy√u −tu e e u

0

∞

√

du sin y u e−tu u

= i2 0

∞ = i4

du

sin yu −tu2 e u

0

∞ = i2

du

−∞

sin yu −tu2 e u

(5.179)

So it appears we have reduced the inverse LT to an integral that we may evaluate using the Fourier Parseval equality! Didn’t see that one coming, huh? ∞ −∞

sin yu −tu2 1 e du = u 2π



π π t

y

dk e−k

2

/(4t)

−y

y √ 2 t

 2 2 = √ π dw e−w π 0   y = π erf √ 2 t

(5.180)

5.2 Laplace Transforms

183

Fig. 5.9 Plots of the temperature distribution in a semi-infinite rod for log10 t ∈ {−2, −1, 0, 1, 2, 3, 4} and κ = 1 and α0 = 1

Now that we have evaluated all of the integrals along the segments, we may now apply Cauchy’s Theorem and state what the inverse Laplace transform and hence the solution to the PDE is. α0 v(y, t) = i2π

τ0+i∞

τ0 −i∞

ds −√sy st e e = α0 − α0 erf s



y √

2 t

 (5.181)

or, rescaling back to the original solution: 

x u(x, t) = α0 erfc √ 2 κt

 (5.182)

where erfc is the complementary error function. It is interesting to see what the temperature curves look like over time; this is shown in Fig. 5.9 for κ = 1. We now have the tools to examine the effect of a non-constant temperature α(t). Recall that we would seek the following inverse Laplace transform: 1 v(y, t) = i2π

τ0+i∞

√

ds A(s)e−y s est

(5.183)

τ0 −i∞

which we know, by the Convolution Theorem, is the convolution of α(t) and some function β(t) which is the following inverse Laplace transform: 1 β(t) = i2π

τ0+i∞

√

ds e−

sy st

e

(5.184)

τ0 −i∞

To evaluate this ILT, we don’t need to go through all of the above analysis again; we can simply recognize that we may simply differentiate the constant-temperature expression with respect to t.

184

5 Integral Transforms

⎤ ⎡ τ0+i∞ ds −√sy st ⎦ ∂ ⎣ 1 β(t) = e e ∂t i2π s τ0 −i∞    ∂ y = 1 − erf √ ∂t 2 t y 2 = √ t −3/2 e−y /(4t) 2 π

(5.185)

The temperature of a semi-infinite rod with a varying initial temperature α(t) is then t x2 x dt  (t − t  )−3/2 e− 4κ(t−t ) α(t  ) (5.186) u(x, t) = √ 2 πκ 0

We will work out a case of nonconstant α in the exercises. Convolution Interlude In Eq. (5.185), we established that 1 i2π

τ0+i∞

√

ds e−

τ0 −i∞

y 2 e = √ t −3/2 e−y /(4t) 2 π

sy st

(5.187)

We can use the convolution theorem to generate some pretty nifty integrals. For example, consider the integral 1

dt [t(1 − t)]−3/2 e−a /t e−b 2

2

/(1−t)

(5.188)

0

for a > 0 and b > 0. By the convolution theorem, this is equal to the inverse Laplace transform 4π 1 ab i2π

τ0+i∞

ds e

√ √ −2a s −2b s s

e

τ0 −i∞

4π 1 e = ab i2π

τ0+i∞

√

ds e−2(a+b) s es

τ0 −i∞

4π a + b −(a+b)2 = √ e ab 2 π

(5.189)

For any value of a, b we just replace those values with absolute values. Accordingly, we can just write down the value of the integral we sought. 1 0

dt [t(1 − t)]−3/2 e−a /t e−b 2

2

/(1−t)

√ =2 π



 2 1 1 + e−(|a|+|b|) |a| |b|

(5.190)

5.2 Laplace Transforms

185

Neat, huh? We will present numerous examples in the exercises of integrals that may be evaluated in a similar manner. But first, let’s go through some examples of inverse Laplace transforms that lend themselves to these kind of integrals. Adventures of the Stretched Exponential Let’s evaluate the following inverse Laplace transform. 1 f (t) = i2π

τ0+i∞

ds s−(1+a) e−s est a

(5.191)

τ0 −i∞

where a ∈ (0, 1). This is going to be an adventure as the section title indicates. The analysis is going to be unconventional and a little creative. At one point, we will seem to perform the impossible, only made possible by a trick of analytic continuation. But as with all cases in which we stretch ourselves, the stretched exponential4 is going to be some messy fun. We begin by writing down the contour integral 

dz z −(1+a) e−z ezt a

(5.192)

C

where C is that illustrated in Fig. 5.8. We can ignore the integrals over γ2 and γ6 as they will vanish as R → ∞. The integral over γ1 is simply i2π times the inverse Laplace transform we seek. The rest of the segments are as follows.  dz z

R

−(1+a) −z a zt

e

e =e

iπ

γ3

dx e−iπ(1+a) x−(1+a) e−e

iπa a

x

e−xt

(5.193)

∞

 dz z

−(1+a) −z a zt

e =e

e

−iπ

γ5

 dz z

∞

dx eiπ(1+a) x−(1+a) e−e

−iπa a

x

e−xt

(5.194)

R

−(1+a) −z a zt

e

γ4

−π

e = iR

d φ eiφ R−(1+a) e−i(1+a)φ e−R e e−tR e a iaφ

iφ

(5.195)

π

Let’s evaluate the integral over γ4 first. We can expand the terms e−tR e e−R e because they are vanishingly small as R → 0. iφ

4

See what I did there?

a iaφ

186

5 Integral Transforms

 dz z

−(1+a) −z a zt

e

e =

iR−a

γ4

−π



d φ e−iaφ 1 − Ra eiaφ + . . . 1 − tR eiφ + . . .

π

= −i2R−a

sin π a + i2π + . . . a

(5.196)

where the ellipses refers to terms that vanish as  → 0. Next, the evaluation of the integral over γ3 + γ5 . We combine and get the following expression. ⎡



dz z −(1+a) e−z ezt = i2 Im ⎣eiπa a

γ3 +γ5

∞

⎤ dx x−(1+a) e−e

−iπa a

x

e−xt ⎦

(5.197)

R

This is a tough situation: the integrand clearly diverges as R → 0. The divergence should balance out with the divergence of the integral over γ4 but it is not clear how to get there. In a case like this it helps to rescale the integral. In this case, it makes sense to define u = xt; the result is as follows. ⎡ ⎤ ∞  a −iπa −a a dz z −(1+a) e−z ezt = i2t a Im ⎣eiπa du u−(1+a) e−u e−e t u ⎦ (5.198) γ3 +γ5

R t

This is where things get tricky. In the integral over γ4 , both exponentials were −iπa −a a of small arguments and could be expanded. Here, the exponential term e−e t u −u may be expanded because it is subdominant to the term e , meaning its variation inside the integrand is far less significant. Then the integral is as follows upon the expansion. ∞ du u

−(1+a) −u −e−iπa t −a ua

e e

∞ =

R t



du u−(1+a) e−u 1 − e−iπa t −a ua + . . .

R t

∞ =

du u

−(1+a) −u

e

−e

R t

−iπa −a

∞

t

du R t

e−u +(5.199) ... u

It is still unclear how to proceed from here without running headlong into a divergent integrand at the lower limit. And this is where analytic continuation comes in handy. Let’s be a bit bold and rewrite the first integral as follows. ∞ du u R t

−(1+a) −u

e

∞ =

du u 0

−(1+a) −u

e

R t − 0

du u−(1+a) e−u

(5.200)

5.2 Laplace Transforms

187

What we did here seemingly makes no sense, as we have now written in an explicit difference between two infinities. But we can argue that the infinities cancel and what we have is really an effective Cauchy principal value. Moreover, the first term corresponds to a formal definition of (−a). By the  reflection formula, it turns out that we may write the following analytic continuation of the integral. ∞

du u−(1+a) e−u = −

0

1 π sin π a (1 + a)

(5.201)

Again, this is a formal statement because it makes no sense numerically: we have an integral over a positive quantity being equal to a negative number. We may pull a similar trick in the second integral and simply evaluate it only at the upper limit (as the lower limit divergence “cancels” the lower limit divergence from the first integral) and we now have ∞

du u−(1+a) e−u = −

R t

1 t −a −a π +  sin π a (1 + a) a R

As for the second integral in Eq. (5.199), e−iπa t −a

(∞ R t

(5.202)

−u

du eu , note that the factor

e−iπa renders the term real, and as we are only interested in the imaginary part, this term contributes nothing to the integral. So we finally arrive at the following place. 

dz z −(1+a) e−z ezt = a

γ3 +γ5

⎡

i2t Im ⎣eiπ a

∞

a

du u

−(1+a) −u −e−iπa t −a ua

e

e

−1+e

0

=

i2R−a

−iπ a −a a

t

u

!

