Advanced Calculus Explored: With Applications in Physics, Chemistry, and Beyond 0578616823, 9780578616827

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1

Advanced Calculus Explored With Applications in Physics, Chemistry, and Beyond

2

ADVANCED CALCULUS EXPLORED WITH APPLICATIONS IN

PHYSICS, CHEMISTRY, AND BEYOND

Hamza E. Alsamraee

3

This page is intentionally left blank.

4 Copyright © 2019 by Hamza E. Alsamraee. All rights reserved. No part of this book may be reproduced or distributed in any form, stored in any data base or retrieval system, or transmitted in any form by any means—electronic, mechanical, photocopy, recording, or otherwise—without prior written permission of the publisher, except as provided by United States of America copyright law. For permission requests, e-mail the publisher at curiousmath. [email protected]. First edition published November 2019 Book cover design by Ayan Rasulova Typeset using LATEX Printed on acid-free paper

ISBN 978-0-578-61682-7 www.instagram.com/daily_math_/

5

To my parents.

6

To numerically confirm the many evaluations and results throughout this book, various integration and summation commands available in software produced by Wolfram Research, Inc. were utilized. Moreover, virtually all illustrations and graphs were produced by software produced by Wolfram Research, unless specified otherwise. Specifically, Wolfram Desktop Version 12.0.0.0 running on a Windows 10 PC. As of the time of the release of the book, this is the latest release of Wolfram Desktop. The commands in this book are standard and are likely to continue to work for subsequent versions. Wolfram Research does not warrant the accuracy of the results in this book. This book’s use of Wolfram Research software does not constitute an endorsement or sponsorship by Wolfram Research, Inc. of a particular pedagogical approach or particular use of the Wolfram Research software.

Contents

About the Author

15

Preface

23

I

27

Introductory Chapters

1 Differential Calculus 1.1

1.2

1.3

29

The Limit . . . . . . . . . . . . . . . . . . . . . . . . .

30

1.1.1

L’Hopital’s Rule . . . . . . . . . . . . . . . . .

36

1.1.2

More Advanced Limits . . . . . . . . . . . . . .

40

The Derivative . . . . . . . . . . . . . . . . . . . . . .

45

1.2.1

Product Rule . . . . . . . . . . . . . . . . . . .

46

1.2.2

Quotient Rule . . . . . . . . . . . . . . . . . . .

48

1.2.3

Chain Rule . . . . . . . . . . . . . . . . . . . .

49

Exercise Problems . . . . . . . . . . . . . . . . . . . .

56

7

8

CONTENTS

2 Basic Integration

59

2.1

Riemann Integral . . . . . . . . . . . . . . . . . . . . .

60

2.2

Lebesgue Integral . . . . . . . . . . . . . . . . . . . . .

63

2.3

The u-substitution . . . . . . . . . . . . . . . . . . . .

65

2.4

Other Problems . . . . . . . . . . . . . . . . . . . . . .

82

2.5

Exercise Problems . . . . . . . . . . . . . . . . . . . .

99

3 Feynman’s Trick

101

3.1

Introduction . . . . . . . . . . . . . . . . . . . . . . . . 102

3.2

Direct Approach . . . . . . . . . . . . . . . . . . . . . 103

3.3

Indirect Approach . . . . . . . . . . . . . . . . . . . . 128

3.4

Exercise Problems . . . . . . . . . . . . . . . . . . . . 131

4 Sums of Simple Series

135

4.1

Introduction . . . . . . . . . . . . . . . . . . . . . . . . 136

4.2

Arithmetic and Geometric Series . . . . . . . . . . . . 136

4.3

Arithmetic-Geometric Series . . . . . . . . . . . . . . . 141

4.4

Summation by Parts . . . . . . . . . . . . . . . . . . . 146

4.5

Telescoping Series . . . . . . . . . . . . . . . . . . . . . 152

4.6

Trigonometric Series . . . . . . . . . . . . . . . . . . . 159

4.7

Exercise Problems . . . . . . . . . . . . . . . . . . . . 163

9

CONTENTS

II

Series and Calculus

5 Prerequisites

165 167

5.1

Introduction . . . . . . . . . . . . . . . . . . . . . . . . 168

5.2

Ways to Prove Convergence . . . . . . . . . . . . . . . 172

5.3

5.2.1

The Comparison Test . . . . . . . . . . . . . . 172

5.2.2

The Ratio Test . . . . . . . . . . . . . . . . . . 173

5.2.3

The Integral Test . . . . . . . . . . . . . . . . . 176

5.2.4

The Root Test . . . . . . . . . . . . . . . . . . 181

5.2.5

Dirichlet’s Test . . . . . . . . . . . . . . . . . . 184

Interchanging Summation and Integration . . . . . . . 185

6 Evaluating Series

191

6.1

Introduction . . . . . . . . . . . . . . . . . . . . . . . . 192

6.2

Some Problems . . . . . . . . . . . . . . . . . . . . . . 193 6.2.1

6.3

Harmonic Numbers . . . . . . . . . . . . . . . . 204

Exercise Problems . . . . . . . . . . . . . . . . . . . . 214

7 Series and Integrals

215

7.1

Introduction . . . . . . . . . . . . . . . . . . . . . . . . 216

7.2

Some Problems . . . . . . . . . . . . . . . . . . . . . . 216

7.3

Exercise Problems . . . . . . . . . . . . . . . . . . . . 230

10

CONTENTS

8 Fractional Part Integrals

III

233

8.1

Introduction . . . . . . . . . . . . . . . . . . . . . . . . 234

8.2

Some Problems . . . . . . . . . . . . . . . . . . . . . . 235

8.3

Open Problems . . . . . . . . . . . . . . . . . . . . . . 255

8.4

Exercise Problems . . . . . . . . . . . . . . . . . . . . 255

A Study in the Special Functions

9 Gamma Function

257

261

9.1

Definition . . . . . . . . . . . . . . . . . . . . . . . . . 262

9.2

Special Values . . . . . . . . . . . . . . . . . . . . . . . 262

9.3

Properties and Representations . . . . . . . . . . . . . 264

9.4

Some Problems . . . . . . . . . . . . . . . . . . . . . . 271

9.5

Exercise Problems . . . . . . . . . . . . . . . . . . . . 275

10 Polygamma Functions

277

10.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . 278 10.2 Special Values . . . . . . . . . . . . . . . . . . . . . . . 279 10.3 Properties and Representations . . . . . . . . . . . . . 280 10.4 Some Problems . . . . . . . . . . . . . . . . . . . . . . 282 10.5 Exercise Problems . . . . . . . . . . . . . . . . . . . . 294

CONTENTS 11 Beta Function

11 295

11.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . 296 11.2 Special Values . . . . . . . . . . . . . . . . . . . . . . . 297 11.3 Properties and Representations . . . . . . . . . . . . . 297 11.4 Some Problems . . . . . . . . . . . . . . . . . . . . . . 302 11.5 Exercise Problems . . . . . . . . . . . . . . . . . . . . 309 12 Zeta Function

311

12.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . 312 12.2 Special Values . . . . . . . . . . . . . . . . . . . . . . . 312 12.3 Properties and Representations . . . . . . . . . . . . . 317 12.4 Some Problems . . . . . . . . . . . . . . . . . . . . . . 326 12.5 Exercise Problems . . . . . . . . . . . . . . . . . . . . 335

IV Applications in the Mathematical Sciences and Beyond 339 13 The Big Picture

341

13.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . 342 13.2 Goal of the Part . . . . . . . . . . . . . . . . . . . . . 343 14 Classical Mechanics

345

14.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . 346

12

CONTENTS 14.1.1 The Lagrange Equations . . . . . . . . . . . . . 346 14.2 The Falling Chain . . . . . . . . . . . . . . . . . . . . 347 14.3 The Pendulum . . . . . . . . . . . . . . . . . . . . . . 353 14.4 Point Mass in a Force Field . . . . . . . . . . . . . . . 357

15 Physical Chemistry

363

15.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . 364 15.2 Sodium Chloride’s Madelung Constant . . . . . . . . . 370 15.3 The Riemann Series Theorem in Action . . . . . . . . 371 15.4 Pharmaceutical Connections . . . . . . . . . . . . . . . 377 15.5 The Debye Model . . . . . . . . . . . . . . . . . . . . . 378

16 Statistical Mechanics

381

16.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . 382 16.2 Equations of State . . . . . . . . . . . . . . . . . . . . 383 16.3 Virial Expansion . . . . . . . . . . . . . . . . . . . . . 385 16.3.1 Lennard-Jones Potential . . . . . . . . . . . . . 386 16.4 Blackbody Radiation . . . . . . . . . . . . . . . . . . . 388 16.5 Fermi-Dirac (F-D) Statistics . . . . . . . . . . . . . . . 393

17 Miscellaneous

401

17.1 Volume of a Hypersphere of Dimension N . . . . . . . 402

CONTENTS

13

17.1.1 Spherical Coordinates . . . . . . . . . . . . . . 402 17.1.2 Calculation . . . . . . . . . . . . . . . . . . . . 404 17.1.3 Discussion . . . . . . . . . . . . . . . . . . . . . 407 17.1.4 Applications . . . . . . . . . . . . . . . . . . . . 409 17.1.5 Mathematical Connections . . . . . . . . . . . . 410

V

Appendices

413

Appendix A

415

Appendix B

421

Acknowledgements

425

Answers

427

Integral Table

435

Trigonometric Identities

439

Alphabetical Index

443

14

CONTENTS

About the Author Hi! My name is Hamza Alsamraee, and I am a senior (12th grade) at Centreville High School, Virginia. I have always had an affinity for mathematics, and from a very young age was motivated to pursue my curiosity. When I entered a new school in 7th grade after moving, I encountered some new mathematics I was unequipped for. Namely, I did not know what a linear equation even was! I was rather low-spirited, as I was stuck in an ever-lasting loop of confusion in class. My mother and father soon began teaching me to the best of their ability. Fortunately, their efforts were effective, and I got a B on my first linear equations test! It was a huge improvement from being totally lost, but I wanted to know more. I did not care much about the grade, but I did care that I did not completely master the material. I soon entered in a period of rapid learning, delving into curriculums significantly beyond my coursework simply for the sake of mastering higher mathematics. It seemed to me that the more I explored the field, the more beautiful the results were. I quickly got bored of the regular algebra and geometry problems, and wanted to know if there was more to mathematics. I browsed the web for "hard math problems," and stumbled across this integral:

15

16

CONTENTS

Z

1

−1

r

1+x dx = π 1−x

(1)

Well, I knew what π is, but what is that long S symbol on the left side? So, I browsed the web again, but this time asking what that "long S symbol" is. It is an integral! Cool, I thought, but what does that even mean? Soon started a period where I was almost obsessed with those mathematical creatures! The evaluation of integrals and series quickly became a hobby, leading me to challenge myself everyday with a new integral or series that "looked" like it might yield a nice closed form. As I delved deeper into the subject matter, I discovered the wellknown special functions such as the gamma and zeta functions. Being a physics enthusiast, I was fascinated by their applications in physics as well as in other scientific disciplines. I found my love for mathematics and science converge, and was determined to cultivate this passion. Ever since then, I began collecting results and solutions to various problems in the evaluation of integrals and series. It was only about a year ago, due to a suggestion of one of my close friends, that I thought about compiling my results into a cohesive curriculum. I remembered my early days in doing these sorts of problems, and my frustration at the lack of quality resources. It was then that I became determined to write this book! After I began writing the book, I began wondering how many individuals were genuinely interested in this type of mathematics. As an experiment, I set up a mathematics Instagram account by the name of daily_math_ to test the reaction to some of the book’s problems. After an overwhelmingly positive response, I was more motivated than ever to finish this book! At the time of the publishing of this book, the account has garnered over 40,000 followers

17

CONTENTS globally, from middle school students to mathematics PhD’s.

Proof of (1) This integral is truly special to me, as it marked the beginning of a period of immense investment into my passion, mathematics. Thus, I have compiled a list of proofs that I derived over the years. Even though all these proofs were arrived at independently, they exist in the literature since this integral is rather common.

Figure 1: Graph of the integrand of (1) Proof 1. We will begin by splitting this integral from −1 to 0 and from 0 to 1 to get: Z

1

I= −1

Z

0

= −1

r

r

1+x dx 1−x

1+x dx + 1−x

Z 0

1

r

1+x dx 1−x

18

CONTENTS

Using the substitution x → −x on the first integral, Z

0

Z 1r 1−x 1+x dx + dx 1+x 1−x 0 ! r r 1−x 1+x + dx 1+x 1−x

r

I=− 1 1

Z = 0

1

(1 − x) + (1 + x) p dx (1 − x)(1 + x) 0 Z 1 1 √ dx =2 1 − x2 0

Z =

(2)

Since d arcsin x 1 =√ dx 1 − x2 We have: I = 2[ arcsin x]10

= π An alternate way to compute (2) is by the substitution x = sin u, dx = cos u du: Z

arcsin 1

cos u p du arcsin 0 1 − sin2 u Z arcsin 1 = du arcsin 0

19

CONTENTS = arcsin 1 − arcsin 0 =

π 2

 √ 1+x Proof 2. Consider multiplying (1) by √ 1+x

Z

1

I= −1

Z

1

= −1

1+x √ dx 1 − x2

1 √ dx + 1 − x2

Z

1

√

−1

x dx 1 − x2

For the second integral, we let u = 1 − x2 , − du 2 = x dx to get:

1 I =π− 2

Z |0

0

1 √ du u {z } =0

Which can be easily argued from the fact that the integrand in the second integral is an odd function on the interval (−1, 1).

20

CONTENTS

Figure 2: Graph of y = √

x on (−1, 1) 1 − x2

We therefore obtain: I=π Proof 3. Consider the integral reflection property, Z

b

Z

b

f (a + b − x) dx

f (x) dx = a

a

Applying that to (1) gives: Z

1

I= −1

Adding (3) and (1) gives:

r

1−x dx 1+x

(3)

21

CONTENTS

Z

1

r

2I = −1

Z

1−x + 1+x

1

= −1

√

r

1+x 1−x

! dx

2 dx 1 − x2

= 2( arcsin(1) − arcsin(−1)) = 2π Therefore, I=π

22

CONTENTS

Preface What is the value of this integral? Z

1

x2 dx

0

Hopefully, you calculated the correct value of 13 . Now, what about this one? Z

π 2

ln(sin x) dx

0

Well, that was quite a jump in difficulty. If your pencil is already out, trying out your every tool, then this book is for you! However, if you are perplexed as to why anyone cares about the evaluation of this integral, then this book is for you as well! The complete solution to the integral above can be found in (2.2). However, the value has little importance. Rather, it is the techniques and methods that are employed that are worth attention. If problems like these give you a kick, then you are in for a good ride. If not, then by part 4 of this book, you will see the importance of the techniques used to answer such questions! I have written this book with two types of readers in mind: 1) mathematical enthusiasts who love a challenging problem, and 2) physics, 23

24

CONTENTS

chemistry, and engineering majors. In this book, I aim to use the methods introduced in a standard two-semester calculus course to develop both problem solving skills in mathematics and the mathematical sciences. The examples given often have very differing solutions, some of marvelous ingenuity and some that are rather standard. This is done on purpose, as it ultimately benefits readers to see multiple perspectives on similar problems. From u−substitutions to clever interchanges of integration and summation, various methods will be presented that can be used to solve the same problems. In order to make this book accessible to a larger base of students, contour integration is not included in the book. It is worth noting that this is not an elementary calculus book, although the first chapters are there as a refresher for those who need it. A key difference between this book and other mathematics books that attempt to address a similar topic is that it is written with the reader in mind. Rarely would you need to struggle through proving a non-trivial statement that was previously declared as "trivial and left to the reader." Moreover, instead of the normal theorem-focused advanced mathematics book, I aim to minimize the number of techniques and methods and instead focus on examples. In writing this book, I intended to make it as self-contained as possible. The various identities and theorems used in this book are often proved in the book, and the scientific concepts behind each application are explained and elaborated upon. Moreover, for theorems that need an extensive mathematics background, I aimed to simplify and translate their statements into the book’s area of concern as best as possible without losing key details. Even though this book is heavily mathematical, almost 100 pages are dedicated to applications of the techniques explored in the book. The applications are broad and include various topics of concern in the sciences and engineering. Many of the problems in this book, particularly those in chapter 8, are included simply because they

25

CONTENTS

are elegant results and develop the problem-solving skills of the reader. However, a few of the integrals and series, and certainly all of the methods employed, have wide applications in many science and engineering fields. In the first two chapters, I introduce Mathematica as a way to numerically or symbolically verify the correctness of a result. This is done to familiarize the reader with Mathematica syntax so they are able to employ it on their own in later chapters. In addition to almost one hundred fully worked out examples, there are exercise problems for the reader at the end of each chapter. Generally, these problems get progressively harder by number. However, all these problems involve the same techniques used in the chapter they are included in! A few challenge problems are scattered throughout the book to engage the more experienced reader. The answers to these exercise problems are all included at the end of the book. There is a high likelihood that I will compile a list of solutions to these problems in a solutions manual to be published a few months after the book’s launch. Enjoy!

Hamza Alsamraee, Centreville High School

A Note on the Originality of the Results I have tried to cite results attributed to well-known mathematicians to the best of his ability. However, it is virtually impossible to check the originality of all the results in this book. Many are classic results, and a few are more unusual. To the best of my knowledge, many of the integrals and series in this book will be exposed to the literature for the first time. I do not claim originality, but I

26 do claim authenticity.

CONTENTS

Part I

Introductory Chapters

27

Chapter 1

Differential Calculus

29

30

CHAPTER 1. DIFFERENTIAL CALCULUS

This chapter will serve as a review of differential calculus, which will be used throughout the book, especially in chapter 3. In this chapter, we will also delve into elementary as well as advanced limits, giving a glimpse into the later chapters of the book. We will begin with the epsilon-delta definition of the limit and transition into the evaluation of limits. After all, what better way is there to start a calculus book other than to define the limit?

1.1

The Limit

We shall define the limit as the following (epsilon-delta definition): Definition Let f (x) be a function defined on the interval (a, b), except possibly at x0 , where x0 is in the interval i.e. a < x0 < b then the limit is lim f (x) = L if for every number ε > 0 x→x0

there exists a δ > 0 such that f (x) − L < ε whenever 0 < |x − x0 | < δ

(1.1)

For all x. This is a formalization of the limit which turns our rather informal notion of the limit to a rigorous one. Instead of using broad terms such as f (x) gets "close" to L as x gets "close" to x0 , this definition allows us to rigorously discuss limits. The theorem originated from the French mathematician and physicist Augustin-Louis Cauchy and was modernized by the German mathematician Karl Weierstrass. This marked an interesting time in the history of mathematics, representing its move towards rigor.

31

1.1. THE LIMIT y

L

ε ε

δ

δ x0

x

Figure 1.1: A visualization of the epsilon-delta definition of the limit

"...Since Newton the limit had been thought of as a bound which could be approached closer and closer, though not surpassed. By 1800, with the work of L’Huilier and Lacroix on alternating series, the restriction that the limit be onesided had been abandoned. Cauchy systematically translated this refined limit-concept into the algebra of inequalities, and used it in proofs once it had been so translated; thus he gave reality to the oft-repeated eighteenth-century statement that the calculus could be based on limits." - American mathematician Judith Grabinera a

Grabiner, Judith V. (March 1983), "Who Gave You the Epsilon? Cauchy and the Origins of Rigorous Calculus" , The American Mathematical Monthly, 90 (3): 185–194, doi:10.2307/2975545, JSTOR 2975545

This formulation will not be used extensively in this book, but is nonetheless widely used in analysis. Specifically, it is employed heavily in proving the continuity of a function.

32

CHAPTER 1. DIFFERENTIAL CALCULUS Definition A function f is said to be continuous at x0 if it is both defined at x0 and its value at x0 equals the limit of f (x) as x approaches x0 , i.e. lim f (x) = f (x0 )

x→x0

Consequently, f (x) is said to be continuous on some interval (a, b) if it is continuous for every x0 belonging to that interval. Let us begin with an easy first example! Example 1: Prove that lim x2 = 0 using the epsilon-delta definition of a limit.

x→0

Solution In this case both L and x0 are zero. We start by letting ε > 0. According to the (ε, δ) definition of the limit, if limx→0 x2 = 0 we will need to find some other number δ > 0 such that 2 x − 0 < ε whenever 0 < |x − 0| < δ Which gives us x2 < ε whenever 0 < |x| < δ Starting with the left inequality and taking the square root of both sides we get √ |x| < ε This looks very similar to the right inequality, which drives us to √ set ε = δ. We now need to prove that our choice satisfies 2 x < ε whenever 0 < |x| < √ε

33

1.1. THE LIMIT Starting with the right inequality with the assumption that 0 < √ |x| < ε √
2 2 2 ε =ε x = |x|
0 we can find a δ > 0 such that

2 x − 0 < ε whenever 0 < |x − 0| < δ

Which proves our original supposition that

lim x2 = 0

x→0

sin (x) = 1 using the epsilon-delta x→0 x

Example 2: Prove that lim definition of a limit. Solution

To start off, we will make use of the unit circle to derive some identities. Consider figure 1.2.

34

CHAPTER 1. DIFFERENTIAL CALCULUS y

1

1 2

tan α

sin α α −1

cos α

− 12

x 1

− 12

−1

Figure 1.2: The unit circle. Figure generated using Tik Z software

Figure 1.3: Points B, C lie on the unit circle

35

1.1. THE LIMIT

Now, consider figure 1.3. Using the basics of the unit circle, we can say that point C has a y−coordinate equal to sin x and point A has a y−coordinate equal to tan x. Now, consider triangle 4OCB and triangle 4OAB where O is the origin. 4OCB has area is 12 sin x while 4OAB has area 12 tan x. Moreover, the area of the sector formed by x is radians). We now have the inequality 1 x 1 sin x ≤ ≤ tan x 2 2 2 Dividing (1.2) by

1 2

x 2

(x is measured in

(1.2)

sin x,

1≤

x 1 ≤ sin x cos x

Taking the reciprocal,

cos x ≤

sin x ≤1 x

Since sinx x and cos x are even functions,
this inequality holds for any non-zero x on the interval − π2 , π2 . This leads us to sin (x) < 1 − cos (x) − 1 x

(1.3)

We can then use the trigonometric identity cos(2x) = 1 − 2 sin2 x
2 To obtain that 1 − cos (x) = 2sin2 x2 < x2 where the inequality holds because of (1.2). (1.3) is then transformed into sin (x) x2 x − 1 < 2

(1.4)

36

CHAPTER 1. DIFFERENTIAL CALCULUS

We proceed to let δ =

√

ε which gives us |x − 0| < δ =

√

ε

Plugging back into (1.4), √ sin (x) x2 ε2 − 1 < 0 we can find a δ > 0 such that sin (x) x − 1 < ε whenever 0 < |x − 0| < δ

1.1.1

L’Hopital’s Rule

Theorem L’Hopital’s rule is one of the most widely known limit properties. It states that if two functions f and g are differentiable on an open interval I, with an exception made for a, and the limit of their quotient takes an indeterminate form such as 00 or ∞ ∞ , the limit of their quotient can be expressed as f 0 (x) f (x) lim = lim 0 x→a g (x) x→a g (x) Given that g 0 (x) 6= 0 and the RHS exists. The rule is named after the 17th-century French mathematician Guillaume de L’Hopital. Even though the rule is often attributed to L’Hopital, the theorem was first introduced to him in 1694 by the Swiss mathematician Johann Bernoulli. This rule simplifies many limits and sometimes is needed more than once. Again, we can start with an easy problem to get the gist of how to use L’Hopital’s rule.

37

1.1. THE LIMIT Example 3: Find the value of lim xe−x x→∞ Solution We will first need to establish a non-constant denominator. This is because if our denominator is constant, its derivative will be 0. There are two simple choices e−x 1 x

Or

x ex

We can see that the latter choice is better as the first will produce an even more complicated result when L’Hopital’s rule is applied. Remember, when one key does not work, one must proceed to try another one! As the problems in this book get harder, this idea becomes vital. Going back to our problem, L = lim xe−x = lim x→∞

x→∞

x ex

After applying L’Hopital’s rule we obtain: L = lim

x→∞

1 ex

Now it is easy to see that L = lim xe−x = 0 x→∞

To numerically "confirm" our result, we can use Wolfram Desktop (Or Mathematica):

In[1]:=

Out[1]=

Limit[x/E^x, x -> Infinity]

0

38

CHAPTER 1. DIFFERENTIAL CALCULUS 

1  1+ n Example 4: Find lim  n→∞ e

n n  

Figure 1.4: The graph shows limiting behavior approaching a value near .6

Solution Define 

1  1+ n L = lim  n→∞ e

n n  

We then have 

 1+ ln L = lim n ln  n→∞ e

1 n

n 

"

  = lim

n→∞

#   1 n ln 1 + −n n 2

We will proceed to transform the limit into one we can apply L’Hopital’s

39

1.1. THE LIMIT rule on, i.e. a fraction with a non-constant denominator:

ln L = lim

  n ln 1 + n1 − 1 1 n

n→∞

After applying L’Hopital’s we obtain:   n· 1   n2 ln 1 + n1 − 1 1 ln 1 + 1+ n n − ln L = lim = lim n→∞ n→∞ − n12 − n12

1 n+1

Applying L’Hopital’s rule again and simplifying we obtain: ln L = lim = n→∞

1 − n(n+1) 2 2 n3

n3 n→∞ 2n(n + 1)2

= − lim

We can already see that expanding the denominator would result in a first term of 2n3 . Thus, ln L = −

1 2

By using the ratio of the coefficients. 1 =⇒ L = √ e Mathematica gives:

In[2]:=

Out[2]=

Limit[((1 + 1/x)^x/E)^x, x -> Infinity]

1/Sqrt[E]

40

1.1.2

CHAPTER 1. DIFFERENTIAL CALCULUS

More Advanced Limits

∞ X 1 Example 5: Define the Riemann zeta function as ζ(s) = = ns n=1
1 1 1 1 1 + s + s + s ... Find lim ζ(s) − 1 s s s→∞ 1 2 3 4


1 Figure 1.5: A graph of y = ζ(x) − 1 x

Solution We know that the first term of the zeta function will always be 1 =1 1s Therefore,
1 L = lim ζ(s) − 1 s s→∞

41

1.1. THE LIMIT  = lim

s→∞

1 s 1 1 1 1 + s + s + s ··· − 1 2 3 4 

= lim

s→∞

We proceed to factor out

1 1 1 + + ··· 2s 3s 4s

1 s

1 2s  s  s ! 1s 2 2 + ··· 1+ 3 4

1 L = lim s→∞ 2

As s → ∞, all terms inside the parentheses with s as an exponent will approach zero. Therefore, 1 1 · 1s s→∞ 2

L = lim Now it is easy to see that

lim ζ(s) − 1


1

s→∞

s

=

1 2

Testing our result with Mathematica,

In[3]:=

Out[3]=

(Zeta[x]-1)1/x 1 2

Example 6: Define the double factorial, which is usually denoted ! !, as n! ! = n · (n − 2) · (n − 4) · · ·

42

CHAPTER 1. DIFFERENTIAL CALCULUS

For even numbers, the last number to be multiplied by is 2, while for odd numbers it is 1. For example,

6! ! = 6 · 4 · 2 = 48 5! ! = 5 · 3 · 1 = 15 √

(2n − 1)! ! n for n ∈ N n→∞ (2n)! !

Find lim

Figure 1.6: Using Mathematica’s ListLinePlot function, we can get √ (2n−1)!! n a graph of the sequence an = (2n)!!

Solution A double factorial , n! !, can be expressed using normal factorials. The expressions for our case are as follows (2n)! ! = 2n · n! (2n − 1)! ! =

(2n)! 2n · n!

43

1.1. THE LIMIT By substituting the above expressions into our limit we obtain: √ (2n)! n L = lim n→∞ (n)!2 ·4n Now, we will introduce Stirling’s approximation.

Theorem Stirling’s approximation, or Stirling’s formula, is one of the most common approximations for factorials. The general formula is ln k! = k ln k − k + O(ln k) (1.5) Where Big O notation indicates that the LHS (Left-hand side) describes the RHS’s (Right-hand side) limiting behavior, i.e. as k → ∞. A more precise approximation, which is derived from (1.5), is  n √ n n! ∼ 2πn (1.6) e

Using (1.6) we have: √

4πn(2n)2n · e2n √ n n→∞ e2n · (2πn) · n2n · 22n

L = lim

After simplifying, we are only left with √ n · 4πn L = lim n→∞ 2πn √ 2n π = lim n→∞ 2πn 1 =⇒ L = √ ≈ .56419 π

44

CHAPTER 1. DIFFERENTIAL CALCULUS

We can test our result with Mathematica. Unfortunately, Mathematica is unable to predict a value due to the computational heaviness of the factorial. However, evaluating our sequence for n = 1, 000, 000 gives us ≈ .56419, so we can be confident in our result. Example 7: Define the nth Fibonacci number as Fn = Fn−1 + Fn−2 Fn+1 . where F1 = F2 = 1. Find lim n→∞ Fn Solution We start by noting that through the definition of the Fibonacci sequence we have Fn+1 = Fn + Fn−1 =⇒ Define an =

Fn+1 Fn−1 =1+ Fn Fn

Fn+1 Fn

Then,

an = 1 +

1 an−1

Taking the limit of both sides,  lim an = lim 1 +

n→∞

n→∞

1



an−1

Since both the limits of the RHS and LHS exist we can let limn→∞ an = x to get 1 x=1+ x 2 x −x−1=0 Using the quadratic formula and eliminating the negative solution we obtain that √ Fn+1 1+ 5 lim = =φ n→∞ Fn 2

45

1.2. THE DERIVATIVE

Where φ denotes the famous golden ratio. To check our result, we can use Mathematica’s built in Fibonacci sequence function:

In[4]:=

Out[4]=

1.2

Limit[Fibonacci[n+1]/Fibonacci[n],n->Infinity] √ 1 (1+ 5) 2

The Derivative

Definition The derivative of a function f (x), denoted by f 0 (x), gives us the instantaneous rate of change at a point (x, f (x)) by using the concept of a limit. This is done by measuring the "slope" over an infinitesimal interval, i.e. [x, x + h]. f (x + h) − f (x) h→0 h

f 0 (x) = lim

Example 8: What is the derivative of the function f (x) = xn , where n is a constant?

Solution Many have already memorized the power rule, but let us prove it here. Using the definition of a derivative, d (x + h)n − xn f (x) = lim h→0 dx h

46

CHAPTER 1. DIFFERENTIAL CALCULUS

We now recall the binomial theorem which states that n   X n n−k k n x a (x + a) = k k=0

Therefore, d f (x) = lim h→0 dx

Pn

n k=0 k




xn−k hk − xn h

By expanding this expression, we can see that the first term, xn , cancels out and we are left with nxn−1 h + d f (x) = lim h→0 dx

n(n−1) n−2 2 h 2! x

+ ...

h

Which simplifies to d n (n − 1) n−2 1 f (x) = lim nxn−1 + x h + ... h→0 dx 2! Notice that the degree of h increases. As h → 0, all terms but the first approach 0. Thus, d f (x) = nxn−1 dx

1.2.1

Product Rule

Theorem For two differentiable functions, the derivative of their product can be stated as follows d d d uv = u v + v u dx dx dx

47

1.2. THE DERIVATIVE

Example 9: Find lim

x→0



 x 1 − . sin3 x x2

Solution The easiest way to go about this problem is using Maclaurin series, however we will save that for later chapters. We will attempt a more basic approach by factoring the above expression. 

 x 1 L = lim − 2 x→0 sin3 x x ! x3 − sin3 x = lim x→0 x2 sin3 x Now we will attempt to convert this into a product of multiple limits. It is easy to see that x3 −sin3 x = (x−sin x)(x2 +x sin x+sin2 x). Hence, !   x − sin x x2 + x sin x + sin2 x L = lim · lim x→0 x→0 x2 sin3 x Both limits are 00 cases so we will proceed to apply L’Hopital’s rule.     1 − cos x 2x + sin x + x cos x + 2 sin x cos x L = lim · lim x→0 3 sin2 x cos x x→0 2x Applying the rule again to the first limit,  L = lim

x→0

 sin x 6 sin x cos2 x − 3 sin3 x   2x + sin x + x cos x + 2 sin x cos x · lim x→0 2x

48

CHAPTER 1. DIFFERENTIAL CALCULUS

One last round to both limits!   cos x L = lim x→0 6 cos3 x − 21 sin2 x cos x · lim

x→0

2 + 2 cos x − x sin x + 2(cos2 x − sin2 x) 2

!

Now the limit is simplified to an easy plug-in. Our limit is therefore: 1 1 ·3= 6 2 Evaluating our limit using Mathematica, L=

In[5]:=

Out[5]=

1.2.2

Limit[x/Sin[x]^3 - 1/x^2, x -> 0] 1 2

Quotient Rule

Theorem g(x) Let f (x) = h(x) where both h and g are differentiable and h(x) 6= 0. The derivative of f (x) is then

f 0 (x) =

g 0 (x)h(x) − h0 (x)g(x) [h(x)]2

This rule can be found from the product rule by setting u = g(x) 1 and v = h(x) . However, it is helpful to know on its own. Example 10: Find

d dx

tan x

49

1.2. THE DERIVATIVE Solution We know that tan x =

sin x cos x

Applying the quotient rule, (tan x)0 =

(sin x)0 cos x − (cos x)0 sin x cos2 x

cos2 x + sin2 x cos2 x 1 = cos2 x We then have by the definition of sec x, =

d tan x = sec2 x dx

1.2.3

Chain Rule

Theorem If g(x) is differentiable at x = a and f (x) is differentiable at x = g(a), then the derivative of f (g(x)) at x = a is: d f (g(x)) = f 0 (g(a)) · g 0 (a) dx x=a After a basic review, we now can start delving into more advanced derivatives. Let us get started with an interesting derivative! x

Example 11: Find the derivative of f (x) = xx at x = 1. Solution We will approach this problem using logarithmic differentiation. We first set h (x) = xx and then take the natural logarithm of both

50

CHAPTER 1. DIFFERENTIAL CALCULUS

Figure 1.7: Graph of y = xx

x

sides, ln h (x) = x ln x Differentiating both sides using the chain rule, h0 (x) = ln x + 1 → h0 (x) = xx (ln x + 1) h (x) We can now return to our original function and take the natural logarithm of both sides to get: ln f (x) = xx ln x Differentiating both sides, f 0 (x) d = (xx ) ln x + xx−1 f (x) dx = xx



1 + ln2 x + ln x x



51

1.2. THE DERIVATIVE Plugging in x = 1, f 0 (1) = 1 → f 0 (1) = 1 1 Checking with Mathematica, In[6]:=

Out[6]=

f[x_]:=xx f0 [1]

x

1

Example 12: Evaluate

dπ π x at x = π dxπ

Solution You might be wondering what a π th derivative even is, or if it is even valid mathematically! This notion of generalizing the derivative operator is the basis of an entire branch of mathematics, fractional calculus. The first appearance of a fractional derivative is in a letter written to l’Hôpital by the German polymath Gottfried Wilhelm Leibniz in 16951 . The theory and foundations of the subject were introduced by the likes of the French mathematician Joseph Liouville in the 19th century. In the late 19th century, the electrical engineer Oliver Heaviside introduced their practical use in electrical transmission line analysis2 . To solve this problem, let’s look at the function f (x) = xa . By the power rule, we can deduce that the first derivative is f 0 (x) = axa−1 1 Katugampola, Udita N. (15 October 2014). A New Approach To Generalized Fractional Derivatives. Bulletin of Mathematical Analysis and Applications. 6 (4): 1–15. arXiv:1106.0965. Bibcode:2011arXiv1106.0965K. 2 Bertram Ross (1977). The development of fractional calculus 1695-1900. Historia Mathematica. 4: 75–89. doi:10.1016/0315-0860(77)90039-8

52

CHAPTER 1. DIFFERENTIAL CALCULUS

Applying differentiation repeatedly gives us a pattern, namely that the nth derivative is

dn a a! x = xa−n n dx (a − n) !

We can extend this definition using the Gamma function. Definition The Gamma function, usually denoted by Γ(·), extends the domain of the factorial function into C with the exception of non-positive integers. Γ(n) = (n − 1)! It is mainly defined by a convergent improper integral. However, the product definition due to Weirstrass is also useda . Z Γ (z) =

 ∞  e−γz Y z −1 z/k e dt = e 1+ z k

z−1 −t

t 0

a

∞

(1.7)

k=1

Davis, P. J. (1959). Leonhard Euler’s Integral: A Historical Profile of the Gamma Function". American Mathematical Monthly. 66 (10): 849–869. doi:10.2307/2309786. JSTOR 2309786

53

1.2. THE DERIVATIVE Definition The Euler-Mascheroni constant, usually denoted as γ, is defined as the limiting difference between the harmonic sum and the natural logarithm:   k X 1 ≈ 0.577216 (1.8) γ = lim − ln k + k→∞ n n=1

P Note: The partial harmonic sum kn=1 written as Hk (Harmonic numbers).

1 n

is sometimes also

Since the argument of the Gamma function is shifted down by 1, we can rewrite the nth derivative as dn a Γ (a + 1) x = xa−n n dx Γ (a − n + 1) After plugging in a = n = π we get: Γ (π + 1) 0 dπ π x = x = Γ (π + 1) ≈ 7.188 dxπ Γ (1) Where we used Mathematica to obtain the numerical approximation. Unfortunately, Mathematica has no built in fractional derivative function to date. Example 13: Let f (x) =

x x+

x

. Find lim f 0 (x). x→∞

x x+ x+..

Solution We will first find an expression for the derivative and then take the limit. Notice that x x f (x) = y = = x x + x+ x x+y x+..

54

CHAPTER 1. DIFFERENTIAL CALCULUS

Therefore, xy + y 2 = x y 2 + xy − x = 0 This is a quadratic equation in y. By the quadratic formula we obtain: √ −x ± x2 + 4x y= 2 Since we know that y > 0 for all x > 0 we will ignore the negative value. Hence, √ −x + x2 + 4x y= 2 To get our desired derivative expression we will differentiate to get:   dy 1 2x + 4 √ = −1 dx 2 2 x2 + 4x   1 x+2 √ = −1 2 x2 + 4x Taking the limit we obtain:   x+2 1 lim f (x) = lim √ −1 x→∞ 2 x→∞ x2 + 4x   1 x+2 1 = lim √ − 2 x→∞ 2 2 x + 4x ! 1 x+2 1 = lim p − 2 x→∞ 2 (x + 2)2 − 4 0

We can see that the denominator tends to x + 2 as x → ∞. Our limit is then simply lim f 0 (x) =

x→∞

1 1 − =0 2 2

55

1.2. THE DERIVATIVE Example 14: Find

Solution Define

d2 x dy 2

in terms of

d2 y dx2

and

dy dx .

g(x) = f −1 (x)

With g(x) = y. Then by the basic properties of inverse functions we have f (g(x)) = x Differentiating both sides with respect to x and using the chain rule, f 0 (g(x)) · g 0 (x) = 1 f 0 (g(x)) =

1 g 0 (x)

Differentiating once again we get: f 00 (g(x)) · g 0 (x) = −

∴ f 00 (g(x)) = −

g 00 (x)
2 g 0 (x)

g 00 (x)
3 g 0 (x)

Expressing this in different notation, d2 y

d2 x 2 = −  dx 3 dy 2 dy dx

Example 15: Let a > 1 be a constant. What is the minimum value of a where the curves f (x) = ax and g(x) = loga x are tangent? Solution

56

CHAPTER 1. DIFFERENTIAL CALCULUS

We know that f and g are inverses, which means they are reflections of each other on the line y = x. Therefore, if they were to be tangent their derivatives have to be both 1. We then have a system of equations ax1 ln a =

1 =1 x1 ln a

ax1 = loga x1

(1) (2)

Where (x1 , x1 ) is the point of intersection. (1) and (2) are true since f 0 = g 0 = 1 at (x1 , x1 ) and f = g at (x1 , x1 ), respectively. Substituting (2) into (1) we obtain: loga x1 ln a = 1 Using base conversion, ln x1 = 1 x1 = e Thus,

1 =1 e ln a e ln a = 1 1

a = ee

1.3

Exercise Problems

1) Prove the product rule using the limit definition of the derivative.

57

1.3. EXERCISE PROBLEMS

2) Prove the chain rule using the limit definition of the derivative.

3) Prove the quotient rule using logarithmic differentiation (See Example 11).

4) Find lim n1/n n→∞

5) Evaluate lim xx x→0+ x

6) Find lim xx − xx x→0+

  x2 sin x1 + x 7) Evaluate lim √ x x→0 1+x−e

8) Let f 0 (x) = f −1 (x). If f (2019) = 2019, what is f 00 (2019)?

9) Evaluate lim

ln





k

k→∞

 10) Evaluate lim  n→∞

kk k!


1/n

4n 3n 
2n n

x2 11) Find lim sin x→∞ 2x + 1 squeeze theorem)

(Bonus question: Generalize it!)

  π (Hint: See example 2 and use the x

58

CHAPTER 1. DIFFERENTIAL CALCULUS Theorem Let (a, b) be an interval such that x0 ∈ (a, b). Let g, f , and h be functions defined on (a, b), except possibly at x0 . Suppose that for every x ∈ (a, b) not equal to x0 , we have: g(x) ≤ f (x) ≤ h(x) Also, suppose that lim g(x) = lim h(x) = L

x→x0

x→x0

Then lim f (x) = L

x→x0

Chapter 2

Basic Integration

59

60

CHAPTER 2. BASIC INTEGRATION

In this chapter, we will explore and review basic integration techniques such as u-substitution, integration by parts, partial fraction decomposition, etc. We will try to apply these methods creatively to a collection of elementary yet unique problems. With that in mind, let’s get started!

Figure 2.1: The notation for the indefinite integral,

Z

, was intro-

duced by Leibniz in 1675. He adapted it from an archaic form of the letter s, standing for summa (Latin for "sum" or "total").

2.1

Riemann Integral

Definition For a function f (x) that is continuous on the interval [a, b], we can divide the interval into n subintervals, each with length ∆x, and from each interval choose a point xk . The definite integral is then Z

b

f (x) dx = lim a

n→∞

n X

f (xk )∆x

k=1

This definition, provided by the German mathematician Bernhard Riemann, expresses area as a combination of infinitely many verticallyoriented rectangles, a technique known as Riemann sums. As the rectangles get thinner (∆x → 0), the total area of the rectangles

2.1. RIEMANN INTEGRAL

61

approaches the value of the integral. Perhaps the most advantageous aspect of the Riemann definition is its easy visualization! Consider the graph of y = x2 in figure 2.2. We will begin by estimating the integral, or area under the curve.

Figure 2.2: Graph of y = x2 on x ∈ [0, 1]

Since it is a nonstandard shape, we will approximate it using standard shapes, namely rectangles. We will start our approximation with two subintervals, or two rectangles.

62

CHAPTER 2. BASIC INTEGRATION

Figure 2.3: Graph of the midpoint Riemann sums for y = x2 on [0, 1] with only two subintervals

Note that this is a midpoint Riemann sum. Our thought experiment here will work for any Riemann sum, since as the rectangles get thinner and thinner, the difference between f (xk ) of different Riemann sums will approach 0. Now, back to our thought experiment. We can see that there are areas that our rectangles do not cover and some areas that our rectangles add. To minimize these errors, we have to add more rectangles! Let us now increase the number of subintervals from 2 to 10. As one can see in figure 2.4, we now have a much better estimate of integral, or the area under the curve. As we let the number of subintervals approach ∞ (and consequently their width approach 0), this approximation will become an exact value for area!

2.2. LEBESGUE INTEGRAL

63

Figure 2.4: Graph of the midpoint Riemann sums for y = x2 on [0, 1] with ten subintervals

However, the Riemann integral has its limitations, which are often worked around by invoking the Lebesgue definition of integration.

2.2

Lebesgue Integral

The Lebesgue integral is another definition of the integral proposed in 1904 by the French mathematician Henri Lebesgue. The formal definition of the Lebesgue integral involves measure theory, which is beyond the scope of the book. A helpful intuitive picture of the Lebesgue integral can be given by thinking of the Lebesgue integral as taking the areas of horizontal rectangles instead of the vertical rectangles one encounters in the Riemann definition.

64

CHAPTER 2. BASIC INTEGRATION

In describing his method, Lebesgue said:1 I have to pay a certain sum, which I have collected in my pocket. I take the bills and coins out of my pocket and give them to the creditor in the order I find them until I have reached the total sum. This is the Riemann integral. But I can proceed differently. After I have taken all the money out of my pocket I order the bills and coins according to identical values and then I pay the several heaps one after the other to the creditor. This is my integral. In essence, the Lebesgue integral approximates the total area by dividing it into horizontal strips instead of the vertical strips in the Riemann definition. The Lebesgue integral asks "for each y-value, how many x-values produce this value?" The Lebesgue integral more or less became the "official" integral of research mathematics. However, rarely do engineers and scientists employ this definition. In most scenarios, the Riemann definition suffices. A funny quote is given by the American electrical engineer and computer scientist Richard Hamming who declared confidently ...for more than 40 years I have claimed that if whether an airplane would fly or not depended on whether some function that arose in its design was Lebesgue but not Riemann integrable, then I would not fly in it. Would you? Does Nature recognize the difference? I doubt it! See footnote for the full article2 .

1

Gowers, T., Barrow-Green, J., Leader, I. (2008). The Princeton companion to mathematics. Princeton, NJ: Princeton University Press. 2 Hamming, R. W. (1998). Mathematics on a Distant Planet. The American Mathematical Monthly, 105(7), 640. doi: 10.2307/2589247

65

2.3. THE U -SUBSTITUTION

2.3

The u-substitution

We start by introducing an extremely useful identity. Proposition. Let f (x) be an integrable function. We then have Z

b

b

Z

f (a + b − x) dx

f (x) dx = a

(2.1)

a

Proof. Let Z I=

b

f (x) dx a

Also, let x = a + b − t, dx = −dt. This then transforms I into: a

Z

f (a + b − t) (−dt)

I= b

b

Z

f (a + b − t) dt

= a

Rewriting using x, Z

b

f (a + b − x) dx

I= a

This formula is akin to "integrating backwards." It will be used extensively throughout this book, and is especially useful for trigonometric integrals. To start the chapter off, we will begin with an elegant result involving (2.1). Example 1: Evaluate

Z 0

π

x sin x dx 1 + cos2 x

66

CHAPTER 2. BASIC INTEGRATION

Figure 2.5: Graph of y =

x sin x 1+cos2 x

Solution Using (2.1) gives: Z I= 0

π



 (π − x) sin (π − x) dx 1 + cos2 (π − x)

Adding our original integral with the transformed integral above, π

 x sin x (π − x) sin (π − x) 2I = + dx 1 + cos2 x 1 + cos2 (π − x) 0  Z  1 π x sin x (π − x) sin (π − x) I= + dx 2 0 1 + cos2 x 1 + cos2 (π − x) Z



We can simplify by using the identities sin (π − x) = sin x,

67

2.3. THE U -SUBSTITUTION cos2 (π − x) = cos2 x: 1 I= 2

Z

π

0



 π sin x dx 1 + cos2 x

We now proceed to make the u-substitution u = cos x, du = − sin x dx π 2

−1

Z 1

 1 −du π π2 = arctan u = 1 + u2 2 4 −1

Evaluating our integral with Mathematica,

In[7]:=

Out[7]=

Z π x Sin[x] dx 0 1+Cos[x]2 π2 4

Example 2: Evaluate

Z

π 2

ln (sin x)dx

0

Solution We will use a unique approach to this problem utilizing our reflection identity, (2.1).  ! Z π Z π 2 2 π −x dx = ln sin ln (cos x) dx I= 2 0 0 Adding our original and transformed integral, Z 2I = 0

π 2

Z ln (sin x) dx + 0

π 2

ln (cos x) dx

68

CHAPTER 2. BASIC INTEGRATION

Figure 2.6: Graph of y = ln (sin x) π 2

Z =

ln (cos x sin x) dx

0

Using 2 sin x cos x = sin (2x), we have:  Z π  Z π 2 2 2 cos x sin x 2I = ln dx = ln (sin 2x) − ln 2 dx 2 0 0 Z =

π 2

ln (sin 2x) dx −

0

π ln 2 2

We proceed to make the u-substitution u = 2x , dx = transforms the above expression to: Z 1 π π ln 2 2I = ln (sin u) du − 2 0 2 The period of sin 2x is 1 2

Z

2π 2

which

= π from which we can deduce that:

π

Z ln (sin u) du =

0

du 2

0

π 2

ln (sin u) du

69

2.3. THE U -SUBSTITUTION Therefore, Z 2I =

π 2

π ln 2 ln (sin u) du − 2 {z }

|0

I

=⇒ 2I = I − Z

π 2

π ln 2 2

ln(sin x) dx = −

0

π ln 2 2

Mathrmatica gives the same answer, Z π 2 Log[Sin[x]]dx 0

In[8]:=

-

Out[8]=

1 π Log[2] 2

Example 3: Evaluate

Z

π

x sin x cos2 x + 1

0


2 dx

Solution Using (2.1), π

(π − x) sin (π − x)
2 dx 0 cos2 (π − x) + 1 Z π π sin (x) x sin (x)
2 dx −
2 dx 2 0 cos (x) + 1 cos2 (x) + 1 {z } | Z

I=

Z = 0

π

I

=⇒ I =

π 2

Z 0

π

sin (x) cos2 (x) + 1


2 dx

(2.2)

70

CHAPTER 2. BASIC INTEGRATION

Figure 2.7: Graph of y =

x sin x 2

(cos2 x+1)

We apply the intuitive u-substitution u = cos x , du = − sin x dx. In doing so we get: Z π 1 du I=
2 −1 u2 + 1 2 Substituting again with u = tan x, du = sec2 (x) dx gives π 4

π 2

Z

π = 2

Z

I=

− π4 π 4

cos2 x =

π I= 4

Z

π 4

− π4

cos2 x dx

− π4

Using Then gives:

sec2 x dx (sec2 x)2

1 + cos 2x 2 Z

dx +

π 4

− π4

! cos(2x) dx

71

2.3. THE U -SUBSTITUTION Which can be easily evaluated to I=

π 4

I=



π +1 2



π2 π + 8 4

Let’s see what Mathematica gives, Z π In[9]:=

Out[9]=

xSin[x] dx 0 (1+Cos[x]2 )2

1 π (2+π) 8

Which, if expanded, equals

Example 4: Evaluate

Z

π2 8

+ π4 .