⎤ π 1 t −a −a ⎦  − + a R sin πa (1 + a)

⎡ ⎤ ∞ ! i2πt a sin πa a iπ a −(1+a) −u −e−iπa t −a ua −iπ a −a a ⎦ ⎣ − + i2t Im e du u e e −1+e t u a (1 + a) 0

(5.203) Whew! Note that we subtracted out the two Taylor expanded terms in the integral. This results in a finite integral in the limit as R → 0, so that the fairly wonky way we got here is legitimized. It is also a fairly messy-looking result but an example will illustrate how clean this actually is. Combining with the integral over γ4 , we see the divergent terms cancel as they must. We can now write down an expression for the inverse Laplace transform of the stretched (and scaled) exponential.

188

5 Integral Transforms

1 i2π =

τ0+i∞

ds s−(1+a) e−s est

τ0 −i∞

a

⎡

a

t Im ⎣eiπa π

∞

du −u −e−iπa t −a ua e e u1+a

⎤ ! − 1 + e−iπa t −a ua ⎦

0

ta + −1 (1 + a)

(5.204)

To illustrate that the above expression is actually fairly useful, let’s examine the √ case a = 1/2 in detail. That is, we are taking the inverse LT of s−3/2 e− s . Then plugging a = 1/2 into everything else, we get the following. 1 i2π

τ0+i∞

√

ds s−3/2 e− s est

τ0 −i∞

⎤ ⎡ ∞ √  ! t du −1/2 1/2 −u it u −1/2 1/2 ⎦ e = e − 1 − it u Im ⎣i π u3/2 0 √ t + −1 (3/2)

(5.205)

Substitute u = x2 and take the imaginary part of the integral, we will be gobsmacked to find that we end up with the following beautiful and compact expression: ⎡ Im ⎣i

∞

⎤ ∞ ! 2 du −u it −1/2 u1/2 −a a ⎦ −x2 sin βx e = 2 e − 1 − it u dx e u3/2 x2

(5.206)

−∞

0

1 . (The details are left as an exercise.) And we know how to handle where β = 2√ t integrals like this …right?

∞ 2 −∞

dx e

−x2

sin2 βx 2 = x2 2π √

2β dk −2β

2β

= 2 πβ

  √ −k 2 /4 |k| πe πβ 1 − 2β

e−k

2

/4

  k 1− 2β

0

! √ 2 = 2πβ erf β − 2 π 1 − e−β

(5.207)

5.2 Laplace Transforms

189

Fig. 5.10 Bromwich contour for computing inverse Laplace transform with two branch points

It is relatively simple to put everything back together for the final result; putting 1 , we get back β = 2√ t 1 i2π

τ0+i∞

ds s τ0 −i∞

  2 √ −1 1 4t e =√ t e − erfc √ π 2 t

√ −3/2 − s st

e

(5.208)

Admittedly, other values of a ∈ (0, 1) may not work out so nicely but it is good to see that the general expression works out to something utterly familiar. Putting our Bessel foot forward √

In this example, let’s take the inverse Laplace transform of e− to compute τ0+i∞ √ 1 2 f (t) = ds e− s +s est i2π

s2 +s

. That is, we wish

(5.209)

τ0 −i∞

What’s interesting about this inverse LT is that the contour integral we will use to recover this inverse LT has an integrand with two branch points, at z = 0 and z = −1. That contour integral takes the following form.  √ 2 dz e− z +z ezt (5.210) C

where C is the contour shown in Fig. 5.10. In this case, we should be able to see that the integrals over γ4 , γ6 , and γ8 vanish as R → 0, so we will skip these. The integrals over γ2 and γ10 , in contrast, are interesting here and we will have a look at them. The integral over γ1 will, again,

190

5 Integral Transforms

simply be i2π times the inverse Laplace transform we seek as R → ∞. So without further ado, the integrals over the segments of the contour integral in the limits as R → 0 are as follows.  dz e



3π/2+arcsin

√ − z 2 +z zt

τ0 R

√

d θ eiθ e−

e = iR

γ2 +γ10

dz e

1

√ − z 2 +z zt

e =e

√

iπ

dx, e

γ3

(5.211)

x(x−1) −xt

e

(5.212)

R

 dz e

0

√ − z 2 +z zt

e =e

dx e−i

iπ

γ5



e

τ0 R

π/2−arcsin



R2 ei2θ +Reiθ tReiθ

√

x(1−x) −xt

e

(5.213)

1 √

dz e−

z 2 +z zt

e = e−iπ

 01 dx ei

√

x(1−x) −xt

e

(5.214)

γ7

 dz e

√ − z 2 +z zt

e =e

−iπ

R

√

dx, e

γ9

x(x−1) −xt

e

(5.215)

1

To deal with the integral over γ2 + γ10 , let’s take R → ∞ and sub ζ = eiθ .  dz e

−i

√ − z 2 +z zt

e =R

γ2 +γ10

dζ e

1 −Rζ 1+ Rζ

!1/2

etRζ

i 

−i =R

dζ e

R(t−1)ζ

e

−Rζ

1 1+ Rζ

!1/2

 −1

i

∼e

−1/2

−i d ζ eR(t−1)ζ (R → ∞)

R i

= −i2π e−1/2 We note here that the term sin (R(t−1)) π(t−1)

sin (R(t−1)) π(t−1)

sin (R(t − 1)) π(t − 1)

(5.216)

= δ(t − 1) in the limit as R → ∞. Why?

Because as R → ∞, = 0. Moreover, at t − 1, the term is equal to R/π , which increases without bound as R → ∞. And finally, because R > 0,

5.2 Laplace Transforms

191

∞ dt −∞

Accordingly,



√

dz e−

sin (R(t − 1)) =π (t − 1)

(5.217)

z 2 +z zt

e = −i2π e−1/2 δ(t − 1)

(5.218)

γ2 +γ10

It is interesting that the integral along γ2 + γ10 does not completely vanish, at least at t = 1. Note that we will find that the inverse LT is zero for t < 1. We note that the contributions from the left branch point to the outer circle along the branch cuts, i.e., along γ3 and γ9 , cancel each other out. Accordingly, the only other contribution to the inverse LT comes from between the branch points at z = 0 and z = 1, i.e., along γ5 and γ7 .  dz e

1

√ − z 2 +z zt

e = −i2

dx sin

γ5 +γ7

#

x(1 − x)e−xt

(5.219)

0

√ This is an unusual integral. The form x(1 − x) would suggest a substitution of the form x = cos2 (θ/2); let’s see where that takes us. (Hint: this substitution wouldn’t be suggested if it would take us nowhere.) 1 dx sin

#

x(1 − x)e

−xt

1 = e−t/2 2

0

 1 sin θ e(t/2) cos θ d θ sin θ sin 2 

π 0

1 = e−t/2 Im 4

π

d θ sin θ ei 2 sin θ e(t/2) cos θ 1

(5.220)

−π

We may evaluate the integral on the right by expressing it as a complex integral and taking it from there … π d θ sin θ e

i 21 sin θ (t/2) cos θ

e

 = −i

−π

=−

1 2

|z|=1

dz z



|z|=1



z − z −1 i2

 e

(t/2)

z+z −1 2

!

e

(i/2)

z−z −1 i2

!

1 dz 1 −1 z − z −1 e 4 (t+1)z+ 4 (t−1)z (5.221) z

We know from previous experience that the contour integral is equal to i2π times the sum of the residues of the poles inside of the unit circle. In this case, the integrand has an essential singularity at z = 0. To compute the residue at z = 0, we expand the

192

5 Integral Transforms

exponential in a Laurent series and we need only determine the coefficient of z −1 in that expansion. In short, we wish to compute the coefficient of z 0 in the Laurent expansion of

−1 z − z −1 eaz+bz where a = 41 (t + 1) and b = 41 (t − 1). This expansion looks like

∞



k 1 −1 az + bz −1 z − z −1 eaz+bz = z − z −1 k! k=0 ∞ k  



k−m 1 k −1 (az)m bz −1 = z−z k! m=0 m k=0

= z − z −1

∞

k=0

bk

a !m 2 m−k 1 z (5.222) m!(k − m)! b m=0

k

To simplify things a bit, we note that only odd values of k will result in a nonzero coefficient of z 0 . We can then rewrite that last sum as a !m 2 m−2k−1 1 z m!(2k + 1 − m)! b m=0 k=0 (5.223) We have nonzero coefficients of z 0 when m = k or m = k + 1. Therefore, by the residue theorem, the integral above is equal to ∞ k

−1 z − z −1 eaz+bz = z − z −1 b2k+1

π −π

∞

(ab)k √ ! 1 1 b−a d θ sin θ ei 2 sin θ e(t/2) cos θ = − i2π(b − a) = −iπ √ I1 2 ab 2 k!(k + 1)! ab k=0

(5.224)

Plug in a = 41 (t + 1) and b = 41 (t − 1) and we find that 1 dx sin 0

#

! √ ! I1 21 t 2 − 1 π x(1 − x) e−tx = e−t/2 √ 2 t2 − 1

(5.225)

where I1 is the modified Bessel function of the first kind of order one. Putting this altogether, we finally have, for t > 1,

5.3 Two-Sided Laplace and Mellin Transforms

1 i2π

τ0+i∞

ds e

193

! √ 1 2 1 −t/2 I1 2 t − 1 + e−1/2 δ(t − 1) e = e √ 2 t2 − 1

√ − s2 +s st

τ0 −i∞

(5.226)

and 0 for t < 1. Alternatively, if we use a Heaviside function and note that limt→1 δ(t − 1) (t − 1) = 0, then we have 1 i2π

τ0+i∞

ds e

√ − s2 +s st

e = e−t/2

τ0 −i∞

I1

! √ t2 − 1

(t − 1) √ 2 t2 − 1 1 2

(5.227)

It is always nice to see work we have done being applied elsewhere. All said, we hope this was a solid introductory foray into Laplace transforms and their inverses. We go into slightly less familiar territory next as an investment into our integration quiver.