1


ln Γ(x) dx

0

Solution It is worth noting that this result is due to the well-known Swiss polymath Leonhard Euler. We start our solution by using (2.1), Z I= 0

1


ln Γ(x) dx =

Z

1


ln Γ(1 − x) dx

0

We will now introduce Euler’s reflection formula:

72

CHAPTER 2. BASIC INTEGRATION

Figure 2.8: Graph of the y = ln Γ(x) Theorem The infamous Euler’s reflection formula is given by: (2.3)

Γ(x)Γ(1 − x) = π csc πx Where x ∈ / Z. See (9.3) for proof.

Taking the natural logarithm of both sides in (2.3) gives ln Γ(x) + ln Γ(1 − x) = ln π + ln csc (πx) = ln π − ln sin (πx) Therefore, Z 1 Z

2I = ln Γ(x) + ln Γ(1 − x) dx = 0

1

ln π − ln sin (πx) dx

0

Z

1

Z ln π dx −

2I = 0

1

ln sin (πx) dx 0

73

2.3. THE U -SUBSTITUTION We proceed to substitute u = πx, du = πdx to get: Z 1 π ln sin (u)du 2I = ln π − π 0 It is easy to see that ln (sin u) is symmetric around u = this symmetry, we can write Z

π

Z ln sin (u) du = 2

0

π 2

π 2.

Using

ln sin (u) du

0

We have already evaluated the latter integral in (2.2). Hence, 2I = ln π −

2 −π ln 2 · π 2

= ln π + ln 2 Finally we obtain Z

1

0


ln 2π ln Γ(x) dx = 2

(2.4)

Evaluating with Mathematica, Z 1 Log[Gamma[x]]dx

In[10]:=

0

Out[10]=

1 Log[2 π] 2

Example 5: Define I1 = I2 . Find . I1 Solution

Z 0

∞

1 dx and I2 = 1 + x13

Z 0

∞

1 dx 13 + x13

74

CHAPTER 2. BASIC INTEGRATION

Figure 2.9: Graph of the integrands in I1 and I2 Since

Z

∞

e−ax dx =

0

1 a

We can write I1 as a double integral. Z ∞ Z ∞Z ∞ 1 13 I1 = dx = e−y(1+x ) dy dx 13 1+x 0 0 0 Let t = yx13 , dt = 13x12 y dx, which then transforms I1 into: Z ∞ Z ∞ 1 1 −t − 12 13 I1 = e t dt · e−y y − 13 dy 13 0 0

(2.5)

We now will use the same substitution on I2 . In doing so we obtain: Z

∞Z ∞

13

I2 = e−y(13+x ) dy dx 0 0 Z ∞ Z ∞ 1 1 −t − 12 13 = e t dt e−13y y − 13 dy 13 0 0

75

2.3. THE U -SUBSTITUTION Notice that the only difference is the coefficient of y in the exponential. To get our desired ratio without directly evaluating these integrals, we will change the coefficient of y in I2 through the usubstitution u = 13y, du = 13dy.  − 1 Z ∞ Z ∞ 13 12 1 u −t − 13 −u I2 = e t dt e du 13 · 13 0 13 0  1  Z ∞ Z ∞ 1 13 13 1 −t − 12 −u − = e t 13 dt e u 13 du 13 13 0 0 The expression in parentheses is identical to (2.5). Thus, 12

I2 = 13− 13 I1 =⇒

12 I2 = 13− 13 I1

We can see if Mathematica can generate an exact value, Z∞ In[11]:=

13 Z0 ∞13+x 1 0

Out[11]=

1

1+x13

dx dx

1 12/13 13

Example 6: Evaluate

Z 0

1

 ln cos

πx 2

x2 + x


 dx

Solution We will split the integrand using partial fractions since

76

CHAPTER 2. BASIC INTEGRATION

  ln cos ( πx 2 )

Figure 2.10: Graph of y =

x2 +x

x2 + x = (x)(x + 1)

Z

1

∴I= 0

πx 2

ln cos x




1

Z dx − 0

ln cos πx 2 x+1


dx

Using the trigonometric identity 2 sin x cos x = sin 2x we can derive
sin (πx) that cos πx = . Thus, 2 2 sin ( πx 2 )  Z

1

ln

I=

sin (πx) 2 sin ( πx 2 )

x

0

 1

Z dx − 0

ln cos πx 2 x+1


dx

By the reflection property of integration, Z I= 0

1

ln



sin (πx) 2

x

 Z dx − 0

1

 ln sin x

πx 2




Z dx − 0

1

ln cos πx 2 x+1


dx

77

2.3. THE U -SUBSTITUTION

Z = 0

|

1

ln



sin (πx) 2

x {z

(1)



1

Z dx − } |0

   π(1−x) ln sin 2 1−x {z (2)

Z dx − } |0

1

ln cos πx 2 x+1 {z




(3)

dx }

We will proceed to substitute u = −x , du = −dx in (2)       Z 0 ln cos πu
Z 1 ln sin (πx) Z 1 ln cos πx
2 2 2 I= dx − du − dx x u+1 x+1 −1 0 0 {z } | {z } | {z } | (1)

(3)

(2)

|

{z

Same integrand

Combining (2) and (3) we obtain     Z 1 ln cos πx
Z 1 ln sin (πx) 2 2 dx − dx I= x x+1 |0 {z } | −1 {z } (1)

(2*)

Now, we substitute u = 1 + x , du = dx into (2*)     Z 2 ln sin πu
Z 1 ln sin (πx) 2 2 I= dx − du x u 0 0 Now, since our integrands are not defined at 0, we need to take the limit as the lower bound approaches 0. Notice that if we were attempting to evaluate the indefinite integral, this step would not be needed at this time. However, it is not possible to derive the above indefinite integrals (anti-derivatives) in standard mathematical functions, so we have to go with the definite integral route.       Z 1 ln sin (πx) Z 2 ln sin πx
2 2   I = lim  dx − dx h→0 x x h h

}

78

CHAPTER 2. BASIC INTEGRATION 


 πx




Z 1 Z 2 ln sin  ln sin (πx) = lim  dx − h→0  x x | h {z } |h {z (1)

2

(2*)

Substituting again into (2*) with u = 

x 2

, du =

Z dx − }

1

  ln 2 dx  x 

h

dx 2



Z 
Z 1  1 ln sin (πx)
 ln sin (πu)   I = lim  dx − du + ln 2 ln h h  h→0  h x u |  {z } |2 {z } (1)

(2*)

Combining (1) and (2*) we obtain: # " Z
h ln sin (πx) I = lim − dx + ln 2 ln h h h→0 x 2

As x → 0 , sin x ≈ x. Since h → 0, we can use this in the above integral to get " Z I = lim − h→0

" Z = lim − h→0

h h 2

h h 2

ln (πx) dx + ln 2 ln h x

ln (x) dx − x

Z

h h 2

#

# ln π dx + ln 2 ln h x

Our first integral above can be calculated using Z Therefore,  I = lim − h→0

"



ln x (ln x)2 dx = x 2

h 1 (ln h)2 − ln 2 2

2 #

"



− ln π ln h − ln

h 2

#

 + ln 2 ln h

2.3. THE U -SUBSTITUTION

"

1 = lim − h→0 2

79

#    h h ln h + ln ln h − ln + ln 2 ln h − ln π ln 2 2 2

Where in the last line we used the identity a2 − b2 = (a + b)(a − b). Continuing with our solution,   ! 2 1 h I = lim − ln (ln 2) + ln 2 ln h − ln π ln 2 h→0 2 2  1 = lim − (2 ln h − ln 2) (ln 2) + ln 2 ln h − ln π ln 2 h→0 2 

The terms containing h will cancel out so we are only left with: I=

ln2 2 − ln 2 ln π 2

Oh well, that was a lot of work! To be confident in our result, we can check with Mathematica. Unfortunately, Mathematica only provides a numerical answer. However, we can check the difference between our value above and Mathematica’s numerical approximation,

In[12]:=

Out[12]=

Z 1 Log[Cos[ π x ]] 2 N[ dx] 0 x2 +x Log[2]2 -( -Log[ π ] Log[2]) 2 -8.881784197001252*10^-16

Here, N [·] denotes Mathematica’s numerical value function.

80

CHAPTER 2. BASIC INTEGRATION

Example 7: Evaluate

1

Z 0

x ln x √ dx 1 − x2

Figure 2.11: Graph of y =

√x ln x 1−x2

Solution√ x Let u = 1 − x2 , du = − √1−x dx. Our integral is then 2 Z I=−

0

ln

p 1 − u2 du

1

=

1 2

Z

1

  ln 1 − u2 du

0

We can split I into two integrals using basic logarithm rules, 1 I= 2

Z

1

ln (1 + u) + ln (1 − u) du 0

81

2.3. THE U -SUBSTITUTION 1 = 2

"Z

1

Z

ln (1 − u) du

ln (1 + u) du + 0

#

1

0

 i2 h i0  1 h = u ln u − u − u ln u − u 2 1 1

Since limu→0 u ln u = 0,

Z ∴ 0

1

2 ln 2 − 2 x ln x √ dx = = ln 2 − 1 2 1 − x2

(2.6)

Checking with Mathematica,

In[13]:=

Out[13]=

Z 1 xLog[x] q dx 0 2 1-x

-1+Log[2]

We saw how u-substitutions can make seemingly hard integrals simple. In the next section, we will see some other elementary ways to solve integrals.

82

CHAPTER 2. BASIC INTEGRATION

2.4

Other Problems

Theorem Integration by parts, also known as IBP, is one of the most common integration techniques. It is derived from the product rule, (uv)0 = u0 v + v 0 u And is its analogue in integration. The rule states that Z Z u dv = uv − v du (2.7) For two continuous and differentiable functions u and v. For definite integrals, we have Z

b

u dv = a

[uv]ba

Z

b

−

v du a

This result can be extended to three functions, u, v, w: Z a

b

Z b h ib Z b uv dw = uvw − uw dv − vw du a

a

a

This technique was first discovered by the English mathematician Brook Taylor, who also discovered Taylor series3 . Proof. Consider the statement of the product rule: (uv)0 = u0 v + v 0 u 3

TAYLOR, B. (1715). Methodus Incrementorum directa et inversa. Londini: Apud Gul. Innys.

83

2.4. OTHER PROBLEMS

Define u = u(x) and v = v(x), i.e. u, v are functions of x. Integrating both sides with respect to x, Z

0

Z

(uv) =

Z uv =

u0 v + v 0 u

0

Z

v0u

Z

u0 v

uv+

Rearranging this formula gives: Z

v 0 u = uv −

Expressing this in Leibniz notation, Z

Z u dv = uv −

v du

This can be easily applied to definite integrals Z

b

u dv = a

[uv]ba

Z −

b

v du a

By the first fundamental theorem of calculus. Now, let’s apply our skills! Example 8: Find Solution

Z

∞

arcsin e−x dx

0

Sometimes it is more convenient to do a substitution in terms of functions on both sides. In essence, the substitution form is:

84

CHAPTER 2. BASIC INTEGRATION

Figure 2.12: Graph of y = arcsin e−x

f (u) = g(x)
Instead of u = f −1 g(x) . Also, the introduction of another variable, namely u, is not necessary. One can do a direct substitution in the same variable. In this case, we would like to substitute e−x → sin x. For the sake of ease, we can also write this as e−x = sin u, e−x dx = − cos u du =⇒ dx = − cot u du and obtain that we need to substitute dx → − cot x dx. As one gains more practice with such procedures, it will become almost second nature! Z I=

π 2

x cot x dx

0

Using integration by parts with u = x, du = dx and dv = cot x dx, v = ln sin x leads us to π 2

I = [x ln sin x]0 −

Z 0

π 2

ln sin x dx

85

2.4. OTHER PROBLEMS π ln 2 2 See (2.2) for the evaluation of the integral. We can check Mathematica to verify our solution, I=

Z ∞ In[14]:=

ArcSin[e-x ]dx

0 1 π Log[2] 2

Out[14]=

Example 9: Find

Z

sin(2019x) sin2017 x dx

Solution We will show that for any n, Z In =

n−2

sin(nx) sin


sinn−1 x sin (n − 1)x x dx = +C n−1

We will start by splitting up n into n − 1 and 1 Z In =


sin (n − 1)x + x sinn−2 x dx

By using the fact that sin(α + β) = sin α cos β + sin β cos α We obtain: Z

In = sin (n − 1)x cos(x) sinn−2 x+sin x cos (n − 1)x sinn−2 x dx

86

CHAPTER 2. BASIC INTEGRATION Z =



sin (n − 1)x cos(x) sinn−2 x + cos (n − 1)x sinn−1 x dx

We proceed to multiply by

1 In = n−1

n−1 n−1

Z sin (n − 1)x

to get




 (n − 1) cos(x) sinn−2 x
+ (n − 1) cos (n − 1)x sinn−1 x dx

Notice that  And



0 sinn−1 x = (n − 1) cos(x) sinn−2 x



 0 = (n − 1) cos (n − 1)x sin (n − 1)x

Therefore,

In =

1 n−1

Z



0 sin (n − 1)x (sinn−1 x)0 + sin (n − 1)x sinn−1 x dx


0 The integrand is simply sinn−1 x sin (n − 1)x by the product rule. Therefore,

In =

1 n−1

Z 

0
sin (n − 1)x sinn−1 x dx


sin (n − 1)x sinn−1 x = +C n−1 We can now plug in our desired value, n = 2019

87

2.4. OTHER PROBLEMS

Figure 2.13: Graph of the y =

I2019 =

√ n

sinn x + cosn x for n = 1000

sin(2018x) sin2018 x +C 2018

Mathematica gives Z In[15]:=

Out[15]=

Sin[2019 x] Sin[x]2017 dx

Sin[x]2018 Sin[2018 x] 2018

Example 10: Find lim

Z

n→∞ 0

π 2

√ n

sinn x + cosn x dx

88

CHAPTER 2. BASIC INTEGRATION

Solution Let A = lim

n→∞

√ n

sinn x + cosn x = lim

n→∞



 √ cos x n 1 + tann x

Where we factored out cosn x in the last step. Also, let √ B = lim n 1 + tann x n→∞

ln (1 + tann x) n→∞ n

=⇒ ln B = lim



Notice that when x ∈ 0, π4 , tan x ∈ [0, 1). Hence, when x ∈ 0, π4 , we know that ln B = 0 and B = 1
since as n → ∞, tann x = 0. On the other hand, when x ∈ π4 , π2 , tan x ∈ [1, ∞). We can then √ n deduce that B = tann x = tan x. Our integrand is then simply ( 
cos x , x ∈ 0, π4 
B cos x = cos x tan x = sin x , x ∈ π4 , π2 Even though tan x diverges at x = π2 , we can see that lim A = limπ B cos x = 1

x→ π2

x→ 2

Therefore, π 2

Z lim

n→∞ 0 π 4

Z =

√ n

sinn x + cosn x dx

π 2

Z cos x dx +

sin x dx

π 4

0 π

π

= [ sin x]04 − [ cos x] π2 4

89

2.4. OTHER PROBLEMS =

√

2

For n = 1000, Mathematica gives ≈ 1.414.

Example 11: Find an expression for

Z

∞

0

Figure 2.14: Graph of y =

1 1+xn

1 dx for n > 1. 1 + xn

for n = 2, 3, · · · , 7

Solution We will express this integral as a double integral similar to what we did in (2.5). In doing so we obtain: Z ∞Z ∞ n I= e−y(1+x ) dx dy 0

0

Then we substitute u = yxn , du = nxn−1 y dx . Therefore, our double integral expressed as a product of two integrals is Z ∞Z ∞ Z Z ∞ 1 1 ∞ −y − 1 −y −yxn n I= e e dx dy = e y dy u n −1 e−u du n 0 0 0 0

90

CHAPTER 2. BASIC INTEGRATION

  Notice that the first integral is simply Γ 1 − n1 and the second   integral is Γ n1 . Here, Γ denotes the Gamma function as usual (See (1.7)). Using Euler’s reflection formula, (2.3), we can write this as   Z ∞ π π 1 dx = csc (2.8) n 1+x n n 0 Checking with Mathematica, Z ∞ In[16]:=

0

1 dx 1+xn

π Out[16]=

ConditionalExpression[

Example 12: Evaluate the integral

Z 0

constant. Solution We begin by substituting x =

a2 u

∞

Z I(a) = 2 0

Z = 4 ln a 0

∞

ln x2 dx where a is a x2 + a2

2

, dx = − ua2 du

a2 (2 ln a − ln u)   du 4 u2 ua2 + a2

∞

2 ln a − ln u du a2 + u2 0 Z ∞ du ln u −2 du 2 2 2 a +u a + u2 0 | {z }

Z =2

∞

π Csc[ ] n ,Re[n]>1] n

I(a)

91

2.4. OTHER PROBLEMS Therefore,

Z 2I(a) = 4 ln a 0

∞

a2

du + u2

The substitution u = a tan y , du = a sec2 y dy gives: Z π Z π 2 2 ln a 2 a sec2 y
dy = dy I(a) = 2 ln a 2 2 a 0 a tan y + 1 0 π ln a a Here, Mathematica returns the same result. =⇒ I(a) =

In[17]:=

Z ∞ Log[x2 ] dx 0 x2 +a2 √

Example 13: Find

Z

6

q

4x2 +1 x2 −1

x

1

dx

Solutionq 2 +1 Let u = 4x . Therefore, x2 −1 u2 =

4x2 + 1 x2 − 1

4(x2 − 1) + 5 x2 − 1 5 =⇒ u2 − 4 = 2 x −1 =

We now need to obtain an expression for dx in terms of u only. With some algebraic manipulation we obtain: 1 x2 − 1 = u2 − 4 5

92

CHAPTER 2. BASIC INTEGRATION

r

Figure 2.15: Graph of y =

4x2 +1 x2 −1

x

5 u2 − 4 + = x2 u2 − 4 u2 − 4 x2 =

u2 + 1 u2 − 4

(2.9)

Differentiating both sides, 2x dx =

Now we proceed to find

dx x .

−10u du (u2 − 4)2

We start by writing dx 2x = 2 dx x 2x

Plugging in our results from (2.9) and (2.10), dx −10u u2 − 4 = 2 du · x (u − 4)2 2(u2 + 1)

(2.10)

93

2.4. OTHER PROBLEMS =

(u2

Therefore,

∞

Z I=

√

5

−5u du − 4)(u2 + 1) 5u2 du (u2 − 4)(u2 + 1)

Using the method of partial fractions we obtain: Z

∞

4 1 + 2 du 2−4 u u +1 5  Z ∞ h i∞ 1 1 = √ − du + arctan u √ u−2 u+2 5 5 "  # ∞  √ u−2 π = ln + − arctan 5 u+2 √ 2 5 ! √ √ 5−2 π = − ln √ + − arctan 5 2 5+2 I=

√

Simplifying,

 √  π √ I = 2 ln 2 + 5 + − arctan 5 2 Let’s see what Mathematica gives, s √ Z In[18]:=

1

Out[18]=

6

4x2 +1 x2 -1 x

dx

1 2 ArcSinh[2]+ArcTan[ √ ] 5

Which are equivalent since sinh−1 x = ln

p  x2 + 1 + x

94

CHAPTER 2. BASIC INTEGRATION

And arctan x =

Example 14: Evaluate

Z

1

π − arctan x−1 2

arcsin



1+

0

2x 1+x2 x2



Figure 2.16: Graph of y =

dx



2x 1+x2 2 1+x

arcsin



Solution In general, a 1 + x2 term in the denominator is a tell-tale sign that the substitution x = tan u will work, so we begin by letting x = tan u , dx = sec2 u du. Our integral is then transformed to

Z I= 0

π 4

arcsin



1+

2 tan u 1+tan2 u tan2 u

 · sec2 u du

95

2.4. OTHER PROBLEMS Z I=

π 4

 arcsin

0

 2 tan u du 1 + tan2 u

We proceed to apply the trigonometric identity: sin u =

2 tan u2 2 tan u u =⇒ sin 2u = 2 1 + tan 2 1 + tan2 u

Which gives π 4

Z I=

arcsin (sin 2u) du

0

On the interval [0, π4 ], arcsin (sin 2u) = 2u. Therefore, Z I=

π 4

2u du

0

I=

π2 16

Mathematica also gives this value as well 2 x ] 1+x2 dx 1+x2

Z 1 ArcSin[ In[19]:=

0

Out[19]=

π2 16

We can end this chapter with an interesting formula

(2.11)

96

CHAPTER 2. BASIC INTEGRATION Theorem Let f (x) be an even Riemann-integrable function on [−α, α] and g(x) an odd Riemann-integrable function on [−α, α], where α ∈ R. We then have: Z α Z α f (x) dx = f (x)dx (2.12) g(x) −α 1 + b 0 For any b ∈ R+ .

Proof. Notice that we can write (2.12) as: Z

α

−α

f (x) dx = 1 + bg(x)

Z

0

Z α f (x) f (x) dx + dx g(x) g(x) 1+b 0 1+b {z } | {z }

| −α

I1

I2

The substitution u = −x, du = −dx in I1 gives: α

Z I1 = 0

f (−u) du 1 + bg(−u)

Using the fact that f (u) is even and g(u) is odd, we have Z

α

f (u) du 1 + b−g(u)

α

f (u)bg(u) du 1 + bg(u)

I1 = 0

Multiplying by

bg(u) , bg(u) Z I1 = 0

Hence,

97

2.4. OTHER PROBLEMS

Z

α

f (x) dx = I1 + I2 g(x) −α 1 + b Z α Z α f (u)bg(u) f (x) = du + dx g(u) g(x) 0 1+b 0 1+b   Z α 1 + bg(u) f (u) = du 1 + bg(u) 0 α

Z =

f (u) du 0 α

Z =

f (x) dx 0

Example 15: Evaluate

Z

π

−π

sin2 x dx 1 + ex3

Solution Notice that we can use (2.12) since f (x) = sin2 x is an even function on [−π, π] and g(x) = x3 is an odd function on [−π, π]. We then have: π

Z I=

−π

Z =

sin2 x dx 1 + ex3

π

sin2 xdx

0

=

iπ 1h x − sin x cos x 2 0

=

π 2

98

CHAPTER 2. BASIC INTEGRATION

Figure 2.17: Graph of y =

sin2 x 1+ex3

Verifying with Mathematica,

In[20]:=

Out[20]=

Z π Sin[x]2 dx π 1+ex3 -π π 2

And here we see the beauty of generalization!

In the first two chapters we saw how our elementary calculus tools can solve a plethora of problems through creative application of these tools. In nonstandard mathematical problems such as the ones presented in this book, expanding one’s mathematical toolkit

2.5. EXERCISE PROBLEMS

99

is of tremendous help. However, the mastery of one’s toolkit will lead to far more ingenious and creative solutions! In the proceeding chapters, we will explore and master additional tools to add to your toolkit.

2.5

Exercise Problems

1) Evaluate

Z

∞

0

2) Evaluate

Z

1

0

3) Find

Z

tan x dx sin x(sin x + cos x)

0

√

4) Find

ln(1 + x) dx 1 + x2

√

π 2

Z

dx (Hint: u3 + 1 = (u2 − u + 1)(u + 1)) 3 (1 + x) + 1

3

dx 1 + x2

1

5) Evaluate

Z

π

√


3/2

1 − sin x dx

0

6) Evaluate

Z

π/2

0

7) Find

Z 0

∞

x2

dx (Hint: 2019 is a distractor) 1 + tan2019 x

ln x dx (Hint: Let u = x4 ) + 2x + 4

100 8) Find (2.11))

CHAPTER 2. BASIC INTEGRATION Z

2π

0

9) Show that

3 − cos x dx (Hint: Use the substitution u = tan x2 . See 3 + cos x

22 7

> π using the integral

1968 Putnam competition) 10) Define I(a) =

Z 0

π 4

1

Z 0

x4 (1 − x)4 dx (Source: 1 + x2

ex tana x dx. Find lim aI(a) a→∞

Chapter 3

Feynman’s Trick

101

102

3.1

CHAPTER 3. FEYNMAN’S TRICK

Introduction

A straightforward but extremely effective technique, differentiation under the integral sign is simply a clever use of Leibniz’s integral rule. In mathematical pop culture, it is often labeled as "Feynman’s trick" after the late American physicist Richard Feynman (1918-1988). Feynman said this in discussing his "trick": One thing I never did learn was contour integration. I had learned to do integrals by various methods shown in a book that my high school physics teacher Mr. Bader had given me. The book also showed how to differentiate parameters under the integral sign - It’s a certain operation. It turns out that’s not taught very much in the universities; they don’t emphasize it. But I caught on how to use that method, and I used that one damn tool again and again. So because I was self-taught using that book, I had peculiar methods of doing integrals. The result was that, when guys at MIT or Princeton had trouble doing a certain integral, it was because they couldn’t do it with the standard methods they had learned in school. If it was contour integration, they would have found it; if it was a simple series expansion, they would have found it. Then I come along and try differentiating under the integral sign, and often it worked. So I got a great reputation for doing integrals, only because my box of tools was different from everybody else’s, and they had tried all their tools on it before giving the problem to me. The last sentence perhaps best summarizes not only the purpose of this chapter but the entire book as well. Often in engineering and the mathematical sciences, peculiar and nonstandard integrals arise. And unfortunately, few resources are available for solving such integrals! The goal of this book is to provide the reader with a plethora of techniques and methods to go about such problems. So, with

103

3.2. DIRECT APPROACH that in mind, let us get started!

Theorem Let f (x, α) be a differentiable function in α with uous. Then the following equality holds: d dα

Z

b

Z f (x, α) dx =

a

a

b

∂ ∂α f

contin-

∂ f (x, α) dx ∂α

See footnote for a rigorous justification of the above theorem1 .

3.2

Direct Approach

In this method, an integral f (α) is defined with the parameter α such that when the integral is differentiated, a standard integral is found. After deriving a closed form for f 0 (α), one integrates the closed form and obtains an expression for f (α), of course with a +C at the end. After an expression is derived for f (α), an initial is needed to determine the added constant ”C”. In most integrals, α = 0 or α = ∞ is easy or even trivial to compute and therefore chosen as an initial condition. Now, what does this look like? Let us take a look at a well-known example.

Example 1: Find the value of 1

Z 0

∞

sin2 x dx x2

Hijab, O. Introduction to Calculus and Classical Analysis. New York: Springer-Verlag, p. 189, 1997.

104

CHAPTER 3. FEYNMAN’S TRICK

Figure 3.1: Graph of y =

sin2 x x2

Solution We first use IBP with u = sin2 x and dv = Z I= 0

∞

dx . x2

" #∞ Z ∞ sin2 x sin2 x 2 sin x cos x dx = − + dx x2 x x 0 0

∞

sin 2x dx x

∞

sin x dx x

Z =0+ 0

Substituting 2x → x, Z I= 0

This is the famous Dirichlet integral, named after the German mathematician Peter Dirichlet. Now, consider an integral of the

105

3.2. DIRECT APPROACH form Z

∞

f (α) = 0

e−αx sin x dx x

Where α ≥ 0. Our original integral is I = f (0). Differentiating under the integral sign, Z

0

f (α) = −

∞

e−αx sin x dx

0

=−

1 1 + α2

Where we used IBP to get our last result. We can now integrate to get an expression for f (α), Z f (α) = −

1 dα = − arctan α + C 1 + α2

We know that lim f (α) = 0 =⇒ C =

α→∞

Hence f (α) =

π 2

π − arctan α 2

Plugging in our desired value, f (0) =

π 2

This example serves as a perfect example of thinking outside the box!   Z ∞ x ln x3 +1 3 x Example 2: Evaluate dx 3 1 + x 0

106

CHAPTER 3. FEYNMAN’S TRICK



Figure 3.2: Graph of y =

x ln

x3 +1 x3



1+x3

Solution Let

Z

∞

x ln



x3 +α x3 + x3



f (α) = dx 1 0   Z ∞ x ln x3 + α
− ln x3 =

1 + x3 Differentiating under the integral sign,

dx

0

0

Z

∞

f (α) = 0

The substitution u = x3 , dx =

x3 du 2

3u 3

x
dx + α (1 + x3 ) gives

107

3.2. DIRECT APPROACH

f 0 (α) =

1 3

Z

∞

du 1 3

u (α + u)(1 + u)

0

Using the method of partial fractions, "Z # Z ∞ ∞ 1 du du f 0 (α) = − 1 3α − 3 0 u 13 (1 + u) 0 u 3 (α + u)

(3.1)

We can use the fact that Z

∞

du u (u + α)

0

2π 1√

=

1 3

α3 3

The proof is left as an exercise to the reader (Hint: Make the sub1 stitution y = u 3 and then use the method of partial fractions). Plugging in that result into (3.1) gives: " # 1 2π 2π √ − 1√ f (α) = 3α − 3 3 α3 3 0

Simplifying algebraically, 2π f 0 (α) = √ 3 3

!

1 1

2

α3 + α3 + α

(3.2)

Since f (0) = 0, we can take the definite integral of (3.2) from 0 to 1 to get our desired integral, f (1). 1

Z f (1) =

f 0 (α) dα

0

2π = √ 3 3

Z 0

1

dα 1 3

2

α + α3 + α

108

CHAPTER 3. FEYNMAN’S TRICK 2π = √ 3 3

1

Z

dα α

0

1 3



1

2 3

α + α3 + 1



1

The substitution u = α 3 , dα = 3u2 du yields: 2π f (1) = √ 3 2π =√ 3

"Z

1

0

Z 0

1

u2

u du +u+1

2u + 1 du − 2 2(u + u + 1)

Z 0

1

du 2 2(u + u + 1)

#

Both integrals can be easily evaluated (The first by the substitution y = u2 + u + 1, dy = (2u + 1) du and the second by completing the square). Therefore, " √ # 2π ln 3 π 3 f (1) = √ − 18 3 2 √ π 3 ln 3 π 2 = − 3 9

Example 3: Evaluate

Z

π 2

− π2

ln (1 + α sin x) dx for |α| < 1. sin x

Solution Define Z f (α) =

π 2

− π2

We then have

ln (1 + α sin x) dx sin x

109

3.2. DIRECT APPROACH

Figure 3.3: Case when α = .5

f 0 (α) =

Z

π 2

− π2

1 dx 1 + α sin x

A standard substitution here is the half angle tangent substitu
tion, u = tan x2 , dx = u22+1 du. This lets us express sin x as

sin(x) =

2u 1 + u2

Thus, f 0 (α) =

Z

1

−1

Z

1

=2 −1

2 du u2 + 2αu + 1

du ( 1 − α2 )2 + (u + α)2 √

110

CHAPTER 3. FEYNMAN’S TRICK

This integral is in the form: Z a

b

"  #b dx 1 x = arctan x2 + d2 d d

a

Therefore, Z

1

du 2 2 2 −1 ( 1 − α ) + (u + α) "  #1 2 u+α = √ arctan √ 1 − α2 1 − α2 −1 0

√

f (α) = 2

After a tedious round of algebra and simplification, we obtain: f 0 (α) = √

π 1 − α2

Thus, Z f (α) = Z =

√

f 0 (α) dα

π dα 1 − α2

= π arcsin α + C Notice that Z f (0) =

π 2

− π2

ln 1 dx = 0 sin x

=⇒ C = 0 Therefore,

111

3.2. DIRECT APPROACH

Z f (α) =

π 2

− π2

Example 4: Find

Z 1

∞

ln (1 + α sin x) dx = π arcsin α sin x

ln(ln x) dx for α > 1. xα

Figure 3.4: Graph of y =

ln(ln x) xα

Solution Let

Z f (α) = 1

∞

ln(ln x) dx xα+1

for α = 2

112

CHAPTER 3. FEYNMAN’S TRICK

Substituting y = ln x, dy =

dx x ,

Z

∞

f (α) = 0

ln y dy eαy

We used α + 1 instead of α to make the aforementioned substitution much cleaner. This is one of many examples where the parameter choice is not immediately obvious, and some vision is required. Now, we can differentiate under the integral sign to get ∞

Z

0

f (α) = −

ye−αy ln y dy

0

By applying IBP with u = −y ln y, dv = e−αy dy we get: "

y ln ye−αy f (α) = α

#∞

0

0

1 − α

∞

Z

e−αy ln y + e−αy dy

0

We first need to evaluate the first expression: "

y ln ye−αy B= α This is an

∞ ∞

#∞

 = lim

y→∞

0

# "  y ln y ln y − lim eαy α y→0 eαy α y

case in both limits, so we can apply L’Hopital’s rule: ln y + 1 − lim y→∞ α2 eαy y→0

B = lim

= lim

y→∞

1 y α2 yeαy −αeαy y2

ln y + 1 y − lim 2 αy y→0 α ye α2 eαy − αeαy

The latter limit can be easily evaluated to equal 0, but the first limit needs one more round of L’Hopital’s rule as it is still a ∞ ∞ sit-

113

3.2. DIRECT APPROACH uation: B = lim

y→∞

= lim

y→∞

1 y

α3 eαy

1 α3 yeαy

=0 Therefore, Z 1 ∞ −αy e ln y + e−αy dy f (α) = 0 − α 0 Z ∞  Z ∞ 1 −αy −αy =− e ln y dy + e dy α 0 0 0

The first integral is simply f (α) and the latter integral is a standard integral. We then get the differential equation f 0 (α) = −

f (α) 1 − 2 α α

We can rearrange the above differential equation and multiply both sides by α to obtain: αf 0 (α) + f (α) = −

1 α

d 1 αf (α) = − dα α Integrating both sides, αf (α) = − ln α + C f (α) = −

ln α + C α

114

CHAPTER 3. FEYNMAN’S TRICK

Since

∞

Z f (α) =

e−αy ln y dy

0

A standard value would be f (1) = −γ. Thus, f (α) = − Z ∴ 1

∞

ln α + γ α

ln(ln x) ln(α − 1) + γ dx = f (α − 1) = − α x α−1

Example 5: Evaluate

Z

∞

2

e−x cos(5x) dx

0

2

Figure 3.5: Graph of y = e−x cos(5x)

Solution

115

3.2. DIRECT APPROACH Define a function: ∞

Z f (α) =

2

e−x cos(αx) dx

0

Differentiating under the integral sign, Z

0

f (α) = −

∞

2

xe−x sin(αx) dx

0 2

We can now use IBP with u = sin(αx), dv = xe−x dx to obtain: f 0 (α) =

i∞ α 1 h −x2 e sin(αx) − 0 |2 {z } 2 =0−0=0

Z

∞

|0

2

e−x cos(αx) dx {z } f (α)

Notice that we can set up a simple differential equation: α f 0 (α) = − f (α) 2 We can now transition to using Leibniz notation to make solving this a little easier to digest. α df =− f dα 2 Multiplying both sides by

dα , f df α = − dα f 2

Z

df =− f

Z

α dα 2

116

CHAPTER 3. FEYNMAN’S TRICK ∴ ln f = −

α2 +C 4

Exponentiating both sides, f (α) = eC e−

α2 4

The case when α = 0 is half of the famous Gaussian integral: 1 2

Z

∞

e

−x2

∞

Z dx =

−∞

√ −x2

e

dx =

0

π 2

See (9.9) for proof. This integral is core to many disciplines due to its applications in statistics. √ π − α2 =⇒ f (α) = e 4 2 To get our desired integral, we evaluate f (5) to get: √ f (5) = Z

∞

−x2

e

∴

π − 52 e 4 2

√

cos(5x) dx =

0

Example 6: Find

Z 0

π 4


arctan 4 sin(2x) dx sin(2x)

Solution The substitution u = 2x, du = 2 dx gives

π − 25 e 4 2

117

3.2. DIRECT APPROACH

Figure 3.6: Graph of the y =

1 I= 2

π 2

Z 0

arctan(4 sin(2x)) sin(2x)

arctan(4 sin u) du sin u

Now, define a function: π 2

Z f (α) = 0

arctan(α sin u) du sin u

Differentiating under the integral sign,

f 0 (α) =

Z 0

We can multiply by

π 2

1 du 1 + α2 sin2 u

sec2 u to get: sec2 u

118

CHAPTER 3. FEYNMAN’S TRICK

π 2

Z

0

f (α) = 0

sec2 u du sec2 u + α2 tan2 u

Using the fact that sec2 u = 1 + tan2 u we can write: π 2

Z

0

f (α) = 0

sec2 u du tan2 u + 1 + α2 tan2 u

π 2

sec2 u
du α2 + 1 tan2 u + 1 0 Z π 2 1 sec2 u = 2 du α + 1 0 tan2 u + α21+1 Z

=

The substitution y = tan u, dy = sec2 u du then gives: 1 f (α) = 2 α +1 0

∞

Z 0

y2

dy + α21+1

Which is a standard integral that evaluates to π f 0 (α) = √ 2 α2 + 1 Integrating both sides, Z

Z

0

f (α) dα = =

=

π dα 2 α2 + 1 √

π sinh−1 α + C 2

 p π  ln α + α2 + 1 + C 2

119

3.2. DIRECT APPROACH Using f (0) = 0 lets us determine the value of C. Hence,

f (α) =

 p π  ln α + α2 + 1 2

√  1 π  ∴ I = f (4) = ln 4 + 17 2 4

Example 7: Evaluate

Z 0

∞Z ∞ 0

sin x sin y sin(x + y) dx dy xy(x + y)

Figure 3.7: Three-dimensional plot of integrand

Solution Recall that

120

CHAPTER 3. FEYNMAN’S TRICK

Figure 3.8: Contour plot of integrand

sin(x + y) = sin x cos y + cos x sin y Thus, Z

∞Z ∞

I= 0

0

(sin x cos y + cos x sin y) sin x sin y dx dy xy(x + y)

By symmetry, Z

∞Z ∞

I=2 0

0

sin2 x cos y sin y dx dy xy(x + y)

121

3.2. DIRECT APPROACH Using the fact that 2 sin θ cos θ = sin 2θ, we have: Z

∞Z ∞

I= 0

0

sin2 x sin 2y dx dy xy(x + y)

We can turn the above double integral into a triple integral: ∞Z ∞Z ∞

Z I= 0

0

0

Since

Z

sin2 x sin 2y −z(x+y) e dz dx dy xy

∞

1 a

e−ax dx =

0

We will now introduce an elementary theorem.

Theorem Let f (x) and g(y) be integrable functions. Then the following equality holds: Z bZ

d

f (x)g(y)dy dx = a

c

! Z

b

Z

f (x) dx a

!

d

g(y) dy c

Proof. Note that for any constant k (i.e. independent of x), Z

b

b

Z kf (x) dx = k

a

f (x) dx a

Also, Z bZ

d

Z

b

Z

f (x)g(y)dy dx = a

c

!

d

f (x)g(y)dy a

c

dx

122

CHAPTER 3. FEYNMAN’S TRICK b

Z

Z f (x)

g(y)dy

a

Now, since

Rd c

!

d

dx

c

g(y)dy is a constant, we can write ! ! Z d Z b = g(y)dy · f (x)dx c

a

Using this theorem, we can split the integral above so that we have a product of integrals: Z

∞

∞

Z

I= 0

0

! Z  ∞ sin2 x −zx sin 2y −yz e dx e dy dz x y 0

Now, define two functions: Z

∞

sin2 x −zx e dx x

∞

sin 2y −yz e dy y

f1 (z) = 0

Z f2 (z) = 0

Both these integrals can be easily solved using Feynman’s trick (Differentiation under the integral sign). Starting with f1 we get: Z ∞ 0 f1 (z) = − sin2 x e−zx dx 0

=− =−

1 2

Z 0

1 2

∞

Z

∞

(1 − cos 2x) e−zx dx

0

e−zx dx −

Z 0

∞

cos(2x) e−zx dx



123

3.2. DIRECT APPROACH

The second integral can be easily evaluated using IBP. We therefore have: 1 z − + 8 2z

f10 (z) =

2z 2

Thus, Z f1 (z) =

=

1 2

f10 (z) 

1 dz = 2

Z 

z 1 − 2 z +4 z

 dz

  1  2 ln z + 4 − ln z + C 2

It is easy to see that f1 (∞) = 0 =⇒ C = 0. We then have   4 1 f1 (z) = ln 1 + 2 4 z Likewise, we can differentiate f2 to obtain: f20 (z)

Z =−

∞

sin(2y)e−yz dy

0

Evaluating the above integral using IBP, f20 (z) = − Z =⇒ f2 (z) = −

z2

2 +4

2 dz = − arctan 2 z +4

  z +C 2

Using the fact that f2 (∞) = 0, we can determine the value of C:   z C = lim arctan z→∞ 2

124

CHAPTER 3. FEYNMAN’S TRICK π 2

C=

π ∴ f2 (z) = − arctan 2 Since

  z 2

  1 π arctan + arctan (x) = x 2

We have that f2 (z) = arctan

  2 z

Now, back to our original integral: Z I= 0

∞

! Z  ∞ sin 2y −yz sin2 x −zx e dx e dy dz x y 0 0     Z 1 ∞ 4 2 = ln 1 + 2 arctan dz 4 0 z z Z

∞

We proceed to substitute tan t = Z I=

π 2

2 z

=⇒ dz = −2 csc2 t dt

t csc2 (t) ln(sec t) dt

0

Using IBP with u = t ln(sec t) , dv = csc2 (t)dt gives π 2

h iπ Z 2 I = − t ln(sec t) cot t + 0

Z =

π 2

0

Z t dt +

0

π2 = + 8

  cot t t tan t + ln(sec t) dt

π 2

ln(sec t) cot t dt

0

Z 0

π 2

ln(sec t) tan t dt sec2 t − 1

125

3.2. DIRECT APPROACH Substituting eu = sec t, dt =

du we have: tan t

π2 I= + 8

∞

Z

u du −1

e2u

0

By the substitution 2u → u, the integral is transformed to: Z π2 1 ∞ u I= + du 8 4 0 eu − 1 By definition (16.2) of the zeta function, we know that: 1 ζ(s) = Γ(s)

Z 0

∞

xs−1 dx ex − 1

Therefore,

I= Plugging in ζ(2) =

π2 1 + ζ(2) 8 4

π2 6 ,

I=

Example 8: Evaluate

Z 0

Solution Define a function:

∞

π2 π2 + 8 24 2 π = 6

arctan(3x) arctan(2x) dx x2

126

CHAPTER 3. FEYNMAN’S TRICK

Figure 3.9: Graph of y =

Z

∞

f (α, β) = 0

arctan(3x) arctan(2x) x2

arctan(αx) arctan(βx) dx x2

Since f is a multi-variable function, we will use partial differentiation. For the reader who has not taken a course in multivariable ∂ calculus, a partial derivative, ∂x , can be computed by treating all terms not containing x as constants. After differentiating we get ∂ f (α, β) = ∂α

Z 0

∞

arctan(βx) dx x(1 + α2 x2 )

And, ∂2 f (α, β) = ∂α∂β

Z

∞

1 dx 2 x2 )(1 + β 2 x2 ) (1 + α 0 π = 2(α + β)

127

3.2. DIRECT APPROACH

The integral above is a standard one, and can be easily calculated using the method of partial fractions. We can now integrate with respect to β to get: ∂ f (α, β) = ∂α Z =

=

∂2 f (α, β)dβ ∂α∂β π dβ 2(α + β) Z

π ln(α + β) + C(α) 2

∂ The case ∂α f (α, β)|β=0 = 0 is trivial to calculate. We can then calculate the value of C(α):

π C = − ln α 2  π ∂ f (α, β) = ln(α + β) − ln α =⇒ ∂α 2 By symmetry, it is easy to see  ∂ π f (α, β) = ln(α + β) − ln β ∂β 2 We can now integrate with respect to α to obtain an expression for f (α, β). Z

∂ f (α, β) = f (α, β) dα ∂α Z π = ln(α + β) − ln α dα 2 =

 π α ln(α + β) + β ln(α + β) − α ln α + C(β) 2

128

CHAPTER 3. FEYNMAN’S TRICK

Where C(·) denotes the integration constant analogue for multivariable functions. Similarly, Z f (α, β) = π = 2

∂ f (α, β)dβ ∂β

Z ln(α + β) − ln β dβ

 π α ln(α + β) + β ln(α + β) − β ln β + C(α) 2

= Therefore,

f (α, β) =

 π α ln(α + β) + β ln(α + β) − β ln β − α ln α 2

Converting into a single logarithm, π f (α, β) = ln 2

(α + β)α+β αα β β

!

Plugging in our values, we have Z 0

3.3

∞

arctan(3x) arctan(2x) π dx = ln 2 x 2

55 33 · 22

!

Indirect Approach

Quite a few integrals are actually derivatives of integrals that are easier to compute. One advantage of differentiation under the inte-

129

3.3. INDIRECT APPROACH

gral sign is that it can go both ways, i.e. one can integrate a derivative of an expression to obtain a result, or one can take the derivative of a known integral to obtain a result. We can start off with an easy example: Z Example 9: 0

∞

xα ln x dx (1 + xα )2

Figure 3.10: Graph of y =

xα ln x (1+xα )2

Solution Define Z f (α) = 0

∞

dx 1 + xα

Differentiating under the integral sign gives

for α = 2

130

CHAPTER 3. FEYNMAN’S TRICK

f 0 (α) = −

∞

Z 0

xα ln x dx (1 + xα )2

Notice that our desired integral is −f 0 (α). We can now use the result from (2.8),   π π f (α) = csc (3.3) α α Therefore, π csc I = −f (α) = α2 0

Example 10: Evaluate

Z

π 2

π α


−

π 2 csc

π α




cot

π α




α3

√ sin x cos x tan x ln(tan x) dx

0

Solution √ sec2 x dx, A straightforward substitution would be u = tan x, du = 2√ tan x but our integral needs some rearranging first. Notice that we can rearrange our desired integral as:

Z I= 0

π 2

√  4 sec2 x tan2 x ln tan x √ dx 2 sec4 x tan x

Therefore the substitution u = Z I=4 0

√

∞

tan x yields: u4 ln u du (u4 + 1)2

3.4. EXERCISE PROBLEMS

√ Figure 3.11: Graph of y = sin x cos x tan x ln(tan x) This integral has been already evaluated (See (3.3)). Thus, π csc 42

I=4

 !  ! π π π 1 − cot 4 4 4

√ π 2(4 − π) = 16

3.4

Exercise Problems

1) Find the value of

Z 0

1

x2 − 1 dx ln x

131

132

CHAPTER 3. FEYNMAN’S TRICK

2) Evaluate

1

Z 0

3) Find

Z

π/2

ln(1 + x) dx 1 + x2

  ln α cos2 x + β sin2 x dx for α, β > 0

0

4) Use differentiation under the integral sign to evaluate Z

π

−π

cos2 x dx 1 + ax

For a > 0.

5) Evaluate

2π

Z

ecos x cos(sin x) dx

0

Z

6) Prove that

∞

2

e−x dx =

√

π (Hint: Let f (α) =

Z

−∞

7) Find

Z

π/2

8) Find

∞

0

9) Evaluate

0

x dx tan x

0

Z

∞

cos x dx (Hint: Use the indirect method) 1 + x2

Z 0

π/2

dx α cos2 x


2 for α, β > 0. + β sin2 x

10) Prove Frullani’s theorem

2

2

e−α (1+x ) dx) 1 + x2

3.4. EXERCISE PROBLEMS Theorem Let a, b > 0 and let f be continuously differentiable on [0, ∞). Suppose that lim f (x)

x→∞

Exists and is finite. Then the following equality holds:   Z ∞
f (ax) − f (bx) a dx = f (∞) − f (0) · ln x b 0

133

134

CHAPTER 3. FEYNMAN’S TRICK

Chapter 4

Sums of Simple Series

135

136

4.1

CHAPTER 4. SUMS OF SIMPLE SERIES

Introduction

In this chapter, we will explore the evaluation of simple series. We will start with finite series and later invoke the concept of a limit to extend our formulas to infinite series. This will be seen in action as we focus on telescoping series later in the chapter! To introduce our chapter, we can begin with the definition of summation. Definition Summation is the addition of a sequence of numbers, called addends or summands. The result is their sum or total. A series is similar to a sum, but is often used to refer to summations of infinite sequences.

4.2

Arithmetic and Geometric Series

We will start by a famous example provided by the story of the German mathematician Carl Freidrich Gauss in his mathematics class1 . One day, Gauss’ teacher asked his class to add together all the numbers from 1 to 100, a task the teacher thought would occupy the students for quite some time. He was shocked when young Gauss quickly wrote down the correct answer! The teacher could not understand how his student had calculated the sum so quickly, but the eight-year old Gauss pointed out that the problem was way simpler than the teacher imagined! Example 1: Evaluate S =

100 X

n

n=1

Solution 1

Waltershausen, W. S. von, Gauß, C. F. (1994). Gauss zum Gedächtnis. Vaduz: Sändig Reprint Verlag.

4.2. ARITHMETIC AND GEOMETRIC SERIES

137

Many reading this book are familiar with the formula to solve such a problem and perhaps the background of the aforementioned story as well. We begin by expanding S, S = 1 + 2 + 3 + 4 · · · + 96 + 97 + 98 + 99 + 100 Gauss started by recognizing that one can pair the 1st and the 100th term, the 2nd and the 99th term, and so on to make a sum of 101. S = (1 + 100) + (99 + 2) + (98 + 3) + (97 + 4) + · · · Now, there are our answer is:

100 2

such sums equivalent to 101 in S. Therefore,

 S = 101

100 2

 = 5050

The reader might recognize this as the formula k X

n=

n=1

Example 2: Evaluate

10 X

k(k + 1) 2

2n

n=0

Solution We begin by expanding our sum,

S=

10 X n=0

2n = 1 + 2 + 4 + · · · + 210

138

CHAPTER 4. SUMS OF SIMPLE SERIES

Consider multiplying S by 2, or the common ratio. 2S = 2 + 4 + 8 + · · · 211 Subtracting 2S from S, −S = 1 − 211

S = 211 − 1

Example 3: Evaluate

k X

rn for r 6= 1 and finite k.

n=0

Solution We can apply the same logic from our last example. Consider

S=

k X

rn = 1 + r + r2 + · · · + rk

(4.1)

n=0

Multiplying (4.1) by r gives rS = r + r2 + · · · + rk+1

(4.2)

As we have done previously, we can subtract (4.2) from (4.1) to get S − rS = 1 − rk+1

(1 − r)S = 1 − rk+1

4.2. ARITHMETIC AND GEOMETRIC SERIES

139

Therefore, k X

rn =

n=0

∞ X

Example 4: Evaluate

1 − rk+1 1−r

(4.3)

rn for |r| < 1.

n=0

Solution By definition, ∞ X

rn = lim

k→∞

n=0

k X

rn

n=0

Plugging in our result from (4.3), ∞ X

1 − rk+1 k→∞ 1 − r

rn = lim

n=0

Since |r| < 1, as k → ∞, rk+1 = 0. Thus, ∞ X n=0

rn =

1 1−r

It is now a good time to introduce some standard results. The sum of the first N squares is: N X n=1

n2 =

N (N + 1)(2N + 1) 6

140

CHAPTER 4. SUMS OF SIMPLE SERIES

P 1 Figure 4.1: A geometric visualization of ∞ n=1 2n = 1. We begin with a rectangle with area 12 and divide it in half with each iteration. The resulting figure is a square of area 1! Also, N X

 n3 = 

n=1

=

N X

2 n

n=1

N 2 (N + 1)2 4

Both formulas can be easily proved by mathematical induction. The proofs for these formulas are left as an exercise for the reader. We also have a general formula for such sums known as Faulhaber’s formula2 . 2

Donald E. Knuth (1993). Johann Faulhaber and sums of powers. Mathematics of Computation. 61 (203): 277–294. arXiv:math.CA/9207222. doi:10.2307/2152953. JSTOR 2152953.