5.3 Two-Sided Laplace and Mellin Transforms We have gone through quite a lot of difficult Fourier and Laplace transforms, and the hope is that we have seen enough that we will know what to do when we encounter other difficult problems involving those transforms. But now we move to a slightly more esoteric transform, one which we will need for our complex integration schemes later. We begin in familiar territory though. Two-sided Laplace transforms Recall that the Laplace transform ∞

dt f (t)e−st

(5.228)

0

converges for all Re s > s0 when f has a leading exponential behavior of es0 t , i.e., f (t) = h(t)es0 t where h has no exponential factor. In the complex plane, this translates to a region of convergence that is a half-plane to the right of Re s > s0 . To compute the inverse Laplace transform, the line of integration at Re s = τ0 can be anywhere in that half-plane. The Laplace transform we have been considering up until now in Eq. (5.228) is also known as a one-sided Laplace transform which obeys a causality relation. There are also two-sided Laplace transforms defined as follows. ∞ −∞

dt f (t)e−st

(5.229)

194

5 Integral Transforms

In this case, f need not obey a causality relation. We can express the two-sided transform in terms of two one-sided transforms. ∞

∞

dt f (t)e−st =

−∞

dt f (t)e−st +

∞ dt f (−t)est

0

0

= F+ (s) + F− (s)

(5.230)

If the integral defining F+ is convergent when Re s > s0 , then analogously the integral defining F− is convergent when Re s < s1 for some s1 . In this case—and this is true in general—the two-sided Laplace transform is convergent when s lies in the strip s0 < Re s < s1 in the complex plane. As a simple example, define  f (t) = Then

∞ F+ (s) =

eat t > 0 ebt t < 0

dt eat e−st =

0

∞ F− (s) =

1 s−a 1 s−b

(5.232)

1 1 − s−a s−b

(5.233)

dt e−bt est = −

0

(5.231)

where a < Re s < b. Moreover, F(s) = F+ (s) + F− (s) =

The original f (t) is recovered by the inverse transform. 1 f (t) = i2π

τ0+i∞

ds τ0 −i∞



 1 1 − est s−a s−b

(5.234)

In this case, however, a < τ0 < b. Moreover, we close to the left when t > 0 and to the right when t < 0. In this way, we recover the original value of f (t).

5.3 Two-Sided Laplace and Mellin Transforms

195

Mellin Transforms In Eq. (5.229), we substitute t = − log x, g(x) = f (− log x) and we end up with ∞ G(s) =

dx g(x)xs−1

(5.235)

0

Equation (5.235) is a Mellin transform. The integral converges in the strip s0 < Re s < s1 in the complex plane. Its inverse is as follows. 1 g(x) = i2π

τ0+i∞

ds G(s)x−s

(5.236)

τ0 −i∞

The Mellin transform is a rescaled two-sided Laplace transform. Its use to us is not apparent yet, but let’s go through osme examples which will be very useful to us later. Let’s compute the Mellin transform of the following function.  g(x) =

log x x ∈ (0, 1) 0 x>1

(5.237)

1 G(s) =

dx xs−1 log x 0

∞ =−

dt te−st

0

=−

1 s2

(5.238)

This in and of itself is not a very exciting example because it is just a rescaling of a one-sided Laplace transform. Other examples, however, will better distinguish the

−1/2 Mellin transform. Let’s find the Mellin transform of h(x) = 1 + x2

196

5 Integral Transforms

∞ H (s) = 0

xs−1 dx √ 1 + x2

π/2 tans−1 θ = d θ sec2 θ sec θ 0

π/2 d θ sins−1 θ cos−s θ = 0

π/2 = d θ cos θ sins−1 θ cos−s−1 θ 0

π/2

−(s+1)/2 = d θ cos θ sins−1 θ 1 − sin2 θ 0

1 =



−(s+1)/2 du us−1 1 − u2

0

1 = 2

1

dv v−1/2 v(s−1)/2 (1 − v)−(s+1)/2

0

=

1 2

1

dv v 2 −1 (1 − v)− s

s−1 2 −1

(5.239)

0

This is a beta function, so we may conclude that  

1−s 1 s! 2 −1/2  =⇒ H (s) = √  h(x) = 1 + x 2 2 2 π

(5.240)

The presence of  functions should not surprise us as the Mellin transform definition lends itself to generating  functions, but the sheer amount of algebraic manipulation in this example is a bit surprising. It should also be suspicious that we will see this example again, as it is so specific. On that note, let’s do another: find the Mellin transform of h(x) = log (1 + x). ∞ dx xs−1 log (1 + x) 0

(5.241)

5.3 Two-Sided Laplace and Mellin Transforms

197

Fig. 5.11 Contour for computing Mellin transform in Eq. (5.241)

This one is a surprisingly tricky but elegant integral to evaluate in the complex plane. Of course we have the tools to do the evaluation but as we will see the contour integral will have two branch points and we must treat each separately. So we consider the contour integral  dz z s−1 log (1 + z)

(5.242)

C

Note here that in this case we do not need the extra factor of log because the factor z s−1 provides the multivaluedness we can exploit about z = 0. We also have to reckon with the branch point at z = −1 because of the log (1 + z) factor. Accordingly, the contour C is as in Fig. 5.11. It should be clear that the Mellin transform converges when Re s ∈ (−1, 0); this is what the analysis of γ2 , γ4 , γ6 , and γ8 in the limits of R → ∞ and R → 0 would tell us. Accordingly, we will only need to consider the integrals along γ1 , γ3 , γ5 , and γ7 in those limits. ∞

 dz z

s−1

log (1 + z) =

dx xs−1 log (1 + x)

(5.243)

  dx xs−1 log (x − 1) + iπ

(5.244)

γ1

0



1 dz z s−1 log (1 + z) = eiπ(s+1)

γ3

∞

 dz z γ5

s−1

log (1 + z) = e

−iπ(s+1)

∞ 1

  dx xs−1 log (x − 1) − iπ

(5.245)

198

5 Integral Transforms



0 dz z

s−1

log (1 + z) = e

dx xs−1 log (1 + x)

i2πs

γ7

(5.246)

∞

Let’s pause here and note that there is a branch cut for x > 0 for the term xs−1 , while there is a branch cut along x < −1 for the term log (x + 1). Moreover, it is important to note that we applied these branch cuts separately. For example, Arg z s−1 ∈ [0, 2π ), so that at the branch cut for the log (1 + z) term, Arg z s−1 = π . Moreover, Arg [log (1 + z)] ∈ (−π, π ], so that at the branch cut for z s−1 , Arg [log (1 + z)] = 0. We combine the above integrals and find that the contour integral is equal to the following.  dz z

s−1

log (1 + z) = 1 − ei2πs

C

∞

∞ dx x

s−1

log (1 + x) + i2π e

iπs

0

dx xs−1 1

(5.247) As is usually the case, the integral that we end up evaluating is far simpler than the integral we seek. We apply Cauchy’s Theorem and after sifting through the algebra, the result is as follows. ∞ dx xs−1 log (1 + x) = 0

π s sin π s

(5.248)

OK, one more Mellin transform: h(x) = log2 (1 + x). For this, we may use the same contour we used previously in Fig. 5.11. Again, the Mellin transform converges when Re s ∈ (−2, 0) and we need only deal with the contributions of the integrals along γ1 , γ3 , γ5 , and γ7 in the limits R → ∞ and R → 0. ∞

 dz z

s−1

log (1 + z) =

dx xs−1 log2 (1 + x)

(5.249)

 2 dx xs−1 log (x − 1) + iπ

(5.250)

2

γ1

0



1 dz z

s−1

log (1 + z) = e 2

γ3

∞

 dz z γ5

iπ(s+1)

s−1

log (1 + z) = e 2

−iπ(s+1)

∞ 1

 2 dx xs−1 log (x − 1) − iπ

(5.251)

5.3 Two-Sided Laplace and Mellin Transforms

199



0 dz z

s−1

log (1 + z) = e 2

dx xs−1 log2 (1 + x)

i2πs

γ7

(5.252)

∞

This time, we will have to deal with a log term in the integrand in the integral we evaluate: 

∞ dz z s−1 log2 (1 + z) = i4π eiπs

γ3 +γ5

dx xs−1 log (x − 1) 1

1 = i4π e

iπs

  dx x−(1+s) log (1 − x) − log x (5.253)

0

Now, 1 dx x

−(1+s)

log (1 − x) = −

 ∞ 1 k=1

0

k

1

dx xk−1−s

0

∞

1 = k(s − k) k=1  ∞  1 1 1 = − s k k −s k=1

1 = (ψ(1 − s) + γ ) s where ψ(z) =

1 d (z) (z) dz

(5.254)

(5.255)

is the digamma function and γ is the Euler-Mascheroni constant. Moreover, 1

dx x−(1+s) log x = −

1 s2

(5.256)

0

This term is canceled when we apply the digamma identity ψ(1 − s) = ψ(−s) −

1 s

Putting this altogether and applying Cauchy’s Theorem, we get

(5.257)

200

5 Integral Transforms

∞ dx xs−1 log2 (1 + x) = − 0

2π (ψ(−s) + γ ) s sin π s

(5.258)

Finally, let’s compute an inverse Mellin transform to see what that looks like: let’s reverse the Mellin transform of log (1 + x) and compute 1 i2π

τ0+i∞

ds τ0 −i∞

π x−s s sin π s

(5.259)

To compute this integral, we set τ0 ∈ (−1, 0) and close the contour to the right. The integral is then equal to the negative sum of the residues of the poles at z = 0, 1, 2, . . .. The residue at z = 0 is π x−z = log x (5.260) − Res z=0 z sin π z The sum of the rest of the residues puts us in familiar territory. −

∞ k=1

∞

Res z=k

(−1)k 1 π x−z = − z sin π z k xk k=1   1 = log 1 + x

(5.261)

Then we have 1 i2π

τ0+i∞

ds τ0 −i∞

  π 1 x−s = log x + log 1 + = log (1 + x) s sin π s x

(5.262)

as expected. Convolution, or San Andreas’s Faltung Like the Fourier and Laplace transforms, the Mellin transform satisfies a convolution relation. Before we determine what that is, let’s derive an expression for the delta function. 1 g(x) = i2π

τ0+i∞

τ0 −i∞

1 ds x G(s) = i2π −s

τ0+i∞

ds x τ0 −i∞

∞ =

dy g(y) 0

−s

∞ dy ys−1 g(y) 0

1 i2π

τ0+i∞

ds ys−1 x−s

τ0 −i∞

(5.263)

5.3 Two-Sided Laplace and Mellin Transforms

201

So we have our delta function: 1 i2π

τ0+i∞

ds ys−1 x−s = δ(x − y)

(5.264)

τ0 −i∞

We now consider the convolution of two Mellin transforms F and H . ⎡∞ ⎤ τ0+i∞ τ0+i∞  1 1  ds F(s )H (s − s ) = ds ⎣ dx f (x )xs −1 ⎦ i2π i2π 0 τ0 −i∞ τ0 −i∞ ⎤ ⎡∞   ⎣ dx h(x )xs−s −1 ⎦ 0

∞ =





∞

dx f (x ) 0

0

∞

0

dx f (x )



dx h(x )x

∞ =



s−1

1 i2π

τ0+i∞



ds xs −1 x−s

τ0 −i∞

dx h(x )xs−1 δ(x − x )

0

∞ =

dx f (x )h(x )xs−1

(5.265)

0

Accordingly, we have our Mellin convolution Theorem: 1 i2π

τ0+i∞







∞

ds F(s )H (s − s ) = τ0 −i∞

dx f (x)h(x)xs−1

(5.266)

0

Because of the structure of the Mellin transform, by plugging in s = 1, we are able to get something like a Mellin Parseval equality. 1 i2π

τ0+i∞

∞

ds F(s)H (1 − s) = τ0 −i∞

dx f (x)h(x) 0

This will come in handy for us in the next chapter.

(5.267)



202

5 Integral Transforms

Problems 5.1 Find the Maclurin series expansions for the following functions, using the convolution series. (a) f (z) = log2 (1 − z) (b) f (z) =

log (1 − z) 1−z

5.2 Find the Fourier transforms of the following functions. (a)

f (x) = xn e−x

(b) f (x) = (c)

2

/2

x − sin x x3

−(1+ν)/2 f (x) = 1 + x2

(d) f (x) =

1 2 − 2x − x2

5.3 (a) Find the Fourier transforms of  fn (x) =

sin x x

n

for n ∈ {3, 4, 5, 6, 7, 8}. Express as an explicit even function in each case. (b) Determine exactly  ∞  sin n 8 S8 = n n=−∞ 5.4 Verify Eq. (5.39). 5.5 Determine the convolution of f and g for the following, both directly and using the Convolution Theorem. (a) f (x) =

a 1 1 2 2 g(x) = √ e−x /b π a2 + x2 b π

5.3 Two-Sided Laplace and Mellin Transforms

(b)

203

⎧ ⎪ ⎨T + t t ∈ [−T , 0] f (t) = T − t t ∈ [0, T ] ⎪ ⎩ 0 t∈ / [−T , T ]

⎧ ⎪ ⎨T + t t ∈ [−T , 0] g(t) = T − t t ∈ [0, T ] ⎪ ⎩ 0 t∈ / [−T , T ]

5.6 (a) Derive Eq. (5.87) directly via the Residue Theorem. (b) Derive Eq. (5.87) by finding the Fourier transform of f (x) =

sin (x + iy) x + iy

and then using the Parseval equality. 5.7 In computing the error function of complex argument, we considered the path  that runs from 0 to x along the real axis, then from x to z = x + iy parallel to the imaginary axis. This time, perform a similar computation of the error function using another path   that runs from 0 to iy along the imaginary axis, then from iy to x + iy parallel to the imaginary axis. 5.8 Solve the following ODE using Laplace transforms. x¨ + 2a˙x + a2 x = sin ωt x˙ (0) = x(0) = 0 5.9 Solve the heat equation as described in the text, but with α(t) = cos ωt in Eq. (5.158). This is a difficult but rewarding computation. You are better off working in the Laplace transform domain than using the direct convolution result. 5.10 Solve the following heat equation in a radially symmetric geometry. ∂u ∂ 2 u 1 ∂u = + 2 ∂r r ∂r ∂t u(1, t) = 1 lim u(r, t) = 0

r→∞

u(r, 0) = 0 r > 1 The answer should be qualitatively similar to the erfc solution we derived in the text.

204

5 Integral Transforms

5.11 Evaluate the following integrals. (a)

1

dt [t(1 − t)]−1/2 e−a /t e−b 2

2

/(1−t)

0

(b)

1

dt t −3/2 (1 − t)−1/2 e−a /t e−b 2

2

/(1−t)

0

(c)

1 dt log t log (1 − t) 0 √

5.12 Derive the inverse Laplace transform of s−3/2 e− to the stretched exponential we derived.

s

directly without reference

5.13 Compute the inverses of the following Laplace transforms. (a) For b ∈ R (b) For b ∈ R

√

F(s) = e−

s(s+b)

√

F(s) = s−3/2 e−

s(s+b)

5.14 Verify the Mellin transform result in Eq. (5.258) using the Mellin Convolution Theorem. 5.15 Show that, for f , g, h ∈ L2 (−∞, ∞) ∞ −∞

1 dx f (x)g(x)h(x) = 4π 2

∞

∞ dk F(k)

−∞

−∞

dk  G(k  )H (k + k  )

(5.268)

Chapter 6

Asymptotic Methods

Abstract In this final chapter, we will review methods for analyzing the behavior of functions defined by integrals in various neighborhoods, e.g., infinity. We will review some asymptotic expansions and provide their meaning through log-log plots. Methods of generating asymptotic expansions including integration by parts, Watson’s Lemma, and Laplace’s Method will be reviewed. This is all preparation for the presentation of Bleistein and Handelsman’s h-transform, which generalizes Laplace’s Method for any nonmonotonic kernel and makes good use of the Mellin transforms we discussed in Chap. 5. We will apply this h-transform in the service of one of the key topics in this book: the evaluation of difficult integrals using complex integration. Finally, we will present the Euler-Maclurin sum formula which provides an asymptotic relation between the integral and sum of a function. This will provide new perspectives on some of the results we derived in Chap. 3. More importantly, we will see that judicious use of these approximate methods are needed occasionally in derivations of exact results of integrals and sums.