141

4.3. ARITHMETIC-GEOMETRIC SERIES Theorem In general, N X n=1

 α  1 X α+1 Bn N α+1−n n = n α+1 α

n=0

Where Bn denotes the nth Bernoulli number. Definition The Bernoulli numbers, usually denoted as Bn , are a sequence of real numbers that satisfy the generating function ∞

X Bn xn x = ex − 1 n!

(4.4)

n=0

These numbers have a deep relationship with many of the special functions discussed later in the book. Remarkably, these numbers can also be used to evaluate even zeta function values: ζ(2n) = (−1)n−1

4.3

2π 2n B2n 2(2n)!

Arithmetic-Geometric Series

We now have formulas for both geometric and arithmetic series, but what about series that are both? For example, the series N X

n · 4n

n=1

This type of series is known is an arithmetic-geometric series. Fortunately, there is a general formula for this series!

142

CHAPTER 4. SUMS OF SIMPLE SERIES

Theorem Let {an } = [a + (n − 1)b]rn−1 be defined as an arithmeticgeometric progression. Also, denote N X

SN =

an = a+[a+b]r +[a+2b]r2 +· · ·+[a+(N −1)b]rN −1

n=1

As the N th partial sum. We can then express SN as: SN =

br(1 − rN −1 ) a − [a + (N − 1)b]rN + (1 − r)2 1−r

(4.5)

(4.6)

Proof. Multiplying SN by r gives:

rSN = r

N X

an = ar+[a+b]r2 +[a+2b]r3 +· · ·+[a+(N −1)b]rN (4.7)

n=1

Subtracting (4.7) from (4.5) gives: SN = a+ [a + b]r+ [a + 2b]r2 + · · ·+ [a + (N − 1)b]rN −1 −

SN r = 0+ (1 − r)SN

ar+

[a + b]r2 + · · ·+

[a + (N − 1)b]rN

=a +br +br2 + · · · +brN −1 −[a + (N − 1)b]rN

Notice that br + br2 + · · · + brN −1 is a geometric series. We can therefore write: (1 − r)SN = a +

∴ SN =

br(1 − rN −1 ) − [a + (N − 1)b]rN 1−r

br(1 − rN −1 ) a − [a + (N − 1)b]rN + (1 − r)2 1−r

4.3. ARITHMETIC-GEOMETRIC SERIES

143

What about infinite arithmetic-geometric series? Before we try to derive the general case, we can get some intuition with an example. ∞ X n Example 5: Evaluate 3n n=1

Figure 4.2: Plot of the partial sums

Solution Consider S= Multiplying S by

1 3

gives

1 2 3 + + + ··· 3 9 27

(4.8)

144

CHAPTER 4. SUMS OF SIMPLE SERIES

S 1 2 3 = + + + ··· 3 9 27 81

(4.9)

Subtracting (4.9) from (4.8),

−

S=

1 + 3

S = 3

0+

2 S= 3

1 + 3

2 + 9 1 + 9 1 + 9

3 + 27 2 + 27

4 + ··· 81 3 + ··· 81 1 1 + + +··· 27 81

Notice that 23 S is a geometric series with common ratio 13 . Hence, ∞

1 X 1 2 3 S= = 3 3n 1− n=1

∴S=

1 3

=

1 2

3 4

Notice that we can also use (4.6) to obtain a general form. We have the following theorem: Theorem ∞ X

[a + (n − 1)b]rn−1 =

n=1

br a + 2 (1 − r) 1−r

For |r|< 1. If |r| ≥ 1, then: ∞ X n=1

Diverges.

[a + (n − 1)b]rn−1

(4.10)

4.3. ARITHMETIC-GEOMETRIC SERIES

145

Proof. Notice that for |r| ≥ 1, the sequence {an } = [a+(n−1)b]rn−1 does not converge to 0, lim [a + (n − 1)b]rn−1 = ∞

n→∞

Therefore, the sum of the terms of this sequence does not converge. However, for |r|< 1, the series converges by the ratio test, which will be discussed in the next chapter. To begin, consider the equation for the partial sums of an arithmeticgeometric series, SN =

N X

[a + (n − 1)b]rn−1 =

n=1

br(1 − rN −1 ) a − [a + (N − 1)b]rN + (1 − r)2 1−r

By taking the limit as N → ∞, we can obtain an expression for S:

br(1 − rN −1 ) a − [a + (N − 1)b]rN + N →∞ (1 − r)2 1−r

S = lim SN = lim N →∞

rN → 0 as N → ∞ because we are only considering |r|< 1. Also, lim [a + (N − 1)b]rN = 0

N →∞

Which can be easily shown by L’Hopital’s rule. Therefore,

S=

br a + (1 − r)2 1 − r

Now, what about any series that is the product of multiple sequences? Is there a general formula? Unfortunately, there is no

146

CHAPTER 4. SUMS OF SIMPLE SERIES

"general" formula for any series. If that were to exist, many mathematicians would lose their job! However, we can get something almost as good: summation by parts.

4.4

Summation by Parts

By this point of this book, the reader should be familiar with integration by parts, which is proved in Chapter two. There is a lesser-known discrete analogue to integration by parts called summation by parts, which is for sequences instead of functions. Theorem ∞ Let {an }∞ n=1 and {bn }n=1 be two sequences. Also, denote

SN =

N X

an

n=1

As the N th partial sum of the series N X

an bn = SN bN − Sk−1 bk −

n=k

P∞

N −1 X

n=1 an .

We then have:

Sn (bn+1 − bn )

(4.11)

n=k

This formula is useful not only in the evaluation of series, but also in proving the convergence or divergence of series.

Proof. Define SkN

=

N X

an bn

n=k

First, notice that: Sn − Sn−1 = an

147

4.4. SUMMATION BY PARTS Hence,

SkN

=

N X

an bn =

n=k

=

N X

N X

(Sn − Sn−1 ) bn

n=k

Sn bn −

n=k

N X

Sn−1 bn

n=k

Expanding the expression above

SkN = Sk bk + Sk+1 bk+1 + Sk+2 bk+2 + · · · + SN bN − (Sk−1 bk + Sk bk+1 + Sk+1 bk+2 + · · · + SN −1 bN ) We can pull the underlined terms to the outside to obtain

SkN = [SN bN − Sk−1 bk ] + Sk bk + Sk+1 bk+1 + Sk+2 bk+2 + · · · SN −1 bN −1 − (Sk bk+1 + Sk+1 bk+2 + · · · + SN −1 bN ) (4.12) Notice that combining the expressions from the second and third line in (4.12) gives

S = [SN bN − Sk−1 bk ] + Sk (bk − bk+1 ) + Sk+1 (bk+1 − bk+2 ) + · · · + SN −1 (bN −1 − bN )

= [SN bN − Sk−1 bk ] +

N −1 X n=k

Sn (bn − bn+1 )

148

CHAPTER 4. SUMS OF SIMPLE SERIES

= [SN bN − Sk−1 bk ] −

N −1 X

Sn (bn+1 − bn )

n=k

Hence proved.

Example 6: Evaluate

N X

n · 2n

n=1

Figure 4.3: Plot of the partial sums when N = 10 Solution We will apply (4.11) here, N X

an bn = SN bN − Sk−1 bk −

n=k

N −1 X n=k

Where SN =

N X n=1

an

Sn (bn+1 − bn )

149

4.4. SUMMATION BY PARTS Let an = 2n and bn = n. We therefore have:

SN =

N X

2n = 2(2N − 1)

n=1

And S0 = 0. Thus,

SN =

N X

*0 S0 b − n · 2 = N · 2(2 − 1) −  1 n

N

N −1 X

2(2n − 1) · (n + 1 − n)

n=1

n=1

= 2N (2N − 1) −

N −1 X

2(2n − 1)

n=1



N −1 X

= 2N (2N − 1) − 2 

(2n ) −

n=1

N −1 X

 1

n=1

h i = 2N (2N − 1) − 2 (2N − 2) − N + 1 = N · 2N +1 − 2N − 2N +1 + 4 + 2N − 2

= 2N +1 (N − 1) + 2 Notice that we could have also applied the arithmetic-geometric progression formula in (4.6)! Example 7: Given a fair standard 6-sided die, what is the expected number of rolls before getting a 6? Solution In this problem, we will consider the notion of an "expected value". Although many readers will be familiar with this concept already, it is conveniently defined below:

150

CHAPTER 4. SUMS OF SIMPLE SERIES

Definition Let X be a random variable, or a variable whose value depends on the outcome of a random phenomenon, with a finite number of possible values x1 , x2 , · · · , xN with probabilities p1 , p2 , · · · , pN , respectively. The expected value of X is defined as: E[X] =

N X

xn p n

n=1

For a random variable X with a countably infinite number of values x1 , x2 , · · · with P probabilities p1 , p2 , · · ·, respectively, such that the series ∞ n=1 |xn | pn is convergent, the expected value of X is: ∞ X E[X] = xn p n (4.13) n=1

The P∞ stipulation for the absolute convergence of the series n=1 xn pn is extremely important because of the Riemann rearrangement theorem (See (5.2)). There is much more to the notion of expected value, but this basic introduction will suffice for our purposes. Let X be the number of dice rolls till the number 6 is obtained. Notice that the probability that the number 6 is obtained on any given roll of dice is 16 and the probability of getting any number but 6 is 1 − 16 = 56 . The probability of the first roll yielding the number 6 is 16 , but the probability of the second roll yielding the number 6 is 65 · 16 , as there is only a 56 chance that the second roll will happen. This is because for the second roll to happen, the first roll must not yield a 6. We can continue this pattern such that for the N th roll,

pN =

1 · 6

 N −1 5 6

151

4.4. SUMMATION BY PARTS

Note that X has a countably infinite number of values, which are 1, 2, 3 · · · (1 roll, 2 rolls, etc.). Therefore,

1 1 E[X] = 1 · + 2 · · 6 6

 1  2 5 5 1 +3· · + ··· 6 6 6

∞

1X n = 6 n=1

 n−1 5 6

By (4.13). Notice that this an arithmetic-geometric progression, which we have generalized in (4.10): ∞ X

[a + (n − 1)b]rn−1 =

n=1

br a + 2 (1 − r) 1−r

We can evaluate our desired sum by setting a = 1, b = 1, r = 56 . Thus, ∞

E[X] =

1X n 6 n=1

=

 n−1 5 6

1 (30 + 6) 6 =6

This means that the expected number of rolls till one obtains a 6 is six. This is intuitive as rolling the dice a large number of times, say 600 times, will yield about 600 6 = 100 6’s. Since the "gaps" between the occurrences of the number 6 sum up to 600 and there are about 100 6’s, the average "gap" is 6 rolls.

152

4.5

CHAPTER 4. SUMS OF SIMPLE SERIES

Telescoping Series

Definition A telescoping series is a series whose partial sums only have a fixed number of terms after cancellationa . This method of cancelling terms in partial sums is known as the method of differences. For infinite series, one must take the limit of the partial sums. More formally, let {an }∞ n=1 be a sequence of real numbers, then N X

an − an−1 = aN − a0

n=1

If limn→∞ an = 0, we can write: ∞ X

an − an−1 = −a0

n=1 a Thomson, B. S., Bruckner, J. B., Bruckner, A. M. (2008). Elementary real analysis. Upper Saddle River, NJ: Prentice-Hall.

What does this look like? Let’s take a dive! Example 8: Find

∞ X n=1

1 n(n + 1)

Solution Consider the partial fraction decomposition: 1 1 1 = − n(n + 1) n n+1 We therefore have:

153

4.5. TELESCOPING SERIES

Figure 4.4: Plot of the partial sums

S=

∞ X n=1

∞

X 1 = n(n + 1)



n=1

1 1 − n n+1



For telescoping series, it is often helpful to expand the partial sums of the series:

 SN =

1 1 −  1 2



 +

1 1  −  3 2



 +

1 1  −  4 3

SN = 1 −



 + ··· +

1
1
− N +1
N

1 N +1

Note that SN denotes the N th partial sum. Therefore,

S = lim SN = 1 N →∞



154

CHAPTER 4. SUMS OF SIMPLE SERIES

Example 9: Define a sequence {Fn }∞ n=0 such that the sequence obeys Fn = Fn−1 + Fn−2 and F0 = F1 = 1. This sequence is 1, 1, 2, 3, 5 · · · and is the infamous Fibonacci sequence. Evaluate ∞ X 1 Fn−1 Fn+1 n=1

Figure 4.5: Plot of the partial sums. Notice that the series converges very fast Solution Consider multiplying the summand by

S=

∞ X n=1

Fn , Fn

Fn Fn−1 Fn Fn+1

Since Fn+1 = Fn + Fn−1 =⇒ Fn = Fn+1 − Fn−1 , we can write:

S=

∞ X Fn+1 − Fn−1 n=1

Fn−1 Fn Fn+1

155

4.5. TELESCOPING SERIES By partial fraction decomposition,

S=

∞  X n=1

1 1 − Fn Fn−1 Fn Fn+1



Now, consider the partial sums of S, SN =

1 1 − F0 F1 F1 F2

=

! +

! 1 1 − + ··· F1 F2 F2 F3   1  1 + F − F F FN  −1 N N N +1

1 1 1 =1− − F0 F1 FN FN +1 FN FN +1

Taking the limit gives

S = lim SN = 1 N →∞

Example 10: Define SN =

42

1 1 1 + 2 + ··· + −4 6 −4 (2N )2 − 4

Find S10 and lim SN N →∞

Solution As always, we will try to use the method of partial fraction decomposition for problems like these. Since the problem is asking for both the evaluation of SN for some arbitrary N as well as limN →∞ SN , it would be wise to derive a general expression for any SN . We begin by expressing SN as:

156

CHAPTER 4. SUMS OF SIMPLE SERIES

Figure 4.6: Plot of the partial sums

SN =

N X n=2

=

N X n=2

=

1 (2n − 2)(2n + 2)

 N X 1 n=2

4

N

=

1X 8

n=2

1 (2n)2 − 4

1 1 − 2n − 2 2n + 2



1 1 − n−1 n+1





157

4.5. TELESCOPING SERIES

1 = 8

"

1 1 −  1 3



 +

1 1 −  2 4



 +

1 1  −  3 5





 1 1 + − + ··· 6 4   1 1 + − N −1 N +1

It is easy to see that this sum telescopes, as only the terms 11 , 12 , − N1 , and − N 1+1 remain. Therefore,   1 1 1 3 − − SN = 8 2 N N +1   1 3 2N + 1 = − 8 2 N (N + 1)

=⇒ S10 =

1 8



3 21 − 2 10 · 11

 =

9 ≈ 0.164 55

And,

lim SN =

N →∞

1 3 3 · = = .1875 8 2 16

Notice that our telescoping sum was not of the form ∞ X

an − an−1

n=1

Instead, it was of the form ∞ X n=1

an − an−2

158

CHAPTER 4. SUMS OF SIMPLE SERIES

1 1 and n−1 are two terms apart. In general, for any conSince n+1 + stant k ∈ N , the series ∞ X

an − an−k

n=1

Telescopes.

N X

Example 11: Find

n · n!

n=1

Solution Notice that:

(n + 1)! = n! (n + 1) = n · n! +n! =⇒ n · n! = (n + 1)! −n! Hence,

S=

N X n=1

n · n! =

N X

(n + 1)! −n!

n=1

= (2! −1! ) + (3! −2! ) + (4! −3! ) + · · · + (N + 1)! −N ! This sum telescopes,

∴ S = (N + 1)! −1




159

4.6. TRIGONOMETRIC SERIES

4.6

Trigonometric Series

Many trigonometric series are also telescoping. The strategy in evaluating these series is to utilize one or more trigonometric identities from the list of trigonometric identities in this book. Example

12:3

Evaluate

∞ X

 arctan

n=1

2 n2



Figure 4.7: Plot of the partial sums Solution Consider the trigonometric identity, 3 This problem was proposed by Anglesio in 1993: J. Anglesio, Elementary problem 10292, Amer. Math. Monthly 100, (1993), 291. It also appears in J. W. L. Glaisher, A theorem in trigonometry, Quart. J. Math. 15, (1878), and im S. L. Loney, Plane Trigonometry, Part II, Cambridge University Press, Cambridge, 1893.

160

CHAPTER 4. SUMS OF SIMPLE SERIES

tan(x ± y) =

tan x ± tan y 1 ∓ tan x tan y

Substituting x → arctan x, y → arctan y gives:

tan(arctan x ± arctan y) =

x±y 1 ∓ xy

Taking the arctangent of both sides,  arctan x ± arctan y = arctan

x±y 1 ∓ xy

 mod π

(4.14)

Letting x = n + 1, y = n − 1 gives:  arctan(n + 1) − arctan(n − 1) = arctan

2 n2



Hence,

S=

∞ X n=1

=

∞ X

 arctan

2 n2




arctan(n + 1) − arctan(n − 1)

n=1

We have to be careful here, as limN →∞ arctan N 6= 0. By inspecting the partial sums, we have:

SN =

N X


arctan(n + 1) − arctan(n − 1)

n=1

= − arctan(0) − arctan(1) + arctan(N ) + arctan(N + 1)

161

4.6. TRIGONOMETRIC SERIES And, S = lim SN N →∞

= lim

h

N →∞

− arctan(0) − arctan(1) + arctan(N ) + arctan(N + 1) =−

π + 2 lim [arctan N ] N →∞ 4 π =π− 4 3π = 4

Challenge Problem Prove ∞ X

arctan

n=1

x2 n2

=

 arctan

n=1

√

2 n2

πx √ 2



π  tanh   − arctan   πx 4 tan √

Example 13: Evaluate

2 gives:



=

∞ X n=0

Solution



2

Note that the case x = ∞ X



!

π = − arctan 4



tanh (π) tan (π)

3π π π + = 4 2 4

 arctan

1 n2 + n + 1





i

162

CHAPTER 4. SUMS OF SIMPLE SERIES

Figure 4.8: Plot of the partial sums From (4.14) we have:

      1 1 − 1 1  arctan − arctan = arctan  n n+1 1 n n+1 1 + n(n+1)   = arctan 

 1 

n(n + 1) 1 + 

= arctan

1 n(n+1)

1 2 n +n+1

 



Thus,

S=

∞ X n=0

 arctan

1 n2 + n + 1



mod π

163

4.7. EXERCISE PROBLEMS

= arctan(1) +

∞ X

 arctan

n=1 ∞

π X = + arctan 4 n=1

π = + 4

1 2 n +n+1



 !   1 1 − arctan n n+1

"

 lim arctan(1) − arctan

N →∞

=

1 N +1

#

π 2

In this chapter, we saw how powerful our algebraic tools are in series. We also got a glimpse into how powerful calculus can be in the evaluation of series through our extensive use of the limit. In the next chapters, we will use the tools we built here to tackle more complex problems.

4.7

Exercise Problems

1) Prove that for any arithmetic series, a1 + a2 + · · · + an =

2) Evaluate

∞ X n=1

n(a1 + an ) 2

1 n(n + 2)(n + 4)

3) Find the value of

2019 X n=1

13

1 + 2 + 3 + ··· + n + 23 + 33 + · · · + 20193

164

CHAPTER 4. SUMS OF SIMPLE SERIES

4) Find the value of

k X

n! (n2 + n + 1)

n=1

5) Prove that k X

cos n
1+ + + H4 = 1 + 2 3 4 2 4 4   1 =1+2 2 .. . We can easily deduce that H2k ≥ 1 + k 1

  1 2

. Since the subsequence

Oresme, Nicole (c. 1360). Quaestiones super Geometriam Euclidis [Questions concerning Euclid’s Geometry].

171

5.1. INTRODUCTION is unbounded, the harmonic series ∞ X 1 = lim H k n k→∞ 2

n=1

Diverges. The same argument can be extended to any Hαk with α ∈ N. A particularly interesting theorem on conditionally convergent series is the Riemann series theorem: Theorem The Riemann series theorem, sometimes also called the Riemann rearrangement theorem, states that if an infinite series of real numbers is conditionally convergent, then its terms can be arranged such that the rearranged series converges to an arbitrary real number and can even diverge. P Equivalently, if the series ∞ n=1 an is conditionally convergent, then there exists some permuation, given by the function p, such that: ∞ X

(5.2)

ap(n) = L

n=1

Where L ∈ R ∪ {−∞, ∞}. Example. Consider the sequence {an } = ing series,

S=

and its correspond-

∞ X (−1)n−1 n=1

=1−

(−1)n−1 n

n

1 1 1 + − + · · · = − ln 2 2 3 4

(5.3)

172

CHAPTER 5. PREREQUISITES

The value ln 2 comes from the Taylor series of ln(1 + x). Consider rearranging the series as:  Srearranged =

1 1− 2



      1 1 1 1 − − + − + ··· 4 3 6 8

Equivalently,  Srearranged =

1 1 − 2k − 1 2(2k − 1)

 −

1 4k

Where k ∈ N. Notice that all elements of the sequence {an } are found in Srearranged , but are in a different order. Also, 1 1 1 Srearranged = − + + · · · 2 4 6   1 1 1 = 1 − + + ··· 2 2 3 ∞

=

1 X (−1)n−1 2 n n=1

=

ln 2 2

Which is indeed a different value from (5.3)!

5.2 5.2.1

Ways to Prove Convergence The Comparison Test

∞ Description. Let {an }∞ n=1 and {bn }n=1 be two P∞sequences of nonnegative numbers. If an ≤ bn for all n and n=1 bn converges, then

5.2. WAYS TO PROVE CONVERGENCE

173

P∞ an converges. On P the other hand, if an ≥ bn for all n and Pn=1 ∞ ∞ n=1 bn diverges, then n=1 an diverges. Proof. The proof is trivial and is left as an exercise to the reader.

5.2.2

The Ratio Test

Description. First published by the French mathematician Jean le Rond d’Alembert, this test is also called d’Alembert’s ratio test2 . Given a sequence of complex numbers {an }∞ n=1 where an 6= 0 for large n, the ratio test is concerned with the limit: an+1 L = lim n→∞ an It states that: • If L > 1, then the series diverges. • If L < 1, then the series converges. • If L = 1, then the ratio test is inconclusive. Proof. Although many variations of the ratio test exist, we will present a proof of the original ratio test defined above3 . Consider: an+1 1 |an+1 | > |an | By (5.5), we know that: lim |an | > 0

n→∞

(5.5)

176

CHAPTER 5. PREREQUISITES

∞ Because the absolute value of the terms in the P∞sequence {an }n=1 gets larger and larger. Therefore, the series n=1 an diverges. The special case when L = 1 is inconclusive, and can be easily shown through the two sequences:

1 n (−1)n = n

{an }∞ n=1 = {bn }∞ n=1

(5.6)

And their corresponding series, ∞ X 1 =⇒ Diverges n

n=1

∞ X (−1)n n=1

n

=⇒ Converges

Since an+1 bn+1 =1 lim = lim n→∞ an n→∞ bn

5.2.3

The Integral Test

Description. Let f (x) be a continuous, positive, monotonically decreasing function on the interval [b, ∞] where b ∈ Z. ThenRthe infiP ∞ nite series ∞ f n=b (n) converges if and only if the integral b f (x) dx is finite. If the integral diverges, then the corresponding series also diverges. Proof. Without loss of generality, let b = 1. Since we can just shift

5.2. WAYS TO PROVE CONVERGENCE

177

the index of a series, proving the case when b = 1 is sufficient to prove the general case for any b ∈ Z. In this proof, we will prove the integral test for strictly decreasing f (x). However, the theorem holds true for any monotonically decreasing f (x) (a monotonically decreasing function does not have to be exclusively decreasing, only non-increasing). Consider the plot below:

Figure 5.1: Plot of the right Riemann sums of an arbitrary strictly decreasing function defined on the interval [1, ∞]. The width of the rectangles is a constant 1.