6.1 Asymptotic Expansions We have been using asymptotic expansions throughout this text, so what we will say here is more or less a review and possibly a slight formalization. There are far more comprehensive treatments in the literature and some of them will be listed in the bibliography. Our interest in the subject here is to provide the language we will need to study the h-transform and in so doing, we will be able to attack real integrals using complex integration in an unexpected way. Expansion and the log-log plot Rather than go into the formalism, let’s take a simple example. One of the basic and most magical relations we are taught in Calculus is the following limit.   1 x 1+ =e x→∞ x lim

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 R. Gordon, Complex Integration, Undergraduate Lecture Notes in Physics, https://doi.org/10.1007/978-3-031-24228-1_6

(6.1)

205

206

6 Asymptotic Methods

Recall the definition ofa limit:   for x > x0 , for all  > 0, there exists an x0 such x that when x > x0 ,  1 + x1 − e < . In this sense, x0 becomes a function of . For example, how big must x0 be so that the difference  is 10−3 ? 10−7 ? 10−16 ?. We answer the question by forming an asymptotic expansion of the function of which we are taking a limit.   x For example, for sufficiently large x, what is a good estimate of  1 + x1 − e?  x We can expand 1 + x1 by expressing it as an exponential.  1+

1 x

x

= e x log (1+ x ) 1

(6.2)

We can Taylor expand the log piece.   ∞  1 (−1)k x log 1 + =1+ x (k + 1)x k k=1

(6.3)

  1 1 1 x 1 1+ = ee− 2x e 3x 2 e− 4x 3 . . . x

(6.4)

Accordingly, we have

We may now Taylor expand each of the exponentials and group like terms.      1 x 1 1 1 1 1 1 1+ + = e 1− + . . . 1 + + + . . . ... x 2x 2! (2x)2 3x 2 2! (3x 2 )2   11 1 + − . . . (6.5) = e 1− 2x 24x 2 The ellipses may be replaced with a more precise notation. All along, we have been using the ellipses to denote higher orders that we are not considering and have no impact on our analysis. In Eq. (6.5), we may instead use something called the big O: 

1 1+ x

x

11e e + =e− +O 2x 24x 2



1 x3

 (6.6)

We can be even a little less precise if we want:     1 x 1 e 1+ +O =e− x 2x x2

(6.7)

We write this even though we know what the next term is. We may now answer the question, how does the difference between the function and the limit behave?

6.1 Asymptotic Expansions

207

Fig. 6.1 Log-log plot of difference between function and limit in Eq. (6.8)

  x      1 + 1 − e = e + O 1   2x x x2

(6.8)

The result is represented in the plot in Fig. 6.1. The scale is log-log so we can see the behavior of the error term. The abscissa is log10 (x) and the ordinate is the log base-10 of the difference between function and limit. The solid curve is the exact difference on the LHS of Eq. (6.8), while the dashed line is the asymptotic approximation on the RHS of Eq. (6.8). Note that the slope of the curves for sufficiently large x is -1 as predicted by the expansion we did leading up to Eq. (6.8), which implies a behavior of x −1 in this limit. The log-log plot is a crucial tool of asymptotic analysis in that it provides no-doubt verification that we got the behavior we seek correct. Note that we have done enough analysis to also answer the question, how does the difference between the above difference and the error term behave? That is, did we get the following correct?   x      1 + 1 − e + e  = 11e + O 1  x 2x  24x 2 x3

(6.9)

This is verified in the log-log plot in Fig. 6.2, where the slope of the line is now −2. We can keep repeating this ad infinitum, one term at a time, and in doing so we will generate an asymptotic series of terms such that      N −1  1 x  an  a N 1  − +O  1+ =  x xn  x N x N +1 n=0

(6.10)

It turns out that the series on the LHS converges. This is not always the case for asymptotic series. For example, consider the behavior of following integral as x → ∞.

208

6 Asymptotic Methods

Fig. 6.2 Log-log plot of difference between difference and error term in Eq. (6.9)

∞ f (x) =

dt 0

e−t x +t

(6.11)

We may integrate by parts to get the following series representation of f : f (x) =

∞  n! (−1)n n+1 x n=0

(6.12)

Note that the Stirling approximation states that n! lim √ =1 2π nn n e−n

n→∞

(6.13)

so that the radius of convergence of the above series, the Steltjes series, is zero. How on earth could the Steltjes series possibly serve as an asymptotic series for the integral in Eq. (6.11)? Despite there being a divergent series, the sum serves as an asymptotic approximation all the same. This is shown in Fig. 6.3, where the following is illustrated.  ∞     −t  1 1 2 6  24 1  dt e = − + − + + O  2 3 4 5 6 x + t x x x x x x  

(6.14)

0

Watson’s Lemma We have assumed that asymptotic series take a particular form, such as   N −1  an 1 f (x) = +O n N x x n=0

(6.15)

6.1 Asymptotic Expansions

209

Fig. 6.3 Log-log plot of difference between integral in Eq. (6.11) to its asymptotic approximation for n = 3

However, we will find that even in the most familiar of situations, asymptotic series can take on other forms, such as those with branch points. Consider the problem of √ finding the Laplace transform of f (t) = sin t. ∞ dt e

−st

sin

√

0

∞ t =2

du u e−su sin u 2

(6.16)

0

The integral may be evaluated via Cauchy’s theorem by considering the contour integral 2 dz z e−sz C

where C is the rectangle having vertices ±R ± i/(2s), and the integral traverses C in a positive sense. Because the integral is zero, we may then show that, when we take the limit R → ∞, we get ∞ 2

du u e−su

2

⎡ ∞ ⎤  1 1/(4 s) i 2 sin u = e Re ⎣ d x e−s (x− 2s ) ⎦ s

(6.17)

−∞

0

which may be evaluated via Cauchy’s theorem, again, this time on the integral 2 dz e−sz = 0 C

where C  is the rectangle having vertices ±R and ±R − i/(2s). By taking the limit as R → ∞, we find that

210

6 Asymptotic Methods

⎡ Re ⎣

⎤

∞ dx e

−s (

x− 2is

)⎦= 2

−∞

Accordingly,

∞



∞ dx e

−sx 2

=

−∞

dt e−st sin

√

t=

π s

1 √ −3/2 −1/(4 s) πs e 2

(6.18)

(6.19)

0

We can express this in terms of a convergent asymptotic series as s → ∞ by Taylor expanding the exponential on the RHS: ∞

dt e−st sin

√

t=

0

∞ 1 √ −3/2  (−1)n 1 πs 2 22n n! s n n=0

(6.20)

That is, the nth term in the asymptotic series in Eq. (6.20) is s −(n+3/2 rather than s −n . Interestingly, it turns out that we can derive the asymptotic series in Eq. (6.20) by Taylor expanding the sine term in the integrand of the Laplace transform. ∞ dt e 0

−st

sin

√

∞ t= 0

∞ √  (−1)n t n dt e−st t (2n + 1)! n=0

 ∞  (−1)n = dt e−st t n+1/2 (2n + 1)! n=0 ∞

0

∞ 1 (−1)n = du e−u u n+1/2 n+3/2 (2n + 1)! s n=0 0   ∞ 3 n  (−1)  n + 2 1 = n+3/2 (2n + 1)! s n=0 ∞ 

(6.21)

  To evaluate  n + 23 , we use the Legendre duplication formula:   1 1 (2z) = √ 22z−1 (z) z + 2 π

(6.22)

We reproduce Eq. (6.20) by plugging in z = n + 1 into the duplication formula. What’s interesting about the expansion of the sine is that a Taylor expansion valid in a neighborhood [0, ] near t = 0 completely predicted the behavior of the integral when s is sufficiently large. The reason this works for Laplace transform integrals is that the exponential decreases in that neighborhood faster than any power of s when s is sufficiently large. It is this reasoning on which Watson’s Lemma is based.

6.1 Asymptotic Expansions

211

Theorem 1 Assume the function f (t) has the following asymptotic behavior in a neighborhood [0, ] near t = 0: f (t) = t

α

N −1 

  an t βn + O t α+β N

n=0

where α, β ∈ R. Then ∞

dt f (t)e−st =

N −1  n=0

0

an

 (α + βn + 1) +O s α+βn+1



1



s α+β N +1

When the asymptotic series defining f is convergent, then we may take N → ∞. In fact, we may take N → ∞ for a divergent series if we understand it as a shorthand. That is, we may write in Watson’s Lemma ∞

dt f (t)e−st =

∞  n=0

0

an

 (α + βn + 1) s α+βn+1

(6.23)

√ For f (t) = sin t, an = (−1)n /(2n + 1)!, α = 1/2, and β = 1. As another example, let’s look at a result we derived earlier. 1 d x sin 0



 √  1 2−1  I t 1 2 π x(1 − x) e−t x = e−t/2 √ 2 t2 − 1

(6.24)

For the LHS, we consider the behavior in a small neighborhood [0, ] near x = 0. We may Taylor expand the sine term in this neighborhood:   1 1 1 x(1 − x) = (1 − x)1/2 − x(1 − x)3/2 + x 2 (1 − x)5/2 + O(x 3 ) √ sin 3! 5! x 2 2 2 (6.25) = 1 − x + x + O(x 3 ) 3 15 We then note that we may move the upper limit of integration back to x = , and then out to ∞, with exponentially small errors. Accordingly, the asymptotic behavior of the integral on the LHS of Eq. (6.24) is, by Watson’s lemma, 1 d x sin 0



          3 2  25 2  27 1 x(1 − x) e−t x = 3/22 − + + O t 3 t 5/2 15 t 7/2 t 9/2    1 1 1 √ −3/2 1 1− + 2 +O 3 (6.26) πt = 2 t 2t t