Define a sequence such that {an }∞ n=1 = f (n). Hence,

f (2) = a2 , f (3) = a3 , f (4) = a4 · · · The approximate area under the curve, or the improper integral on [1, ∞) is:

178

CHAPTER 5. PREREQUISITES

∞

Z

f (x) dx ≈ f (2) + f (3) + · · ·

I= 1

= a2 + a3 + a4 + · · · =

∞ X

an

n=2

However, this underestimates I as f (x) is strictly decreasing on [1, ∞), which implies: Z

N +1

(5.7)

f (x) dx > f (N + 1) N

And, ∞

Z

f (x)dx > 1

∞ X

f (n) =

n=2

∞ X

an

n=2

Since f (x) > 0 on [1, ∞), we know that: Z

N

∞

Z f (x) dx
|an | For some n ≥ N and sufficiently large N . Since r < 1, the series: ∞ X

rn

n=N

Converges. Because rn > |an | for all n ≥ N , the series ∞ X

|an |

n=N

Also converges by the comparison test. Therefore, the series ∞ X n=1

Converges since

PN −1 n=1

an =

N −1 X

∞ X

n=1

n=N

|an | +

|an |

|an | is a finite sum of finite terms.

Now, consider the case when L > 1. For some sufficiently large N , any n ≥ N satisfies: p n |an | > 1

5.2. WAYS TO PROVE CONVERGENCE

183

=⇒ |an | > 1 Hence,

lim |an | > 0

n→∞

∴ lim an 6= 0 n→∞

P We can therefore conclude that the series ∞ n=1 an diverges since ∞ the sequence {an }n=1 does not converge to 0.

The case when L = 1 is inconclusive, similar to what we saw in the ratio test. To show this, we simply need the two sequences we have provided in (5.6): 1 n (−1)n = n

{an }∞ n=1 = {bn }∞ n=1 And their corresponding series,

∞ X 1 =⇒ Diverges n

n=1 ∞ X n=1

Since

(−1)n =⇒ Converges n

184

CHAPTER 5. PREREQUISITES

lim

n→∞

p p n |an | = lim n |bn | = 1 n→∞

We know that the case L = 1 is inconclusive.

Note that the root test is stronger than the ratio test.

5.2.5

Dirichlet’s Test

The reader equipped with a standard knowledge of calculus should already be familiar with the traditional methods of proving convergence: ratio test, integral test, direct comparison, etc. These methods, although powerful, are sometimes obsolete to series evaluated here. One powerful technique is Dirichlet’s test. Dirichlet’s test, named after the German mathematician Peter Dirichlet, is a powerful method for testing the convergence of series4 .

4

Demonstration d’un theoreme d’Abel. Journal de mathematiques pures et appliquees 2nd series, tome 7 (1862), p. 253-255 Archived 2011-07-21 at the Wayback Machine.

5.3. INTERCHANGING SUMMATION AND INTEGRATION 185 Theorem Dirichlet’s test states that if {an }∞ n=1 is a sequence of real numbers and {bn }∞ is a sequence of complex numbers satn=1 isfying that: • lim an = 0 n→∞

• an+1 ≤ an P k • n=1 bn ≤ L for all k ∈ N Where L is a constant, then the series ∞ X

an bn

n=1

Converges. Notice that if bn = (−1)n , then Dirichlet’s test is simply the alternating series test often encountered in introductory calculus courses.

5.3

Interchanging Summation and Integration

Regarding series and integrals, one must be very careful in interchanging the two. In general, for an infinite sequence of functions {fn }, Z X n

fn dx 6=

XZ

fn dx

(5.11)

n

Notice the word "infinite." You might be wondering why finite series get a pass, which is due to the additive property of integrals. However, since ∞ is not a number, when we write an infinite sum we are effectively taking the limit as the bound of the sum

186

CHAPTER 5. PREREQUISITES

approaches infinity. Without loss of generality, let {fn } begin at n = 1. (5.11) is then effectively: N X

Z lim

N →∞

fn dx 6= lim

N →∞

n=1

N Z X

fn dx

n=1

It is the interchange of the limit and the integral we are concerned with. Even though in most cases the interchange is valid, one must still be careful. For example, ∞ Z X

2π

sin(x + n) dx

(5.12)

n=0 0

=

∞ X 

− cos(x + n)

2π 0

n=0 ∞ X

( cos(x) − cos(x + 2π))

n=0

=

∞ X

0=0

n=0

Now, consider interchanging summation and integration in (5.12) to give: Z 0

∞ 2π X

sin(x + n) dx

n=0

Since ∞ X n=0

sin(x + n)

5.3. INTERCHANGING SUMMATION AND INTEGRATION 187 Diverges for all x, the integral diverges as well. Well, this is contradictory! This example serves as one of many in which the interchange is not valid. But, what criteria did our example in (5.12) not meet? Instead of investigating each case of (5.11), we can present a general theorem for any sequence of functions {fn } on any interval I. One such theorem is the Lebesgue dominated convergence theorem. Theorem Let {fn }∞ n=1 be a sequence of Lebesgue integrable functions that converge to a limit function f almost everywhere on an interval I. Suppose there exists some Lebesgue integrable function g such that |fn | < g almost everywhere on I and for all n ∈ N. Then, f is Lebesgue integrable on I and Z Z Z lim fn (x) dx = lim fn (x) dx = f (x) dx n→∞ I

I n→∞

I

This theorem is perhaps the most hefty in the book, so let us break it down. First off, what is meant by a "Lebesgue integrable function"? Although the general definition requires a sizeable amount of measure theory, we will present a definition that will do for all the examples we deal with in this book. We first need to establish the notion of a measurable function. In our case, we are trying to employ the Lebesgue measure. A measure can be intuitively understood as the "size" of a set. It is a way of systematically assigning numbers to sets to represent their sizes. The Lebesgue measure is the measure that coincides most with general notions of "size." For example, the set of all points in [0, 1] has Lebesgue measure equal to its length on the number line, which is 1. Since we are only dealing with intervals, we can simply say that an interval [a, b] has a Lebesgue measure b − a. For two dimensional intervals, [a, b] × [c, d], it coincides with area.

188

CHAPTER 5. PREREQUISITES

If we are using the Lebesgue definition of measure, then we can say a function f is measurable on I if f is continuous on I. Theorem If f (x) is a bounded function defined on some closed interval I such that Z f (x) dx I

Exists in the Riemann sense, then f is also Lebesgue integrable. Now, on to the definition of almost everywhere. Intuitively, a property that holds almost everywhere is simply just that. Formally, this notion can be expressed by using the concept of measure. If a property holds almost everywhere on an interval I, the measure of the set where the property does not hold is 0. This serves us by letting us deal with functions diverging at endpoints.

Now, we will present a series reformulation of the dominated convergence theorem.

5.3. INTERCHANGING SUMMATION AND INTEGRATION 189 Theorem Let {fn }∞ n=1 be a sequence of Lebesgue integrable functions such that each fn is nonnegative on I and the sum ∞ X

fn (x)

n=1

Converges almost everywhere on I to a limit function f . Suppose there exists some Lebesgue integrable function g such that f (x) ≤ g(x) almost everywhere on I. Then, f is Lebesgue integrable on I and Z f (x) dx =

Z X ∞

I

fn (x) dx =

I n=1

∞ Z X

fn (x) dx

n=1 I

To see this in action, consider the integral Z

1

dx 0 1+x This integral trivially converges to ln 2. But, perhaps we want to substitute the power series expansion I=

∞

X 1 = (−1)k xk 1+x k=0

To get Z I= Z =

∞ 1X

(−1)k xk dx

0 k=0 n 1 X

lim

0 n→∞ k=0

(−1)k xk dx

Now, is the interchange here justified? Well, all terms of the sequence

190

CHAPTER 5. PREREQUISITES

{fn } =

n X (−1)k xk k=0

Are nonnegative. Moreover, the sequence does converge almost ev1 . It does not converge at x = 1. Moreerywhere on [0, 1] to 1+x over, f (x) is "dominated" by g(x) = 1, i.e. n X (−1)k xk ≤ 1 k=0

Therefore, we are jusified in interchanging the limit (infinite summation) and integral! In doing so we obtain:

I = lim

n→∞

n Z X k=0

= lim

n→∞

1

(−1)k xk dx

0

n X (−1)k k=0

k+1

Which is the alternating harmonic series that equals the same value of ln 2.

Chapter 6

Evaluating Series

191

192

6.1

CHAPTER 6. EVALUATING SERIES

Introduction

The most widely used series are the power series taught in introductory calculus courses. A classic example is: ∞

X 1 = 1 + x + x2 + · · · = xn 1−x

(6.1)

n=0

Notice that this equation is equivalent to the infinite geometric sum formula. We can do a lot with this series. Try substituting x → −x to get: ∞

X 1 = 1 − x + x2 − x3 + · · · = (−1)n xn 1+x

(6.2)

n=0

We can differentiate both sides to obtain: ∞

X 1 2 − = −1 + 2x − 3x + · · · = n · (−1)n xn−1 (1 + x)2 n=0

We can also integrate! Integrating both sides of (6.1) gives

ln(1 − x) = −

∞ X xn+1 n+1

n=0

Re-indexing, ln(1 − x) = −

∞ X xn n=1

n

We can also substitute x → −x in (6.3) to obtain:

(6.3)

193

6.2. SOME PROBLEMS

ln(1 + x) =

∞ X (−1)n+1 xn n=1

n

(6.4)

It is vital to realize that these powerful tools have their limitations too. It is important to check whether these sums converge before proceeding and using them on a specified interval. For example (6.1), only converges for |x|< 1. Using (6.1) for x = 5, for example, would yield nonsensical results.

6.2

Some Problems

Example 1: Evaluate

∞ X n2 n=1

2n

Figure 6.1: Plot of the partial sums Solution The series converges by the ratio test. Using (6.1), we have:

194

CHAPTER 6. EVALUATING SERIES

∞

X x = xn 1−x n=1

Differentiating both sides, ∞

X 1 = nxn−1 2 (1 − x) n=1

We can multiply both sides by x to get: ∞

X x = nxn 2 (1 − x) n=1

We can differentiate and multiply by x again to get: ∞

X x + x2 = n2 xn 3 (1 − x) n=1

1 Notice that if x = , we get our desired sum! Therefore, 2 ∞ X n2 n=1

2n

=

 2 + 12  3

1 2

1 2

=6 Example 2: Find the value of

∞ X n=1

Solution

2n

1 · n2

195

6.2. SOME PROBLEMS

Figure 6.2: Plot of the partial sums Notice that we can express our desired sum as

S=

∞ X n=1

=

∞ Z X

1 2

n=1 0

2n

1 · n2

xn−1 dx n

Using the dominated convergence theorem, we can switch the order of integration and summation to obtain 1 2

Z = 0

Z =− 0

∞ X xn−1 n=1 1 2

n

dx

ln(1 − x) dx x

Where in the last step we used the power series expansion for ln(1 − x). We can integrate by parts with u = ln(1 − x), dv = dx x to get

196

CHAPTER 6. EVALUATING SERIES

 1 S = − ln x ln(1 − x) 02 +

Z

1 2

0

ln x dx x−1

Since

ln(1 + x) =

∞ X (−1)n xn+1

n+1

n=0

We can substitute x → x − 1 to get:

ln x =

∞ X (−1)n (x − 1)n+1

n+1

n=0

Plugging that series into S,   Z 1X ∞ 2 1 (−1)n (x − 1)n S = − ln + dx 2 n+1 0 2

n=0

We can now use the dominated convergence theorem again,   X ∞ Z 1 2 (−1)n (x − 1)n 2 1 S = − ln + dx 2 n+1 0 n=0

" #1   X ∞ n (x − 1)n+1 2 1 (−1) = − ln2 + 2 n+1 n+1 n=0 0   X ∞ ∞ X 1 1 1 = − ln2 − + 2 2n+1 (n + 1)2 (n + 1)2 n=0

n=0

Re-indexing both sums,   X ∞ ∞ X 1 1 1 S = − ln − + n 2 2 2 ·n n2 n=1 n=1 | {z } | {z } 2

=S

=ζ(2)

197

6.2. SOME PROBLEMS   1 π2 =⇒ S = − ln −S+ 2 6 2

Finally solving for S,   1 π2 2S = − ln + 2 6 2

S=

Example 3: Define f (n) =

Z 0

π 2 ln2 2 − 12 2 1

∞ X ln(1 − xn ) f (n) dx. Evaluate x n n=1

Figure 6.3: Plot of the partial sums Solution Consider the power series expansion of ln(1 − x), (6.3):

ln(1 − x) = −

∞ X xk k=1

k

198

CHAPTER 6. EVALUATING SERIES

1

Z

P∞

k=1

=⇒ f (n) = −

x

0 ∞ 1X

Z =−

(xn )k k

dx

xnk−1 dx k

0 k=1

By the dominated convergence theorem, we can interchange summation and integration to obtain

f (n) = −

∞ Z X k=1

1

xnk−1 dx k

0

In evaluating the integral we obtain: " #1 ∞ X xnk f (n) = − nk 2 =−

k=1 ∞ X

1 n

=−

k=1

0

1 k2

ζ(2) n

Plugging in the value of ζ(2), Z 1 ln(1 − xn ) π2 ∴ dx = − x 6n 0 We can now evaluate our desired series:

S=

=−

∞ X f (n)

n=1 ∞ π2 X 1

6

n=1

n2

n =−

π4 36

199

6.2. SOME PROBLEMS

Example 4: A crazy ant is standing on the origin. It begins by walking 1 unit in the +x direction and then turns 60◦ counterclockwise and walks 12 units in that direction. The ant then turns another 60◦ and walks 13 units in that direction. The ant keeps doing this endlessly. What is the ant’s final position?

Figure 6.4: Visualization of the ant’s path Solution Instead of using the Cartesian coordinate system, we will use the polar coordinate system to make calculations easier. This allows us to express each move the ant makes in the form 1 [cos θ + i sin θ] n Where n1 denotes the distance travelled and θ denotes the angle at which the ant is travelling. The final position of the ant is simply the sum of the displacements. Denote the final position as P ,

200

CHAPTER 6. EVALUATING SERIES

   ! π π cos + i sin 3 3    ! 2π 2π cos + ··· + i sin 3 3


1 P = 1 cos(0) + i sin(0) + 2 +

1 3

Now seems like an opportune moment to use Euler’s formula! (See Appendix A for more.) eix = cos x + i sin x Applying this formula to P gives P =

i(n−1)π ∞ X e 3

n=1

n

This looks very similar to the power series expansion for ln(1 − x). In fact, the power series expansion of ln(1 − x) divided by x is ∞

X xn−1 ln(1 − x) =− x n n=1

iπ

In our desired sum, x = e 3 , therefore,   iπ ln 1 − e 3 P =−

iπ

e3 =

iπ 3√ 1+i 3 2

Simplifying using the complex conjugate of the denominator gives

201

6.2. SOME PROBLEMS

π πi P = √ + 6 2 3 We can translate our result to the Cartesian plane and determine that our ant friend will be at:  P = A whole

π 3

π π √ , 2 3 6



≈ 1.05 units away!

Example 5: Evaluate

∞  X n=1

1 1 − 4n − 1 4n



Figure 6.5: Plot of the partial sums Solution Although this series might first look like it is a telescoping one, it is not. One can easily see this by expanding the series,

202

CHAPTER 6. EVALUATING SERIES

 S=

1 1 − 4−1 4



 +

1 1 − 4·2−1 4·2



 +

1 1 − 4·3−1 4·3

 + ···

No terms cancel out. Therefore, we have to think of a different approach. Consider: Z 1 Z 1 1 1 4n−2 − = x dx − x4n−1 dx 4n − 1 4n 0 0 Z 1 = x4n−2 − x4n−1 dx 0

Using the above equation, we can then write our series as a sum of integrals. We can then use the dominated convergence theorem to interchange the order of summation and integration and obtain a geometric series that is easy to evaluate. We begin with ∞ Z X

S=

1

x4n−2 + x4n−1 dx

n=1 0 ∞ 1X

Z =

x4n−2 − x4n−1 dx

0 n=1

Z




n
n  ∞ ∞ X X x4 x4   dx − x2 x

1

= 0

n=1

n=1

Evaluating the series above, Z S= 0

1

1 x4 1 x4 · − · dx x2 1 − x4 x 1 − x4

Z = 0

1

x2 x3 − dx 1 − x4 1 − x4

203

6.2. SOME PROBLEMS 1

Z = 0

x2 (1 − x) dx 1 − x4

Using

1 − x4 = (1 − x2 )(1 + x2 ) = (1 − x)(1 + x + x2 + x3 ) We can write 1

Z S= 0

x2 dx x3 + x2 + x + 1

Using the method of partial fractions,

S=

1 2

1

Z 0

x−1 1 + dx 2 x +1 x+1

1 1 1 = ln|x + 1| 0 + 2 2 ln 2 1 = + 2 2

Z

1

Z

1

0

x 1 dx − 2 x +1 2

0

x−1 dx x2 + 1 Z 0

1

x2

Both integrals are easy to evaluate. Therefore,

S=

ln 2 ln 2 π + − 2 4 8

=

3 ln 2 π − 4 8

1 dx +1

204

6.2.1

CHAPTER 6. EVALUATING SERIES

Harmonic Numbers

Definition Recall the harmonic numbers, usually denoted as Hn , from the definition of the Euler-Mascheroni constant (See note in (1.8)). They are defined as the partial sum of the harmonic series: n X 1 Hn = (6.5) k k=1

Figure 6.6: Plot of the partial sums of the harmonic series, i.e. the harmonic numbers We can try to define an integral form of the harmonic numbers. Notice that

Z 0

1

xk−1 dx =

1 k

205

6.2. SOME PROBLEMS Therefore,

Hn =

n Z X k=1

1

xk−1 dx

0

Since we have a finite sum, we can safely interchange summation and integration to obtain n 1X

Z Hn =

xk−1 dx

0 k=1

We can apply the geometric sum formula here Z Hn = 0

1

1 − xn dx 1−x

(6.6)

Indeed, this form is the well-known extension of the harmonic numbers to the complex plane other than the negative integers via analytic continuation. Example 6: Prove that Hn = n

∞ X k=1

1 k(n + k)

Note that for any integer n we have

lim Hx − Hn+x = 0

x→∞

We can add Hn to both sides to obtain

Hn = lim Hx − (Hn+x − Hn ) x→∞

206

CHAPTER 6. EVALUATING SERIES

Notice that Hn+x − Hn is simply the sum Therefore,

1 n+1

+

1 n+2

+ ··· +

x x X 1 X 1 Hn = lim − x→∞ k n+k k=1

= lim

x→∞

= lim

x→∞

1 − k n+k

x X k=1

∴ Hn = n

(6.7)

k=1

x  X 1 k=1

1 n+x .



n k(n + k)

∞ X k=1

1 k(n + k)

(6.8)

We can derive more identities about harmonic numbers. Theorem The generating function for the harmonic numbers is given by: ∞ X ln(1 − x) Hn xn = (6.9) x−1 n=1

For any |x|< 1.

Proof. Consider the definition of the harmonic numbers:

Hn = Hn−1 +

1 n

∴ Hn − Hn−1 = Multiplying both sides by xn ,

1 n

207

6.2. SOME PROBLEMS

Hn xn − Hn−1 xn =

xn n

Summing both sides from n = 1 to ∞, ∞ X

Hn xn −

n=1

∞ X

Hn−1 xn =

n=1

∞ X xn n=1

n

Notice that the RHS is − ln(1 − x). Therefore, ∞ X

Hn xn −

n=1

∞ X

Hn−1 xn = − ln(1 − x)

n=1

Factoring out an x from the second sum, ∞ X

n

Hn x − x

n=1

∞ X

Hn−1 xn−1 = − ln(1 − x)

n=1

Since H0 = 0, ∞ X

Hn xn =

n=1

∞ X

Hn−1 xn−1

n=1

And,

(1 − x)

∞ X

Hn xn = − ln(1 − x)

n=1

Therefore, ∞ X n=1

Hn xn =

ln(1 − x) x−1

208

CHAPTER 6. EVALUATING SERIES

Example 7: Evaluate

∞ X n=1

Hn 2n (n + 1)

Figure 6.7: Plot of the partial sums Solution The series converges by the ratio test. Consider (6.9) ∞ X

Hn xn =

n=1

ln(1 − x) x−1

Integrating both sides from x = 0 to x = 12 ,

Z 0

1 2

∞ X n=1

Hn xn dx =

Z 0

1 2

ln(1 − x) dx x−1

209

6.2. SOME PROBLEMS

We can now use the dominated convergence theorem to interchange summation and integration, ∞ Z X

1 2

n=1 0 ∞ X n=1

ln(1 − x) dx x−1

0

" Hn

1 2

Z

Hn xn dx =

xn+1 n+1

#1 2

1 2

Z = 0

0

∞ X

Hn =− n+1 2 (n + 1) n=1 |

Z

1 2

0

ln(1 − x) dx x−1 ln(1 − x) dx 1−x {z }

(6.10)

I

For the integral on the RHS, consider IBP with u = ln(1 − x), dv = dx 1−x : Z −I = 0

1 2

h i 1 Z 21 ln(1 − x) ln(1 − x) 2 dx = ln2 (1 − x) − dx 1−x 1−x 0 0 | {z } −I

Hence, h i1 2 −2I = ln2 (1 − x) 0

∴I=

ln2 2 2

Plugging I back into (6.10) gives: ∞ X n=1

Hn ln2 2 = 2n+1 (n + 1) 2

210

CHAPTER 6. EVALUATING SERIES

∴

∞ X n=1

Hn = ln2 2 + 1)

2n (n

Example 8: Evaluate Sn =

n X (−1)k

n k




k

k=1

Figure 6.8: Plot of Sn values Solution Recall the statement of the binomial theorem: n   X n n−k k (x + y) = x y k n

k=0

We therefore have

(1 − x)n =

n   X n k=0

k

(−x)k

211

6.2. SOME PROBLEMS

n   X n ∴ (−x)k = (1 − x)n − 1 k k=1

Dividing by x then gives n   X n (1 − x)n − 1 (−1)k xk−1 = k x k=1

Integrating both sides from x = 0 to x = 1, Z

n  1X

0 k=1

 Z 1 (1 − x)n − 1 n (−1)k xk−1 dx = dx k x 0

Since our sum on the LHS is a finite sum, we can safely interchange summation and integration n X k=1

 Z 1 Z 1 n (1 − x)n − 1 k−1 dx (−1) x dx = x k 0 0 k

n X (−1)k k=1

n k




k

Z = 0

1

(1 − x)n − 1 dx x

Using (2.1), we can write n X (−1)k k=1

k

n k




Z = 0

1

xn − 1 dx 1−x

By (6.6), this is simply −Hn ! We can then express our sum as

212

CHAPTER 6. EVALUATING SERIES

n X (−1)k

n k




k

k=1

= −Hn = −

n X 1 k k=1

This could also be a nice proof for the divergence of the harmonic series, as the series on the left diverges as n → ∞. Example 9: Evaluate

∞ X Hn n=1

n3

Solution Consider the result in (6.8):

Hn =

∞ X k=1

n k(k + n)

We can then express our sum as

S=

∞ X Hn n=1

=

∞ ∞ X 1 X n 3 n k(k + n)

n=1

=

n3

k=1

∞ X ∞ X n=1 k=1

1 n2 k(k

+ n)

By symmetry, ∞ X ∞ X n=1 k=1

∞

∞

XX 1 1 = n2 k(k + n) nk 2 (k + n) n=1 k=1

213

6.2. SOME PROBLEMS Thus,   ∞ ∞ ∞ X ∞ X 1 X X 1 1  S= + 2 n2 k(k + n) nk 2 (k + n) n=1 k=1

∞

n=1 k=1

∞

1 XX = 2



n=1 k=1

Factoring out a

1 1 + 2 2 n k(k + n) nk (k + n)



1 gives: nk(k + n) ∞

S=

∞

1 XX 1 2 nk(k + n)



n=1 k=1

∞

∞

1 1 XX =  2 nk  (k +n)



1 1 + n k

n + k 



nk

n=1 k=1

∞



∞

1 XX 1 = 2 n2 k 2 n=1 k=1

=

∞ ∞ 1X 1 X 1 2 n2 k2 n=1

k=1

∞

=

1 X ζ(2) 2 n2 n=1

∴

∞ X Hn n=1

ζ(2) = n3 2


2 =

π4 72

(6.11)

214

6.3

CHAPTER 6. EVALUATING SERIES

Exercise Problems

1) Evaluate

∞ X n=1

2) Find mula)

n (Hint: Use the Taylor series of ex ) (2n + 1)!

∞ X sin k ◦ k=1

3) Evaluate

k

(Hint: Convert into radians and use Euler’s for-

∞ X

2n (2n + 1) n=0

(Hint: Use the identity

4) Find the value of




R π/2 0

sin2k+1 x dx =

∞ X (−1)n Hn n=1

5) Evaluate

2n n

n

k X (−1)n
for even k. k n=0

n

22k k!2 (2k+1)! )

Chapter 7

Series and Integrals

215

216

7.1

CHAPTER 7. SERIES AND INTEGRALS

Introduction

In this chapter, we will apply both the standard integration techniques given in chapter 2 as well as the skills we developed with series in the last three chapters. This will serve as our introduction to advanced integrals, as most of these integrals are impossible or very hard to solve using the standard techniques of u−substitution, IBP, etc, alone. It will also allow us to have an easy transition to the much more involved chapters dealing with special functions!