212

6 Asymptotic Methods

Fig. 6.4 Log-log plot of difference between the RHS of Eq. (6.24) to its asymptotic approximation in Eq. (6.26)

We then would like to show that this expansion is equal to an expansion of the   RHS of Eq. (6.24) as t → ∞. This is messy to put it mildly. To do this out to O t −7/2 , we need to do careful bookkeeping. We also note that, as z → ∞, ez I1 (z) = √ 2π z

   1 15 3 +O 3 − 1− 8z 128z 2 z

(6.27)

√ where in this case z = 21 t 2 − 1. Rest assured that if the expansion is done consistently, we indeed reproduce the result on the RHS of Eq. (6.26); this is left as an exercise. This is verified numerically in Fig. 6.4, where the slope of the dashed line (and hence the error curve) is indeed − 9/2. Laplace’s Method How does one derive the expression in Eq. (6.27), that we used to verify the asymptotics in Eq. (6.26)? Well, that modified Bessel function may be defined using an integral representation. 1 I1 (z) = 2π

π

dθ cos θ e z cos θ

(6.28)

−π

Clearly, this is not a Laplace integral so at first it is not clear what this integral representation gets us. But we are interested in z being arbitrarily large, and it is interesting to see what happens to the exponential term as z increases. This is shown in Fig. 6.5. Note that for sufficiently large z, the exponential is concentrated in a small interval [−, ] about θ = 0. In Laplace’s method, we obtain the leading behavior of the integral in Eq. (6.28) at the maximum of the exponential term, i.e., at θ = 0. For the cos θ term outside of the exponential—the “amplitude” term—we replace that term with its value at the maximum, i.e., θ = 0. For the exponential term, however, we replace the cos θ

6.1 Asymptotic Expansions

213

Fig. 6.5 Plots of exponential in integrand of integral in Eq. (6.28) for z ∈ {1, 10, 100}

  argument with its quadratic Taylor expansion, i.e., cos θ = cos 0 − 21 θ 2 + O θ 4 . Accordingly, the leading behavior of the integral is  I1 (z) = e

z

   1 2 4 dθ 1 + O θ 2 e− 2 zθ +O (zθ )

(6.29)

−

Here we have defined  such that z 2 ∼ 1. As with the Laplace integral in Watson’s Lemma, we may take  → ∞ with exponentially small error. Accordingly, we can then state that, as z → ∞,    1 ez (6.30) 1+O I1 (z) = √ z 2π z This is not as precise as the result we used in Eq. (6.27). But we can keep going and get additional terms. For example, we can include the next order in each the exponent and the amplitude terms, as follows: ez I1 (z) = 2π

 −

   4  − 1 zθ 2 + 1 zθ 4 +O (zθ 6 ) 1 2 24 e 2 dθ 1 − θ + O θ 2

The second term in the exponent is about to be Taylor expanded: ez I1 (z) = 2π



 dθ −

1 z 4 24

(6.31)

  = O  2 , which is sufficiently small

    4  − 1 zθ 2  6 1 4 1 2 2 1 + zθ + O zθ e (6.32) 1− θ +O θ 2 24

214

6 Asymptotic Methods

We then take  → ∞. Because ∞

√ dθ θ 2 e− 2 zθ = 1

−∞

and

∞

2 − 21 zθ 2

dθ θ e −∞

2

2π z 3/2

√ 3 2π = 5/2 z

(6.33)

(6.34)

we recover the next term in the expansion: ez I1 (z) = √ 2π z

   3 1 1− +O 2 8z z

(6.35)

We can keep adding terms by expanding each of the amplitude and the exponent terms into additional terms. The next term for the Bessel is left as an exercise for the reader. The next example illustrates how Laplace’s method may be used in integrals that are not quite in the form of an explicit exponential. For large n, find the asymptotic behavior of the following sequence of integrals. π/4 d x tann x In =

(6.36)

0

Note that, as constituted, the maximum of the integrand is at x = π/4. We can shift this to x = 0 for convenience. We also can out the integrand in the form of an exponential as follows. π/4 π d x en log tan ( 4 −x ) In = 0

π/4 1−tan x d x en log ( 1+tan x ) =

(6.37)

0

Now,

 log

1 − tan x 1 + tan x



  4 = −2x − x 3 + O x 4 3

(6.38)

6.1 Asymptotic Expansions

215

Fig. 6.6 Log-log plot of difference between integral in Eq. (6.36) and asymptotic result in Eq. (6.40)

so that with exponentially small error, for sufficiently small , i.e., n ∼ 1, we have  In =

    4 d x e−2nx 1 − nx 3 + O x 4 3

(6.39)

0

Note that n 3 ∼  2 , so we were able to Taylor expand the exponential of that argument. Again, with exponentially small error, we may set  → ∞, and we find that upon evaluating the integrals,   ∞ 4 1 − n d x x 3 e−2nx + O In = d x e 3 n4 0 0   1 1 1 − 3 +O = 2n 2n n4 ∞

−2nx

(6.40)

This result is verified in Fig. 6.6; note that the error has slope −5. For a final example, let’s look at what appears to be a doozy but is pretty par for the course here.   ∞ erf t M (6.41) J (M) = dt t sin (st) t 0

and ask for the behavior of the integral in the limit as M → ∞. Here we will limit ourselves to the leading behavior and leave the next order to the exercises. We rewrite the integral as ∞ erf t (6.42) J (M) = dt t sin (st)e M log ( t ) 0

216

6 Asymptotic Methods

Fig. 6.7 Log-log plot of difference between integral in Eq. (6.41) and asymptotic result in Eq. (6.44)

To analyze this integral using Laplace’s method, we Taylor expand to second order the exponential term about the maximum at t = 0 and to first order the amplitude term. The leading behavior of the integral is then  J (M) = s

2 √ π

 M 

dt t 2 e− 3 Mt 1

2

+O(t 4 )

(6.43)

0

We may then take  → ∞ with exponentially small error and get the following result. 1 J (M) = s 2



2 √ π

 M−1 

M 3

−3/2



+O M

−5/2

 

(6.44)

 −(M−1) The log-log plot of 2J (M) √2π for s = 1 is shown in Fig. 6.7, which demonstrates a slope of −5/2 of the error as expected. Laplace’s method is a fairly straightforward way to obtain the leading asymptotic behavior of an integral, and it is not much harder in principle to get subsequent orders of the behavior although in practice sometimes the bookkeeping can get a little hairy. Laplace’s method also covers a wide range of integrals exhibiting exponential behavior in the asymptotic parameter. Nevertheless, there are a lot of integrals, including many with which we have an interest, that do not exhibit such behavior in the asymptotic parameter. Fortunately, we have a way to deal with those, and that way is chock full of delightful surprises.

6.2 The Monotonic h-Transform and a New Complex Integration Technique

217

6.2 The Monotonic h-Transform and a New Complex Integration Technique An example of an integral with a monotonic, non-exponential behavior is a variant of the Steltjies integral we studied recently. Consider the following integral. ∞ f (x) =

dt 0

e−t 1 + xt

(6.45)

We are interested in the behavior of the integral f (x) as x → ∞. Laplace’s method is not very useful here, as the “kernel” 1/(1 + xt) doesn’t produce exponentially small errors. There needs to be a completely different mindset to attack this. Fortunately, there is a result that will be of enormous value to us. This result, demonstrated by Bleistein and Handelsman, relies on the Mellin transforms we studied in the previous chapter and the results are simple to apply. Let’s dive into the general formulation they provide and then analyze the Steltjies integral based on the formulation. The Monotonic h-transform1 We consider the following integral as an h-transform. ∞ dt f (t)h(λt)

H [ f ; λ] =

(6.46)

0

where the function f and kernel h have respective Mellin transforms F and H . We may then apply the Mellin Parseval equality: ∞

1 dt f (t)h(t) = i2π

0

τ0+i∞

ds F(1 − s)H (s)

(6.47)

τ0 −i∞

This is not the h-transform; we need to rescale the argument of the kernel h. It is then straightforward to show that the h-transform satisfies ∞ dt f (t)h(λt) = 0

1

1 i2π

τ0+i∞

ds λ−s F(1 − s)H (s)

(6.48)

τ0 −i∞

The discussion in this subsection relies on the work of N. Bleistein and R. Handelsman, “Asymptotic Expansions of Integrals,” Sect. 4.4, New York: Dover Publications, Inc. (1986). The author is indebted to Dover Books, Inc. for permission to use this material herein.

218

6 Asymptotic Methods

Equation (6.48) defines our approach to obtaining the asymptotic behavior of the h-transform as λ → ∞. For this to make any sense, we need to assume that F is analytic in the strip Re z ∈ (α, β) and that H is analytic in the strip Re z ∈ (γ , δ). Moreover, the line of integration at Re z = τ0 lies in the strip common to both H (z) and F(1 − z); that is, it is assumed that these strips have an overlap strip. The following is not a rigorous derivation of the asymptotic results we will be using. For a more rigorous analysis, we refer to the Bleistein and Handelsman reference. Let G(z) = F(1 − z)H (z), and suppose G is defined in the right half-plane Re z > τ0 as a meromorphic function (i.e., analytic with a finite number of isolated poles). Also, we assume that the behavior of G is such that we may displace the line of integration to the right to Re z = R. We may then form a rectangular contour of integration between Re z = τ0 and Re z = R. By the Residue Theorem, we may then conclude that ∞

1 dt f (t)h(λt) = − Res λ F(1 − z)H (z) + τ0 0. Now, note that we need only consider, for the leading asymptotic behavior of these integrals, the pole at s = 0 and s = 1. (Poles at s = 2 and beyond will have subdominant behavior—note that it is not trivial that we needed the residue at s = 1.) Thus, we need to consider three pairs of residues as follows.   π R −s = log R − 1 (6.84) − Res s=0 s(1 − s) sin π s

228

6 Asymptotic Methods

 − Res s=1

π R −s s(1 − s) sin π s

=

1 (log R + 1) R

(6.85)

  2π R −s (γ + ψ(−s)) = log2 R − 2 log R + 2 − Res − s=0 s(1 − s) sin π s

(6.86)

   2π R −s (γ + ψ(−s)) 1  2 = log R − 2 − Res − s=1 s(1 − s) sin π s R

(6.87)

 − Res − s=0

 Res − s=1



π R −s s(1 − s)2 sin π s

π R −s s(1 − s)2 sin π s 

1 =− R



 = − log R + 2

(6.88)

 1 π2 2 log R + log R + +1 2 6

(6.89)

Now we plug these residues back into Eqs. (6.82) and (6.83) to get the following. 1 du log(1 + Ru)

i4π R log R 0

1 + i4π R

du log u log(1 + Ru) 0

1 + i2π R

du log2 (1 + Ru) − i2π 3 R 0

  1 = i4π R log R log R − 1 + (log R + 1) R    1 1 π2 log2 R + log R + +1 + i4π R − log R + 2 − R 2 6     1 log2 R − 2 + i2π R log2 R − 2 log R + 2 + R   log R − i2π 3 R + O R   2 = i6π R log R − i12π R log R − i2π π 2 − 6 R + i4π log2 R     π2 log R − i4π 2 + +O 6 R

(6.90)

Note that the log R/R comes from the fact that the poles at s = 2 and beyond are simple and accordingly produce no additional log terms in R. Recall also that these relations are valid in the limit as R → ∞. As expected, the approximate terms we

6.2 The Monotonic h-Transform and a New Complex Integration Technique

229

are keeping diverge in this limit. This is no problem, because we have more terms to evaluate that will cancel the diverging terms and leave us with the result we seek. We now need to evaluate the other integral in this limit as R → ∞. π

    dθ eiθ log Reiθ log2 Reiθ − 1

iR −π

π = iR

   dθ e log3 Reiθ − iθ

−π π

= iR

    iθ  2 log R 2 +O log Re Reiθ R

  dθ eiθ log3 R + i3θ log2 R − 3θ 2 log R − iθ 3

−π

π − i2

  dθ log2 R + i2θ log R − θ 2 + O

−π



log R R

(6.91)



The log3 R term in the first integral and the log R term in the second integral vanish. For the other integrals, we need the following results which are easily derived using, e.g., integration by parts. π dθ θ eiθ = i2π

(6.92)

dθ θ 2 eiθ = −4π

(6.93)

  dθ θ 3 eiθ = i2π π 2 − 6

(6.94)

−π

π −π

π −π

With these results, we may then write down the asymptotic expansion of this integral as R → ∞.

230

6 Asymptotic Methods

π iR

    dθ eiθ log Reiθ log2 Reiθ − 1

−π

= (i R)(i3)(i2π ) log2 R + (i R)(−3)(−4π ) log R + (i R)(−i)(i2π )(π 2 − 6)    3 log R 2π 2 − i2(2π ) log R + 0 + i2 +O 3 R  2  2 = −i6π R log R + i12π R log R + i2π π − 6 R   log R 4π 3 2 − i4π log R + i +O 3 R

(6.95)

And we are finally ready to write down the expression for the contour integral in the limit as R → ∞. Note that all terms dependent on R cancel. We should expect that this happens and the requirement that these terms cancel provides a self-check on the solution. The result for the contour integral is as follows.

1 dz log z log (z − 1) = i4π 2

  π2 (6.96) d x log x log (1 − x) − i4π 2 − 6

0

C

And now we invoke Cauchy’s Theorem: because there are no poles within C, the contour integral is equal to zero. Accordingly, the value of the integral is as follows. 1 d x log x log (1 − x) = 2 −

π2 6

(6.97)

0

Again, while the integral above has been evaluated in the past using real methods, the evaluation of the integral using complex integration is new. This evaluation with complex integration has been enabled using the asymptotic techniques discussed in this section. There will be other examples in the exercises.

6.3 Euler-Maclurin Summation So far we have studied methods of analyzing the asymptotic behavior of some integrals (e.g., those with monotonic kernels). Frequently, we also want to analyze the asymptotic behavior of sums, i.e., with regard to summation limit. For example, we may wish to study how quickly some sums diverge. In some cases, we may use the asymptotic behavior of a sum to evaluate the exact value of another sum.

6.3 Euler-Maclurin Summation

231

Euler-Maclurin summation is a well-known technique that generates an asymptotic series that represents the difference between a sum and an integral. We begin with the following observation for a differentiable function f .  n+1  n+1 1 1  f (x) = [ f (n) + f (n + 1)] − dx x − n − d x f (x) 2 2 n

(6.98)

n

We may then sum both sides from (for example) n = M to N − 1 and rearrange to get the following. N  n=M

N f (n) −

1 d x f (x) = [ f (M) + f (N )] + 2

M



N dx

x − [x] −

1 2



f  (x)

M

(6.99) where [x] is the greatest integer less than n. If f is further differentiable p times, a clever integration by parts scheme produces the following asymptotic series. N 

N f (n) −

n=M

d x f (x) =

1 [ f (M) + f (N )] 2

M

p   B2k  (2k−1) f (N ) − f (2k−1) (M) + R p + (2k)! k=1

(6.100)

where B2k is the (2k)th Bernoulli number and R p is a remainder term. This is the Euler-Maclurin sum formula. Alternatively, we may write N

N 

f (n) −

n=M+1

d x f (x) =

1 [ f (N ) − f (M)] 2

M

p   B2k  (2k−1) f (N ) − f (2k−1) (M) + R p + (2k)! k=1

(6.101)

As a first example, consider a sum we have already encountered in Eq. (3.198), whose exact result we had derived. ∞  n=1

n2

1 1 π coth (πa) − 2 = 2 +a 2a 2a

(6.102)

It was a particularly simple matter to derive the asymptotic behavior of the sum from the exact result. How can we derive it without knowing the exact result? Let’s try Euler-Maclurin: for f (x) = 1/(x 2 + a 2 ), where a → ∞, we have from Eq. (6.100),

232

6 Asymptotic Methods ∞  n=1

1 = n2 + a2

∞ 1

dx 1 + 2 + R1 x 2 + a2 2a

(6.103)

where R1 is the remainder term and is subdominant to the other terms. In this case, we can see that f  (x) = −2x/(x 2 + a 2 )2 so that  R1 = O

1 a4

 (6.104)

Accordingly, seeing that ∞ 1

   x ∞ 1 1 dx 1 π arctan − arctan = = 2 2 x +a a a 1 2a a a

(6.105)

then as a → ∞, we get the following asymptotic behavior, ∞  n=1

1 1 π − 2 +O = 2 2 n +a 2a 2a



1 a4

 (6.106)

which is exactly what we found by expanding the exact result. Recall also we derived the exact result for the following alternating sum in Eq. (3.199). ∞  (−1)n+1 1 π csch (πa) (6.107) = 2− 2 2 n +a 2a 2a n=1 We also wish to get the asymptotic behavior of this sum using Euler-Maclurin. Alternating sums are tricky using Euler-Maclurin. One thing we could do is introduce a factor of cos (π x) into the integral, but in this case there is an even better way which maybe we have seen when evaluating special values of the Riemann zeta function: subtract out twice the even terms and get two non-alternating sums. This works here as follows. ∞  (−1)n+1 n=1

n2 + a2

=

∞  n=1

=

∞  n=1

=

∞  n=1

  1 1 1 −2 2 + 2 + ... n2 + a2 2 + a2 4 + a2 ∞  1 1 − 2 2 2 2 n +a 4n + a 2 n=1 ∞

1 1 1 − n2 + a2 2 n=1 n 2 + (a/2)2

(6.108)

6.3 Euler-Maclurin Summation

233

We get two versions of the non-alternating series. To find the asymptotic behavior as a → ∞, we just plug in what we found in Eq. (6.106) and we find that ∞  (−1)n+1 n=1

n2 + a2

=

1 +O 2a 2



1 a4

 (6.109)

where, again, we found previously using the exact expression. So we have been able to determine the asymptotic behavior without the exact expression. Factorial products An interesting problem that may be treated using Euler-Maclurin is the following: evaluate the following product for large n. Pn = 0!1!2! . . . (n − 1)!n!