7.2

Some Problems

As always, we will begin with an easy problem. Z ∞ ln x Example 1: Evaluate dx 1 − x2 0

Figure 7.1: Graph of y =

ln x 1−x2

217

7.2. SOME PROBLEMS Solution Splitting the integral into two parts, Z I= 0

Letting x =

1

ln x dx + 1 − x2

∞

Z 1

ln x dx 1 − x2

1 , dx = − u12 du in the underlined integral, u Z

∞

1

Z

ln x dx = − 1 − x2

1

Z

1

0

Z

1

Therefore,

ln x dx 1 − x2

I=2 0

By the power series expansion of

I=2

  ln u1   du u2 1 − u12

ln u du 1 − u2

=

Z

0

1 1−x ,

∞ 1X

we can write

x2n ln x dx

0 n=0

=2

∞ Z X

∞

x2n ln x dx

n=0 0

Where we used the dominated convergence theorem in the last step. By IBP with u = ln x, dv = x2n dx, we have "  #1 Z 1 ∞ 2n+1 X 1 ln x  x I= − x2n dx 2n + 1 2n + 1 0 n=0

0

218

CHAPTER 7. SERIES AND INTEGRALS

=−

∞ X n=0

1 (2n + 1)2

This series can be easily evaluated. Consider ∞ X 1 ζ(2) = n2

=

∞ X n=1

=

1 + (2n)2

n=1 ∞ X

n=1

∞

∞

n=1

n=1

1 (2n − 1)2

X 1 1X 1 + 2 4 n (2n − 1)2 ∞

3 π2 X 1 ∴ ζ(2) = = 4 8 (2n − 1)2 n=1

We finally have

I = −2 × =−

Example 2: Evaluate

Z

π 2

π2 8

π2 4

ln (cos x + sin x)dx

0

Solution Recall that sin(x + y) = sin x cos y + sin y cos x Thus,

219

7.2. SOME PROBLEMS

Figure 7.2: Graph of y = ln (cos x + sin x)

      π π π sin x + = cos sin x + sin cos x 4 4 4 √ =

2 (sin x + cos x) 2

=⇒ sin x + cos x =

√

  π 2 sin x + 4

Hence,

Z I=

π 2

ln

0

Splitting the integral in half,

√

 ! π 2 sin x + dx 4

220

CHAPTER 7. SERIES AND INTEGRALS

Z I=

π 4

ln

√

0

|

 !  ! Z π √ 2 π π 2 sin x + dx + ln 2 sin x + dx π 4 4 4 {z } | {z } (1)

(2)

Applying our integral reflection identity to (1) and substituting x → x + π4 in (2) gives Z I=

π 4

ln

√

 2 sin

0

Notice that sin

π 2

!  ! Z π √ 4 π π −x dx + ln 2 sin x + dx 2 2 0



− x = sin x + π2 = cos x. π 4

Z =⇒ I = 2

ln

√

 2 cos x dx

0

π ln 2 = +2 4

Z

π 4

ln (cos x) dx

0

Now, define two integrals: π 4

Z I1 =

ln (cos x) dx

0

Z I2 =

π 4

ln (sin x) dx

0

A creative strategy to crack our desired integral, I1 , is to set up a system of linear equations with I1 and I2 being the variables. We will first add I1 and I2 Z I1 + I2 = 0

π 4

ln (cos x sin x) dx

221

7.2. SOME PROBLEMS π 4

 1 = ln sin 2x dx 2 0 Z π Z π 4 4 ln (sin 2x) dx − ln 2 dx = Z



0

0

Notice that both  π integrands above are symmetric around x = the interval 0, 2 . Thus, π 2

"Z

1 I1 + I2 = 2

Z ln (sin 2x) dx −

0

π 2

π 4

on

# ln 2 dx

0

The bracketed expression has been evaluated previously (See (2.2)). Therefore, π I1 + I2 = − ln 2 2 Moving on, we can compute the difference between I2 and I1 , π 4

Z I2 − I1 =

ln (tan x) dx

0

Substituting u = tan x, dx =

du 1+u2

Z I2 − I1 = 0

Z =

∞ 1X

then gives 1

ln u du 1 + u2

(−1)n u2n ln u du

0 n=0 1 Where we applied the series expansion of 1+u 2 in the last step. By the dominated convergence theorem, we can switch the order of summation and integration to obtain

222

CHAPTER 7. SERIES AND INTEGRALS

I2 − I1 =

∞ X

(−1)n

Z

1

u2n ln u du

0

n=0

Using IBP,

I2 − I1 = −

∞ X n=0

(−1)n (2n + 1)2

= −G Where G is Catalan’s constant. Definition Catalan’s constant, often denoted as G, is named after the French and Belgian mathematician Eugène Charles Catalan. It was originally found in the field of combinatorics, but later found use in analysis and even topologya . It is mainly defined as the infinite series G=

∞ X n=0

(−1)n (2n + 1)2

(7.1)

a

Agol, Ian (2010). The minimal volume orientable hyperbolic 2cusped 3-manifolds. Proceedings of the American Mathematical Society, 138 (10): 3723–3732, arXiv:0804.0043, doi:10.1090/S0002-9939-1010364-5, MR 2661571.

Open Problem It is not known whether Catalan’s constant is irrational, let alone transcendentala . Proving the irrationality or transcendentality of this constant would be a major result. a

Nesterenko, Yu. V. (January 2016). On Catalan’s constant. Proceedings of the Steklov Institute of Mathematics, 292 (1): 153–170, doi:10.1134/s0081543816010107.

223

7.2. SOME PROBLEMS To summarize,

−π ln 2 2 I2 − I1 = −G

I1 + I2 =

Therefore,

π 4

Z I1 = 0

π 4

Z I2 = 0

  π ln 2 G− 2   1 π ln 2 ln sin x dx = − G+ 2 2 1 ln cos x dx = 2

We finally have

(7.2) (7.3)

π ln 2 + 2I1 4 π ln 2 =G− 4

I=

Example 3: Define f (α) =

Z

α

ln x ln(α − x) dx. Find when the

0

minimum of f (α) occurs. Solution

The substitution x = αu, dx = αdu transforms our integral into Z f (α) = α

1


ln(αu) ln α(1 − u) du

0

Z

1

[ln α + ln(1 − u)][ln α + ln u] du

=α 0

Expanding the integrand,

Z f (α) = α 0

1

ln2 α + ln α ln u + ln α ln(1 − u) + ln(1 − u) ln u du

224

CHAPTER 7. SERIES AND INTEGRALS

Figure 7.3: Graph of y = f (α) "Z =α

1

1

Z

2

ln α du + 0

ln α ln u du 0

Z

1

Z

1

ln α ln(1 − u)du +

+ 0

ln u ln(1 − u) du 0

The first three integrals are trivial to evaluate. For the last integral, consider the power series expansion of ln(1 − u):

ln(1 − u) = −

∞ X un n=1

Z ∴I=

1

Z ln u ln(1 − u)du = −

0

n ∞ 1X

0 n=1

un ln u n

By the dominated convergence theorem, we can interchange summation and integration to get:

225

7.2. SOME PROBLEMS

Z ∞ X 1 1 n u ln u du I=− n 0 n=1

=

∞ X n=1

1 n(n + 1)2

Notice that we can break up this sum into:  ∞  X 1 1 1 − − I=− n n + 1 (n + 1)2 n=1

=

∞  X 1 n=1

|

 X ∞ 1 1 − − n n+1 (n + 1)2 n=1 {z }

Telescoping

=2−

π2 6

We then have, "

π2 f (α) = α ln α + 2 − 2 ln α − 6

#

2

Therefore, π2 6 2 ln α f 00 (α) = α

f 0 (α) = ln2 α −

π √

It is now easy to see that the minimum is at α = e 6 . Utilizing our expression for f (α) we can compute the minimum as:



f e

π √ 6



=e

π √ 6

" 2

ln



e

π √ 6





+ 2 − 2 ln e

π √ 6



π2 − 6

#

226

CHAPTER 7. SERIES AND INTEGRALS

=e

π √ 6

"

= 2e

π √ 6

Z

π 2

Example 4: Evaluate

# π2 2π π2 +2− √ − 6 6 6   π √ 1− ≈ −2.038 6

tan x ln sin x dx

0

Figure 7.4: Graph of y = tan x ln sin x Solution Consider rewriting the integral as: Z

sin x ln sin xdx 0 cos x  sin x p ln 1 − cos2 x dx cos x

I= Z = 0

π 2

π 2

Substituting u = cos x, du = − sin x dx gives

227

7.2. SOME PROBLEMS

1 I= 2


ln 1 − u2 du u

1

Z 0

Now, using the power series expansion of ln(1 − x), we can write ∞   X u2n ln 1 − u2 = − n n=1

Therefore,

I=−

1 2

Z

∞ 1X

0 n=1

u2n−1 du n

By the dominated convergence theorem, we can interchange summation and integration. ∞

1X1 I=− 2 n n=1

Z

1

u2n−1 du

0

" #1 ∞ 1 X 1 u2n =− 2 n 2n n=1

=−

We can plug in

P∞

1 n=1 n2

= π 2

Z ∴

π2 6

1 4

0

∞ X n=1

1 n2

in the above expression,

tan x ln sin xdx = −

0

For a proof of

P∞

1 n=1 n2

=

π2 6 ,

see (12.2).

π2 24

228

CHAPTER 7. SERIES AND INTEGRALS

Example 5: Evaluate

Z

ln x ln2 (1 − x) dx x

1

0

Figure 7.5: Graph of y =

ln x ln2 (1−x) x

Solution By (2.1) we can express our integral as 1

Z I= 0

ln(1 − x) ln2 x dx 1−x

We can then use the generating function for the harmonic numbers, (6.9), to get

Z I=−

∞ 1X

0 n=1

Hn xn ln2 x dx

229

7.2. SOME PROBLEMS

By the dominated convergence theorem, we can interchange summation and integration to obtain

I=−

∞ X

1

Z

xn ln2 xdx

Hn 0

n=1

Now, consider the integral Z f (α) =

1

xα dx =

0

1 α+1

For α > −1. Differentiating k times under the integral sign gives 1

Z

xα lnk xdx =

0

(−1)k k! (α + 1)k+1

Then, 1

Z

xn ln2 x dx =

0

And I = −2

∞ X n=1

= −2

∞ X n=1

 = 2

∞ X

n=1

1 (n + 1)3

(−1)2 · 2! (n + 1)3

Hn (n + 1)3

 Hn+1 −

1 n+1



 ∞ X 1 Hn+1  − 4 (n + 1) (n + 1)3 n=1

230

CHAPTER 7. SERIES AND INTEGRALS

 ∞ ∞ X X 1 H n − = 2 n4 n3 

n=1

n=1

∴ I = 2ζ(4) − 2

∞ X Hn n=1

n3

We can plug in our result for the latter sum from (6.11).

=⇒ I = 2ζ(4) − π4 90

Using ζ(4) =

(See proof in (12.3)) we have: 1

Z 0

7.3

2 · π4 72

ln x ln2 (1 − x) π4 dx = − x 180

Exercise Problems

1) Evaluate

1

Z

ln x ln(1 − x) dx 0

2) Find the value of

Z

∞

0

3) Evaluate

1

Z

ln 0

4) Find

Z



1+x 1−x

π/4

ln tan x dx 0

sin2n+1 x dx for n ∈ N. x



dx √ x 1 − x2

231

7.3. EXERCISE PROBLEMS 5) Evaluate

Z

∞

ln2 tanh x dx

0

6) Find the value of

1

Z 0

7) Evaluate lim

Z

n→∞ 0

1

x ln2 x dx 1 − x4

xn − x2n dx 1−x

8) Challenge Problem Evaluate Z

1

arctan x arcsin x dx −1

Hint: Use

Z arctan x = 0

1

x dy 1 + x2 y 2

232

CHAPTER 7. SERIES AND INTEGRALS

Chapter 8

Fractional Part Integrals

233

234

8.1

CHAPTER 8. FRACTIONAL PART INTEGRALS

Introduction

In this chapter, we will be dealing with integrals that contain the fractional part function. Definition The fractional part function, usually denoted as {·} is defined as follows: {x} = x − bxc

x≥0

(8.1)

Where b·c denotes the floor function (See below for definition). For example, {1.5} = .5. The fractional part function is defined for negative x, but the definitions vary. Definition The floor function, usually denoted as b·c, is defined as the greatest integer less than or equal to x. Similarly, the ceiling function, usually denoted dxe, is defined as the least integer greater than or equal to x. Equivalently, bxc = max{n ∈ Z | n ≤ x}

(8.2)

dxe = min{n ∈ Z | n ≥ x}

(8.3)

The results that these integrals give rise to are mind-blowing! The essence in solving problems like the ones in this chapter is converting them to infinite sums of consecutive integrals which we can manage. Although these integrals are not a primary area of research, they have a chapter dedicated to them since they develop the skill of using series to evaluate integrals.

235

8.2. SOME PROBLEMS

8.2

Some Problems

We begin with the most basic of fractional part integrals: Z n Example 1: Evaluate {x} dx for n ∈ N. 0

First, we will take a look at the graph of y = {x}.

Figure 8.1: Case when n = 10 Notice that each line is simply y = x shifted along the x axis. This is rather obvious from the definition of the fractional part function. Consider some x ∈ [k, k + 1), it is easy to see that {x} = x − bxc = x − k Therefore, the integral in our example is equivalent to a sum of integrals with intervals [k, k + 1) from k = 0 to k = n − 1. Translating these words into math we have Z n n−1 X Z k+1 I= {x} dx = (x − k) dx 0

k=0

k

236

CHAPTER 8. FRACTIONAL PART INTEGRALS n−1

1X n = 1= 2 2 k=0

Notice the strategy in solving these integrals. Since the fractional part, floor, and ceiling functions are piece-wise functions, we can think about their definite integrals (or the area under their curves) as a sum of consecutive areas. In other words, the principal aim when cracking these fractional part integrals is to transform them to sums of integrals without fractional parts. Now that we know how to approach such types of integrals, let us delve a little deeper. Z n Example 2: Evaluate {x}bxcdx for n ∈ N. 0

Figure 8.2: Try to decipher the pattern from this graph to solve the integral. Recall the definition of the fractional part:

237

8.2. SOME PROBLEMS

{x} = x − bxc From the previous example, we know that for some x ∈ [k, k + 1), {x} = x − k. We also know that on that same interval, bxc = k. Therefore,

Z

n

1

Z {x}bxcdx =

I=

Z

2

(x − 0)(0) dx +

0

0

=

n−1 X Z k+1

k(x − k) dx

k

k=0

=

(x − 1)(1) dx + · · · 1

n−1 X k=0

"

kx2 − k2 x 2 n−1 X

=

k=0

#k+1 k

k 2

n−1

1X n(n − 1) k= 2 4

=

k=0

Example 3: Evaluate

Z

∞

0

{x}bxc dx dxe

Solution As always, we will break this integral into a sum of integrals. However, now we are dealing with an infinite sum!

I=

∞ Z X

n+1

n=0 n

{x}bxc dx dxe

238

CHAPTER 8. FRACTIONAL PART INTEGRALS

Figure 8.3: Graph of y =

=

∞ Z X

n+1

n=0 n

=

∞ Z X

n+1

n=0 n

{x}bxc dxe

{x}n dx n+1

(x − n)n dx n+1

Notice that we can substitute x − n → x to get:

I=

∞ Z X n=0 0

=

∞ X n=0

∴I=

1

xn dx n+1

1 (n + 1)2

∞ X 1 π2 = ζ(2) = n2 6

n=1

239

8.2. SOME PROBLEMS Example 4: Evaluate

Z 0

1

1 x

 dx

Figure 8.4: Graph of y =

n o 1 x

This integral is part of a well-known and beautiful result in mathematics1 . We begin solving this integral by substituting u = x1 , dx = − du . We then have u2 1

Z I= 0

Z = 1

∞

1 x

 dx

{u} du u2

Notice that we can break down the integral above into a consecu1

Havil, J. Gamma: Exploring Euler’s Constant. Princeton, NJ: Princeton University Press, pp. 109-111, 2003.

240

CHAPTER 8. FRACTIONAL PART INTEGRALS

tive sum of integrals. ∞ Z X

I=

n+1

n=1 n

=

Z ∞ X

n+1

n

n=1

u−n du u2

du −n u

Z

n+1

n

du u2

!

Which evaluates to

I=

∞ X

 ln

k=1

n+1 n



1 − n+1

!

Recall the definition of the Euler-Mascheroni constant (See (1.8)): γ = lim (Hk − ln k) k→∞

We can replace ln k by ln(k + 1) since limk→∞ ln(k + 1) − ln k = 0. Thus, γ = lim Hk − ln(k + 1)




k→∞

Notice that ln(k + 1) can be represented by a telescoping sum:

ln(k + 1) =

k X

ln(n + 1) − ln n =

n=1

k X n=1

 ln

n+1 n



Plugging the sum above into the expression for γ then gives     k k X X n+1  1 − ln γ = lim  k→∞ n n n=1

n=1

241

8.2. SOME PROBLEMS

k X

= lim

k→∞

n=1

1 − ln n



n+1 n

!

This well-known result is due to Euler. The expression we obtained for I is then:

I=

∞ X

 ln

k=1

Example 5: Evaluate

Z 0

n+1 n

1

1 √ 3 x



1 − n+1

! =1−γ

 dx

Figure 8.5: Graph y =



We can generalize this for any root. Define

√1 3

x



(8.4)

242

CHAPTER 8. FRACTIONAL PART INTEGRALS

1

Z Ik = 0

Where k > 1. Substituting x =

1 √ k x

 dx

1 k du gives , dx = − tk+1 tk Z

∞

Ik = k 1

{t} dt tk+1

As usual, we will break this integral up into a sum of consecutive integrals

Ik = k

∞ Z X

n+1

n=1 n

t−n dt tk+1

Evaluating the integral and simplifying a bit,



 ∞  1 1 X 1  Ik = k − 1−k (n + 1)k−1 nk−1 n=1

   ∞ X 1 1 1 + n − k  k (n + 1)k n n=1

Notice that the first sum is telescoping: ∞  X n=1

!  1 1 1 1 − = − k−1 + (n + 1)k−1 nk−1 2k−1 1 ! ! 1 1 1 1 − k−1 + − k−1 + · · · 3k−1 2 4k−1 3   1 1 − k−1 + k−1 n→∞ 1 n

= lim

243

8.2. SOME PROBLEMS = −1 Therefore, ∞

Ik =

X k n + k−1



1 1 − k (n + 1)k n



1 1 − k 2k 1

n=1



Notice that the sum above is

 ∞ X S= n n=1

1 1 − k (n + 1)k n

 =1





 1 1 +2 − k 3k 2   1 1 +3 − k + ··· 4k 3

We can regroup the terms in the sum without changing its value (Since it is absolutely convergent for k > 1):  S= Each

n−1 nk

1 − k 1

gets

n nk



 +

1 1 −2· k k 2 2



  1 1 + 2 · k − 3 · k + ··· 3 3

subtracted from it. We therefore have:  S=−

1 1 1 + k + k + ··· k 1 2 3



= −ζ(k) Plugging S back in we get  Z 1 1 k √ dx = − ζ(k) k k − 1 x 0 To evaluate our integral, we simply substitute k = 3 to obtain:  Z 1 1 3 √ dx = − ζ(3) 3 2 x 0

(8.5)

244

CHAPTER 8. FRACTIONAL PART INTEGRALS

Example 6: Evaluate

∞

Z 1

{x} dx for n > 2. xn

Figure 8.6: Graph of y =

{x} x3

We begin by substituting {x} = x − bxc. ∞

x − bxc dx xn 1 Z ∞ dx bxc − dx n−1 x xn 1 Z

=⇒ I = Z = 1

∞

The first integral is trivial to evaluate. Notice that we can rewrite the second integral as an infinite sum of consecutive integrals ∞

X 1 I= − n−2 k=1

Z k

k+1

k dx xn

We have already evaluated a similar integral in (8.5). Therefore,

245

8.2. SOME PROBLEMS

Z

∞

{x} 1 ζ(n − 1) dx = − n x n−2 n−1

I= 1

Example 7: Evaluate

Z

1

0

1 √ k



k x

dx for k > 1.

Figure 8.7: Graph of y =

n

1 √ 2 x

o

√ uk−1 We begin by substituting u = k k x, dx = k−1 du, k  Z 1 Z k  1 1 1 √ I= dx = k−1 uk−1 du k u k k x 0 0

Substituting y =

1 , du = − dy , y2 u I=

1 k k−1

Z

∞ 1 k

{y} dy y k+1

(8.6)

246

CHAPTER 8. FRACTIONAL PART INTEGRALS

Recall the integral we dealt with in (8.5),  Z 1 1 √ Ik = dx k x 0 Which transforms into ∞

Z Ik = k 1

{t} k − ζ(k) dt = k+1 k−1 t

Through the substitution x = that example to obtain: I=

1 . tk

Z

1 k k−1

We can then use our result from ∞ 1 k

{y} dy y k+1 

 =

1 k k−1

 Z 1 Z ∞  {y} {y}   dy   1 y k+1 dy + y k+1  1  k {z } | Ik /k

=

"Z

1 k k−1

Notice that for y ∈

h

I=

1 1 k

1 k,1

1



{y} 1 ζ(k) − dy + k−1 k y k+1

#

, {y} = y. Therefore, "Z

1

dy 1 ζ(k) + − k k−1 k y

#

1 k k−1 k " # 1 k k−1 − 1 1 ζ(k) = k−1 + − k−1 k−1 k k " # 1 k k−1 ζ(k) − = k−1 k−1 k k

247

8.2. SOME PROBLEMS Which finally evaluates to Z

1

I=



k x

0

Example 8: Evaluate

1 √ k

Z 0

1

{2x} x

dx =

1 ζ(k) − k k−1 k

  1 dx x

Figure 8.8: Graph of y =

{2x} x

  1 x

h  h  Notice that when x ∈ 0, 12 , {2x} = 2x and when x ∈ 12 , 1 , {2x} = 2x − 1. Thus, we can split our desired integral into two separate integrals: Z 1  1 dx I= {2x} x x 0 Z = 0

1 2

    Z 1 1 dx 1 dx (2x) + (2x − 1) 1 x x x x 2

248

CHAPTER 8. FRACTIONAL PART INTEGRALS

Z =2 0

1 2

  Z 1  Z 1  1 1 1 dx dx + 2 dx − 1 1 x x x x 2

2

We can combine the first two integrals to get: 1

Z I=2 0

1 x



1

Z dx − 1 2

1 x



dx x

We have already evaluated the first integral i n oIn regards to  in (8.4). 1 the second integral, notice that when x ∈ 2 , 1 , x1 = x1 − 1 by the definition of the fractional part function. Thus,

Z

1 1 2

1 x



dx = x 1

Z = 1 2

1

Z



1 x

x

1 2

dx − x2

 −1

1

Z 1 2

dx x

= 1 − ln 2 Thus, I = 2 (1 − γ) − (1 − ln 2)

= 1 + ln 2 − 2γ

Example 9: Evaluate

Z 1

∞

{x} − 12 dx x

dx

249

8.2. SOME PROBLEMS

Figure 8.9: Graph of y =

{x}− 12 x

By the definition of the fractional part function, Z

∞

I= 1

x − bxc − 12 dx x

We can transform this integral into a sum of integrals with intervals [k, k + 1] so we can get rid of the floor function

I=

∞ Z X k=1

=

k+1

x − bxc − 12 dx x

k+1

x − k − 12 dx x

k

∞ Z X k=1

k

In evaluating the integral above we obtain

250

CHAPTER 8. FRACTIONAL PART INTEGRALS

I=

" ∞  X k=1

1 k+ 2



#   1 ln k − k + ln(k + 1) + 1 2

Notice that we can add ln k − ln k to get:

I=

" ∞  X k=1

= lim

N X

N →∞

k=1

"

1 k+1−1+ 2



#   1 ln k − k + ln(k + 1) + 1 2

#     1 1 1 + ln k + k − 1 + ln k − k + ln(k + 1) 2 2 (8.7)

We can split (8.7) into two the limit of two summations:  N X I = lim  (1 + ln k) N →∞

k=1

!     N X 1 1 + k−1+ ln k − k + ln(k + 1)  2 2 k=1 | {z } Telescoping Sum



= lim

N →∞

1 N + ln N ! − N + 2

!

 ln(N + 1)

This looks like an opportunity to use Stirling’s formula (See (1.6)). However, we first need to convert our expression into one logarithm.      N N + 21 I = lim ln e + ln N ! − ln (N + 1) N →∞

= lim ln N →∞

!

N ! eN 1

(N + 1)N + 2

251

8.2. SOME PROBLEMS Since N → ∞, we can use (1.6) to get   I = lim ln   N →∞

eN



 N √ N e

(N + 1)

2πN    N + 12

! 1√ N N + 2 2π

= lim ln

1

(N + 1)N + 2

N →∞



√

 = lim ln   N →∞

1+

  2π  N + 1  2

1 N

By the definition of e,   1 x e = lim 1 + x→∞ x We have √ I = lim ln N →∞

√

2π e

= ln We finally obtain that: Z ∞ 1

2π e !

!