(6.110)

We can turn this into the evaluation of a sum by taking logs of both sides. log Pn =

n  k=0

log k! =

n+1 

log (k)

(6.111)

k=1

We evaluate this asymptotically for large n using Euler-Maclurin. n+1  1 log (n + 1) − log (1) + R1 log (k) = d x log (x) + 2 k=1

n+1 

(6.112)

0

where R1 is a remainder having terms subdominant to log (n + 1) as n → ∞. The integral we need to evaluate is known as Raabe’s integral and we will evaluate it here. We will first show in the exercises that k+1 1 d x log (x) = d x log (x) + k log k − k k

(6.113)

0

To evaluate the integral over [0, 1] on the RHS, we begin by using Legendre duplication formula. x  x + 1 √  = 21−x π (x)  (6.114) 2 2

234

6 Asymptotic Methods

so that

      x2  x+1 2 log (x) = log √ 1−x π2

x  1 = − log π − log 2 + log  2 2   x +1 + x log 2 + log  2

(6.115)

Accordingly, 1

1 d x log (x) = − log (2π ) + 2

0

1 d x log 

x  2



1 +

0

d x log 

x +1 2



0

1/2 1 1 = − log (2π ) + 2 d x log (x) + 2 d x log (x) 2 0

1 = − log (2π ) + 2 2

1/2

1 d x log (x)

(6.116)

0

i.e.,

1 d x log (x) =

1 log (2π ) 2

(6.117)

0

We now plug this into the Euler-Maclurin result in Eq. (6.112). k+1 n+1 n   d x log (x) = d x log (x) k=0 k

0

n(n + 1)  n+1 log(2π ) − + k log k 2 2 k=1 n

=

n(n + 1) n+1 log(2π ) − + log (11 22 · · · n n ) 2 2 n(n + 1) n+1 log(2π ) − = 2  2  n!n + log (n − 1)!(n − 2)!2 · · · 2n−2 1n−1

=

 n+1 n(n + 1) log(2π ) − + n log n! − log k! 2 2 k=1 n−1

=

(6.118)

6.3 Euler-Maclurin Summation

235

Putting this all together, we find that 2

n 

log k! =

k=1

n+1 n(n + 1) 3 log(2π ) − + n log n! + log n! + R1 2 2 2

(6.119)

From this we can easily deduce the leading asymptotic behavior of the factorial product we seek. Here, Euler-Maclurin led to an integral that we were able to evaluate analytically. The result was a little messy but in the end, very satisfying. An alternating sum of logs We next seek to evaluate the following sum, exactly. S=

∞  log2 k (−1)k+1 k k=1

(6.120)

This is an alternating sum which we will attack in a similar manner as we did for the alternating sum we attacked earlier in this section. To begin, let’s rewrite the sum as a limit of a finite sum, i.e., S = limn→∞ Sn , where S2n =

2n 2n n    log2 k log2 k log2 (2k) = −2 (−1)k+1 k k 2k k=1 k=1 k=1

(6.121)

Note that we are using an even index for convenience, i.e., to avoid extra factors of (−1)n+1 in the second sum on the RHS. We expand the second sum on the RHS and get the following. 2

n  log2 (2k) k=1

2k

Then S2n =

= (log2 2)Hn + 2 log 2

n  log k k=1

k

+

n  log2 k k=1

k

2n n   log2 k log k − (log2 2)Hn − 2 log 2 k k k=n+1 k=1

(6.122)

(6.123)

We analyze the first sum on the RHS using Euler-Maclurin. In this case, we have the following.   2n 2n  1 log2 2n log2 n log2 k log2 x = dx + − + R1 k x 2 2n n k=n+1 n

 1 3 log (2n) − log3 (n) + R0 3   1 = log3 2 + log2 2 log n + (log 2) log2 n + R0 3 =

(6.124)

236

6 Asymptotic Methods

where R0 and R1 consist of terms of vanish as n → ∞. Putting this back into the equation for S2n , we then have the following.  n  log k 1 1 3 2 − log2 n + R0 S2n = log 2 − (log 2)(Hn − log n) − 2 log 2 3 k 2 k=1 (6.125) Now we know that (6.126) lim (Hn − log n) = γ = γ0 n→∞

where γ ≈ 0.577216 is the Euler-Mascheroni constant we have seen occasionally throughout the book. We will call it γ0 for a reason to be made obvious in a moment. There is a sequence of constants called the Stieltjes constants which are defined as follows.  n  logm k logm+1 n − γm = lim (6.127) n→∞ k m+1 k=1 The Stieltjes constants are coefficients of the Laurent series of the Riemann zeta function about s = 1, as follows: ∞

ζ (s) =

 (−1)n 1 + γn (s − 1)n s − 1 n=0 n!

(6.128)

It turns out that γ1 ≈ 0.0728158. We thus have by taking the limit as n → ∞, ∞  log2 k 1 = γ log2 2 − log3 2 + 2γ1 log 2 (−1)k k 3 k=1

(6.129)

Again, we used an asymptotic relation to evaluate something exactly. It should be clear by now why we needed a chapter on asymptotics!

Epilogue

Going back to the original integral from the Prologue, we are able to manipulate that integral into the following form that is ripe for the techniques we have discussed in the book. ∞ I =8

(v2 − 1)(v4 − 6v2 + 1) log v + 4v6 + 70v4 + 4v2 + 1

dv

v8

0

Consider the following contour integral: 

8(z 2 − 1)(z 4 − 6z 2 + 1) log2 z z 8 + 4z 6 + 70z 4 + 4z 2 + 1

dz C

where C is a keyhole contour about the positive real axis. This contour integral is equal to ∞ − i4π

dv 0

 + 4π 2

8(v2 − 1)(v4 − 6v2 + 1) log v v8 + 4v6 + 70v4 + 4v2 + 1

∞

dv 0

8(v2 − 1)(v4 − 6v2 + 1) + 4v6 + 70v4 + 4v2 + 1

v8

It should be noted that the second integral vanishes; this may be easily seen by exploiting the symmetry about v → 1/v. On the other hand, the contour integral is i2π times the sum of the residues about the poles of the integrand. In general, this requires us to find the zeroes of the eight degree polynomial, which may not be possible analytically. Here, on the other hand, we have many symmetries to exploit, e.g., if a is a root, then 1/a is a root, −a is a root, and a¯ is a root. For example, we may deduce that © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 R. Gordon, Complex Integration, Undergraduate Lecture Notes in Physics, https://doi.org/10.1007/978-3-031-24228-1

237

238

Epilogue

z 8 + 4z 6 + 70z 4 + 4z 2 + 1 = (z 4 + 4z 3 + 10z 2 + 4z + 1) (z 4 − 4z 3 + 10z 2 − 4z + 1) which exploits the a → −a symmetry. Now write 

1 ¯ z− z + 4z + 10z + 4z + 1 = (z − a)(z − a) a 4

3

2



1 z− a¯



Write a = r eiθ and get the following equations:   1 r+ cos θ = −2 r   1 2 r + 2 + 4 cos2 θ = 10 r √ From these equations, √ one may deduce that a solution is r = φ + φ and cos θ = 1/φ, where φ = (1 + 5)/2 is the golden ratio. Thus the poles take the form  √   z k = ± φ ± φ e±i arctan φ Now we have to find the residues of the integrand at these 8 poles. We can break this task up by computing:

 8(z 2 − 1)(z 4 − 6z 2 + 1) log2 z z 8 + 4z 6 + 70z 4 + 4z 2 + 1 k=1 

8 8(z 2 − 1)(z 4 − 6z 2 + 1) log2 z k = Resz=zk 8 6 + 70z 4 + 4z 2 + 1 z + 4z k=1

8

Resz=zk

The result is rather unbelievably simple:

Resz=zk

 8(z 2 − 1)(z 4 − 6z 2 + 1) = sgn[cos (arg z k )] z 8 + 4z 6 + 70z 4 + 4z 2 + 1

That is, if the pole has a positive real part, the residue of the fraction is +1; if it has a negative real part, the residue is −1. Now consider the log piece. Expanding the square, we get 3 terms: log2 |z k | − (arg z k )2 + i2 log |z k | arg z k

Epilogue

239

Summing over the residues, we find that because of the ±1 contributions above, that the first and third terms sum to zero. This leaves the second term. For this, it is crucial that we get the arguments right, as arg z k ∈ [0, 2π ). Thus, we have ∞ I =

dv 0

8(v2 − 1)(v4 − 6v2 + 1) log v v8 + 4v6 + 70v4 + 4v2 + 1

1 sgn[cos (arg z k )](arg z k )2 = 2 k=1   1 = [2(arctan φ)2 + 2(2π − arctan φ)2 2   − 2(π − arctan φ)2 − 2(π + arctan φ)2 ]  = 2π 2 − 4π arctan φ  = 4π arccot φ 8