√  {x} − 12 dx = ln 2π − 1 x

Example 10: Evaluate

Z 0

1

1 x

 x ln x dx

252

CHAPTER 8. FRACTIONAL PART INTEGRALS

Figure 8.10: Graph of y =

n o 1 x

x ln x

Solution Define a function Z

1

f (α) = 0

1 x



xα−1 dx

Differentiating under the integral,

f 0 (α) =

Z 0

1

1 x



xα−1 ln x dx

Our desired integral is then f 0 (2). Let us first evaluate f (α). Consider the substitution u = x1 , Z =⇒ f (α) = 0

∞

{x} dx xα+1

253

8.2. SOME PROBLEMS We have already evaluated this integral (See (8.6)). Therefore,

f (α) =

1 ζ(α) − α−1 α

And,

f 0 (α) = −

=

αζ 0 (α) − ζ(α) 1 − (α − 1)2 α2

ζ(α) 1 ζ 0 (α) − − α2 (α − 1)2 α

Plugging in α = 2 gives

f (2) =

ζ 0 (2) ζ (2) −1− 4 2

The value of ζ 0 (2) is:

ζ 0 (2) = −

∞ X ln k k=1

k2

=−


π2 12 ln A − γ − ln(2π) 6

Where A denotes the Glaisher-Kinkelin constant.

(8.8)

254

CHAPTER 8. FRACTIONAL PART INTEGRALS

Definition The Glaisher-Kinkelin constant, usually denoted as A, is a constant that appears in many sums and integrals. It is be given by: K(n + 1)

A = lim

n→∞ e−n2 /4 nn2 /2+n/2+1/12

≈ 1.282

(8.9)

Where K(·) denotes the K-function: K(n) =

n−1 Y

kk

k=1

(8.8) can be easily shown through Glaisher’s 1894 result:2

∞ Y k=1

1

k k2 =

A12

! π2 6

2πeγ

(8.10)

Thus,

f (2) =

2


π2 π2 −1− 12 ln A − γ − ln(2π) 24 12

Glaisher, J. W. L. On the Constant which Occurs in the Formula for 1 .22 .33 . . . nn . Messenger Math. 24, 1-16, 1894. 1

255

8.3. OPEN PROBLEMS

8.3

Open Problems

Open Problem Let n ≥ 3 be an integer. Calculate      Z x1 x2 xn ··· dx1 · · · dxn x2 x3 x1 [0,1]n Where

Z

1

Z ≡

[0,1]n

|0

Z ··· {z

1

0

n times

}

Open Problem Let n ≥ 3, k ≥ 1 be integers. Calculate 

Z [0,1]n

1 x1 + x2 + · · · + xn

k

dx1 · · · dxn

Where the open problems are due to Ovidiu Furdui’s book "Limits, Series, and Fractional Part Integrals: Problems in Mathematical Analysis"3 . It is worth mentioning that many of the fractional part integrals outlined in this chapter can be found in Furdui’s book, along with solutions.

8.4

Exercise Problems

1) Evaluate

Z

n

{x2 } dx for n ∈ N.

0

3

Furdui, O. (2013). Limits, Series, and Fractional Part Integrals: Problems in Mathematical Analysis. New York, NY: Springer.

256 2) Calculate

CHAPTER 8. FRACTIONAL PART INTEGRALS 1Z 1

Z 0

0

x+y x−y

3) Find a closed form for

 dx dy

1Z 1

Z

(xy) 0

4) Evaluate

Z

1

x 0

2019

0

   x y dxdy y x

   1 1 dx x x

5) Challenge Problem Evaluate Z 0

v n o u 1 x dx u n o t 1 − x1 1 − x

1u

Hint: Use the integral definition of the gamma function.

Part III

A Study in the Special Functions

257

259 In this part, we will use our tools from the previous chapters to derive properties and representations of many special functions. After doing so, we will work through a collection of example problems with the aid of our new powerful tools. This part will involve extensive use of the gamma, polygamma, beta, and Riemann zeta functions. Our aim is to explore their utility in series and integral problems as well as to establish a sense of familiarity with their properties and uses. These tools will then be used in our final and culminating part of the book regarding applications in the mathematical sciences.

260

Chapter 9

Gamma Function

261

262

9.1

CHAPTER 9. GAMMA FUNCTION

Definition

We have already introduced this function back in chapter 1 (See (1.7)). This special function is a very powerful tool in evaluating many integrals and series, and is perhaps one of the most applicable special functions in both mathematics and the mathematical sciences.

Figure 9.1: Graph of y = Γ(x)

9.2

Special Values

9.2. SPECIAL VALUES

263

Figure 9.2: Graph of the log-gamma function, y = ln Γ(x), which arises in various problems in mathematical analysis

Γ (1) = 0! = 1   √ 1 Γ = π 2   √ 1 Γ − = −2 π 2   1 Γ ≈ 2.678 938 534 707 747 6337 3   1 Γ ≈ 3.625 609 908 221 908 3119 4

264

9.3

CHAPTER 9. GAMMA FUNCTION

Properties and Representations

In the properties and representations section, we will derive several identities and equations relating to the special function considered in the chapter. Let us get started with our first! Theorem An interesting equation due to Gauss: n nz Y k Γ(z) = lim n→∞ z z+k

(9.1)

k=1

Proof. Denote

n

Z In =

t

z−1

0

Applying IBP with u = 1 − "

tz In = z |



t 1− n {z


t n , n

n #n 0

t 1− n

n dt

dv = tz−1 dt gives

n + nz

Z

n z



t 0

Z 0

n

n−1 dt

  t n−1 t 1− dt n z

Applying IBP successively with u = 1 − k = 1, 2, 3 · · · , n − 1 gives the formula


t n−k n

n n−1 1 In = · ··· nz n(z + 1) n(z + n − 1) =

t 1− n

}

=0

n = nz



Z

, dv = tz+k−1 dt for

n

tz+n−1 dt

0

n n−1 1 nn+z · ··· · nz n(z + 1) n(z + n − 1) n + z

9.3. PROPERTIES AND REPRESENTATIONS

265

Note that there are n multiples of n in the denominator. We can easily express the above expression as a finite product:

n nz Y k z z+k

In =

k=1

Since

n

Z lim In = lim

n→∞

n→∞ 0

t

z−1

  t n 1− dt n

And −t

e

 = lim

n→∞

t 1− n

n

We can write:

Z

∞

lim In =

n→∞

tz−1 e−t dt = Γ(z)

0

Where the interchange of the limit and integral is justified by the dominated convergence theorem. Thus,

n nz Y k n→∞ z z+k

Γ(z) = lim

k=1

266

CHAPTER 9. GAMMA FUNCTION

Theorem Another infinite product expression for the gamma function is given by Weierstrass:  ∞  e−γz Y z −1 z/k Γ(z) = 1+ e z k k=1

Proof. Consider the result from (9.1): n nz Y k n→∞ z z+k

Γ(z) = lim

k=1

We can rewrite this as n
Y k 1 exp zHn − zHn + z ln(n) n→∞ z z+k

Γ(z) = lim

k=1

Since lim − ln n + Hn = γ

n→∞

We can express our product as n

Y k 1 Γ(z) = lim exp (zHn − zγ) n→∞ z z+k k=1

n

Y 1 1 = lim exp (zHn − zγ) n→∞ z 1+ k=1 n

Y 1 e−γz exp (zHn ) n→∞ z 1+

= lim

k=1

z k

z k

(9.2)

267

9.3. PROPERTIES AND REPRESENTATIONS Notice that 

 n X 1  exp (zHn ) = exp z k k=1

z 1

z 2

= e · e ···e n Y z = ek

z n

k=1

Thus, n n e−γz Y z Y 1 ek n→∞ z 1+

Γ(z) = lim

k=1

k=1

z k

z n e−γz Y e k = lim n→∞ z 1 + kz

k=1

Taking the limit as n → ∞,  ∞  e−γz Y z −1 z/k e Γ(z) = 1+ z k k=1

Theorem The infamous Euler’s reflection formula is given by Γ(z)Γ(1 − z) = π csc πz

(9.3)

This equation has been used multiple times throughout the book, so why not prove it!

268

CHAPTER 9. GAMMA FUNCTION

Proof. Consider equation (9.1): n nz Y k n→∞ z z+k

Γ(z) = lim

(9.4)

k=1

The substitution z → −z gives n 1 Y k n→∞ z · nz z−k

Γ(−z) = lim

(9.5)

k=1

We can then multiply (9.4) and (9.5) to get n 1 Y k2 Γ(z)Γ(−z) = lim 2 n→∞ z z 2 − k2 k=1

= lim

n→∞

1 z2

n Y

1

z2 k=1 k2

−1

(9.6)

Lemma. We can express sin πx as

sin πx = πx

∞ Y k=1

x2 1− 2 k

! (9.7)

Proof. (9.7) was used by Euler to prove the famous result that ζ(2) =

∞ X 1 π2 = n2 6

n=1

The problem of evaluating the above sum is widely known as the Basel problem. The problem was originally posed by Pietro Mengoli in 1650, and gained notoriety after many famous mathematicians failed to attack it, notably the Bernoulli family1 . In this proof, we 1

Ayoub, Raymond (1974). Euler and the zeta function. Amer. Math. Monthly. 81: 1067–86. doi:10.2307/2319041.

269

9.3. PROPERTIES AND REPRESENTATIONS

will present the heuristic approach Euler took, which was only justified 100 years later by Weierstrass. In his infamous solution of the Basel problem, Euler argued heuristically that one can express sinx x as a polynomial of infinite degree based on its roots, similar to how one can factorize a finite polynomial. In doing so, Euler gave sin x = (x + π)(x − π)(x + 2π)(x − 2π) · · · x

= (x2 − π 2 )(x2 − 4π 2 )(x2 − 9π 2 ) · · ·

x2 =A 1− 2 π

!

x2 1− 2 4π

!

x2 1− 2 9π

! ···

Where A is a constant. Since lim

x→0

sin x =1 x

It is easy to see that A = 1. This approach was shown to be valid much later by Weierstrass through the Weierstrass factorization theorem. Theorem The Weierstrass factorization theorem asserts that every complex-valued function that is differentiable at all finite points over the whole complex plane can be represented as a product involving its zeroes. It can be seen as an extension of the fundamental theorem of algebra to complex functions. Continuing our proof,

270

CHAPTER 9. GAMMA FUNCTION

∞ Y sin x = x

x2 1− (kπ)2

k=1

! (9.8)

Substituting x → πx gives ∞ Y sin πx = πx

k=1

∴ sin πx = πx

x2 1− 2 k

∞ Y k=1

!

x2 1− 2 k

!

Now, back to our original proof. Notice that we can write (9.6) as n 1 Y Γ(z)Γ(−z) = lim 2 n→∞ z

k=1

=−

z2 −1 k2

1 πz z 2 sin πz

=−

π z sin πz

Thus,

Γ(z)(−zΓ(−z)) =

π sin πz

=⇒ Γ(z)Γ(1 − z) = π csc πz

!−1

271

9.4. SOME PROBLEMS

9.4

Some Problems

Let us begin with a very common integral. Z ∞ n Example 1: Evaluate e−x dx for n > 0. 0

2

Figure 9.3: The graph of the very famous y = e−x (Gaussian integral) We can substitute u = xn , dx = 1 I= n

Z

u

∞

1−n n

n

du to get 1

e−u u n −1 du

0

Using the integral definition of the Gamma function gives 1 I= Γ n Since xΓ(x) = Γ(x + 1) we have

  1 n

272

CHAPTER 9. GAMMA FUNCTION

 I=Γ

n+1 n



(9.9)

Sure does seem like these special functions make a lot of things easier! Let us transition into another well-known integral. Z α+1 Example 2: Evaluate ln Γ(x) dx α

Solution Define a function Z

α+1

f (α) =

ln Γ(x) dx α

Taking the derivative, f 0 (α) = ln Γ(α + 1) − ln Γ(α)

 = ln

Γ(α + 1) Γ(α)



f 0 (α) = ln α Where in the last step we used the definition of the gamma function. Thus, Z f (α) =

ln α dα

= α ln α − α + C

273

9.4. SOME PROBLEMS

The case when α = 0 was evaluated back in chapter 2 (See (2.4)). This gives us the initial condition required to find the value of C. =⇒ C = ln Z

√

2π



α+1

ln Γ(x) dx = α ln α − α + ln

√

2π



(9.10)

α

This formula is named after the Swiss mathematician Joseph Ludwig Raabe (Raabe’s formula), who derived it in 18402 .    Z 1  x x Example 3: Evaluate Γ 1+ Γ 1− dx 2 2 0

Figure 9.4: Graph of y = Γ 1 +

x 2




Γ 1−

x 2




Solution By definition, xΓ(x) = Γ(x + 1), 2

J. L. Raabe, Angenäherte Bestimmung der Factorenfolge 1 · 2 · 3 · 4 · 5 . . . n = R Γ(1 + n) = xn e−x dx, wenn n eine sehr grosse Zahl ist, J. Reine Angew. Math. 25 (1840), 146-159.

274

CHAPTER 9. GAMMA FUNCTION

1 I= 2

Z 0

1

    x x xΓ Γ 1− dx 2 2

We can now apply Euler’s reflection formula Γ(z)Γ(1 − z) = To get 1 I= 2 Substituting u =

1

Z

π sin (πz)

0

πx
dx sin πx 2

Z

π 2

πx 2 ,

2 I= π

0

u du sin u

Now, consider the double angle formula for sin x: sin 2x = 2 sin x cos x Then,     x x cos sin x = 2 sin 2 2 And Z I= 0

π 2

Z π 2 x dx = sin x 0 2 sin Z π 2 x
= x 2 0 2 tan 2 cos

x 2

x 2

x


cos


dx

x 2


dx

275

9.5. EXERCISE PROBLEMS Substituting x = 2 arctan u, dx = Z

2 du, 1+u2

1

2 arctan u du u 0 Using the power series of arctan u, I=

Z I=2

∞ 1X

0 n=0

(−1)n 2n u du 2n + 1

By the dominated convergence theorem we can interchange the summation and integration and integrate to obtain

I=2

∞ X n=0

(−1)n (2n + 1)2

Which is equal to 2G by definition (See (7.1)).

∴I=

9.5

4G π

Exercise Problems

1. Prove that xΓ(x) = Γ(x + 1) using the integral definition of the gamma function. 2. Find

Z

1

cos2 (πx) ln Γ(x) dx

0

3. Show that Γ(n) =

Z 0

1

 !n−1 1 dx ln x

276

CHAPTER 9. GAMMA FUNCTION

4. Evaluate

Z 0

5. Find

Z

∞

∞

x5 (e3x − ex ) dx (ex − 1)4

x3 e−2x sin x dx (Hint: Use differentiation under the

0

integral sign). 6. Challenge Problem Evaluate

Z

1

ln Γ(x)


2

dx

0


Hint: Use the Fourier series of ln Γ(x) , √ ln Γ(x) − ln 2π   ∞ X 1 ln n sin(2πnx)  + (γ + ln 2π)(1 − 2x) = π n 2 n=2

−

1 ln|2 sin(πx)| 2

Chapter 10

Polygamma Functions

277

278

10.1

CHAPTER 10. POLYGAMMA FUNCTIONS

Definition

Definition The polygamma function of order n, usually denoted as ψ (n) (·), is defined as the (n + 1)th derivative of the natural logarithm of the gamma function: ψ (n) (z) =

dn+1 ln Γ(z) dz n+1

(10.1)

Note: The digamma function, ψ (0) (·), is often expressed as ψ(·).

Figure 10.1: Graph of y = ψ(x)

10.2. SPECIAL VALUES

Figure 10.2: Graph of the trigamma function, y = ψ (1) (x)

10.2

Special Values

ψ(1) = −γ   1 ψ = −2 ln 2 − γ 2   1 π 3 ln 3 ψ =− √ − −γ 3 2 2 3   1 π ψ = − − 3 ln 2 − γ 4 2 2 π ψ (1) (1) = 6   1 π2 ψ (1) = 2 2

279

280

CHAPTER 10. POLYGAMMA FUNCTIONS

10.3

Properties and Representations

The first value of the digamma function in the above section follows from (10.4), which can also be used to compute ψ(n) for any n ∈ N. The fractional values can be computed by Gauss’s digamma theorem. Theorem Let n, k be positive integers with k > n. Then the following finite representation of the digamma function holdsa :     n π nπ ψ = −γ − ln(2k) − cot k 2 k k −1 dX     2e 2πjn πj +2 cos ln sin (10.2) k k j=1

Which follows from the digamma function’s recurrence equation. Notice that if n > k, then the recurrence relation in (10.8) can be used. a

Knuth, D. E. The Art of Computer Programming, Vol. 1: Fundamental Algorithms, 3rd ed. Reading, MA: Addison-Wesley, 1997.

We will now attempt to express the polygamma function as an infinite series. Theorem The polygamma function can be written as: ψ

(k)

k+1

(z) = (−1)

k!

∞ X n=0

1 (n + z)k+1

(10.3)

281

10.3. PROPERTIES AND REPRESENTATIONS

Proof. Recall the Weierstrass definition of the gamma function in (1.7):  ∞  e−γz Y z −1 z/k e Γ(z) = 1+ z k k=1

Taking the natural logarithm of both sides gives us 

 −1 ∞  Y z ln Γ(z) = (−γz − ln z) + ln  1+ ez/k  k k=1

Since ln

Q


P an = ln an , we have

ln Γ(z) = −γz − ln z +

∞ X k=1

 ! z z − ln 1 + k k

Term by term differentiation is valid here since the series on the RHS converges absolutely. Thus, ∞

1 X ψ(z) = −γ − + z



k=1

1 1 − k z+k



Notice that the series above telescopes to Hz . Therefore, ψ(z + 1) = −γ + Hz

(10.4)

We can also write (10.4) as an infinite series using (6.8),

ψ(z + 1) = −γ +

∞ X n=1

z n(n + z)

(10.5)

282

CHAPTER 10. POLYGAMMA FUNCTIONS

We proceed to differentiate both sides of (10.5) to obtain

ψ (1) (z + 1) =

∞ X n=1

1 (n + z)2

Substituting z + 1 → z and re-indexing,

ψ

(1)

(z) =

∞ X n=0

1 (n + z)2

Differentiating again,

ψ (2) (z) = −2

∞ X n=0

1 (n + z)3

In general,

ψ (k) (z) = (−1)k+1 k!

∞ X n=0

1 (n + z)k+1

Which is found by differentiating (10.5) k times.

10.4

Some Problems

Example 1: Evaluate

Z 0

Solution Define a function,

1

1−x 4 ln x dx 1 − x3

283

10.4. SOME PROBLEMS

Figure 10.3: Graph of y =

Z

1

f (α) = 0

By the series representation of

1−x 1−x3

ln4 x

1−x α x dx 1 − x3

1 1−x ,

we have: ∞

X 1 = x3n 1 − x3 n=0

And, Z f (α) =

∞ 1X

xα (1 − x)x3n dx

0 n=0

We can then interchange summation and integration by the dominated convergence theorem

284

CHAPTER 10. POLYGAMMA FUNCTIONS

f (α) =

∞ Z X

1

xα (1 − x)x3n dx

n=0 0

=

∞ Z X

1

xα+3n − xα+3n+1 dx

n=0 0

=

∞  X n=0



1 1 − α + 3n + 1 α + 3n + 2

(10.6)

Differentiating f (α) under the integral sign four times gives

f

(4)

Z

1

1−x α 4 x ln x dx 1 − x3

(α) = 0

Our integral is simply f (4) (0). We can differentiate term-wise in (10.6) since the series is absolutely convergent. Differentiating (10.6) four times and then letting α = 0 gives

I = 4!

∞  X n=0

1 1 − 5 (3n + 1) (3n + 2)5



 =

4! 35

∞ X n=0

  



1 n+

1 3

5 − 

1 n+

2 3

 5  

By (10.3) we can write 

(4) 4!  −ψ I= 5 3 4!

  1 3

−ψ (4) −

4!

  2 3

 

285

10.4. SOME PROBLEMS

ψ (4) =

  2 3

− ψ (4)

  1 3

35

Using the Mathematica call for the polygamma function, we can compute

I=

Example 2: Evaluate

32π 5 √ 243 3

1

Z

Hx dx 0

Figure 10.4: Graph of y = Hx Solution Note that we can use our result in (10.4) to get Hx = ψ(x + 1) + γ

286

CHAPTER 10. POLYGAMMA FUNCTIONS

Therefore, Z

1

I=

Z Hx dx =

0

1

ψ(x + 1) + γ dx 0

Z =γ+

1

ψ(x + 1) dx 0

By the definition of the digamma function, we have  1 I = γ + ln Γ(x + 1) 0

Z

1

(10.7)

Hx dx = γ

∴ 0

Example 3: Evaluate

Z 0

1

ln x dx x2 + x + 1

Figure 10.5: Graph of y =

ln x x2 +x+1

287

10.4. SOME PROBLEMS Solution Recall that 1 − x3 = (x2 + x + 1)(1 − x) Hence, Z I= 0

1

(1 − x) ln x dx 1 − x3

Splitting the integral, 1

Z I= 0

ln x dx − 1 − x3

1

Z 0

x ln x dx 1 − x3

1 and interchanging summa1 − x3 tion and integration using the dominated convergence theorem then gives Substituting the power series of

" ∞ Z X n=0

#

1

x

3n

ln x dx −

0

" ∞ Z X

#

1

x

3n+1

ln x dx

0

n=0

As a standard integral, we may use: Z

1

xα ln x dx = −

0

1 (α + 1)2

Which can be easily shown by IBP. Thus,

I=

∞ X n=0

∞

X 1 1 − 2 (3n + 2) (3n + 1)2 n=0

288

CHAPTER 10. POLYGAMMA FUNCTIONS

Notice that the underlined sum takes the sum of squares of the form (3n+2)2 , and leaves out those of the form (3n)2 and (3n+1)2 . We can then express our integral as

I = ζ(2) −

∞ X

∞

∞

n=0

n=0

X 1 X 1 1 − − 2 2 (3n + 1) (3n) (3n + 1)2

n=0

∞

X 1 ζ(2) −2 = ζ(2) − 9 (3n + 1)2 n=0

∞

=

X 8ζ(2) 1 −2 9 (3n + 1)2 n=0

Factoring out a 9 from the sum above we obtain: ∞

4π 2 2 X −  27 9

I=

n=0

1 n+

1 3

2

We can now use the series expansion of the trigamma function,

ψ (1) (x) =

∞ X n=0

1 (n + x)2

4π 2 2 (1) ∴I= − ψ 27 9

Example 4: Evaluate

Z 0

Solution Denote

1

  1 3

5 ln x − x5 + 1 dx (1 − x) ln x

289

10.4. SOME PROBLEMS

Figure 10.6: Graph of y =

1

Z

5 ln x−x5 +1 (1−x) ln x

α ln x − xα + 1 dx ln x(1 − x)

f (α) = 0

Differentiating under the integral sign, Z

0

1

ln x − xα ln x dx (1 − x) ln x

1

1 − xα dx 1−x

Z

1

f (α) = 0

Z = 0

By (6.6), we know:

Hα = 0

1 − xα dx 1−x

290

CHAPTER 10. POLYGAMMA FUNCTIONS

Therefore, f 0 (α) = Hα We can now obtain an expression for f (α) by integrating. Z f (α) =

f 0 (α) dα

Z =

Hα dα

Z =

ψ(α + 1) + γ dα

= ln Γ(α + 1) + γα + C Where we used (10.4). Since f (0) = 0, we can write C=0 1

Z

xα

α ln x − +1 dx = ln Γ(α + 1) + γα ln x(1 − x)

∴ 0

Plugging in α = 5, Z

1

0

5 ln x − x5 + 1 dx = ln 120 + 5γ (1 − x) ln x

Example 5: Evaluate

1

Z

xHx dx 0

Solution By (10.4), we have: Hx = ψ(x + 1) + γ

291

10.4. SOME PROBLEMS

Figure 10.7: Graph of y = xHx Therefore, Z I=

1

x(ψ(x + 1) + γ)dx 0

We can use the recurrence relation

ψ(x + 1) = ψ(x) +

Z =⇒ I =

1 x

(10.8)

1

xψ(x) + γx + 1 dx 0

γ = +1+ 2

Z

1

xψ(x)dx 0

We can apply IBP with u = x, dv = ψ(x)dx on the integral above,

292

CHAPTER 10. POLYGAMMA FUNCTIONS

Z

1



xψ(x)dx = x ln Γ(x) 0

Z

1

− 0

1

ln Γ(x) dx 0

We have already evaluated this integral back in chapter 2 (See (2.4)). √  γ ∴ I = + 1 − ln 2π 2 Example 6: Evaluate the series

∞ X n=0

n2

1 +1

Solution We will introduce the following theorem: Theorem Let x be a complex number. The following equality then holds: ∞ X 1 1 + πx coth xπ = (10.9) k 2 + x2 2x2 k=1

Proof. Recall the product definition of the sine function given in (9.8):

sin x = x

∞ Y k=1

x2 1− (kπ)2

!

Substituting x → ix we get:

sin ix = ix

∞ Y k=1

sinh x = x

∞ Y k=1

x2 1+ (kπ)2

!

x2 1+ (kπ)2

!

(10.10)

293

10.4. SOME PROBLEMS

Where in the last step we used the definition of the hyperbolic sine. Taking the natural logarithm of both sides,

ln sinh x = ln x +

∞ X k=1

x2 ln 1 + (kπ)2

!

We proceed to differentiate both sides with respect to x to obtain:  0 x2 ∞ 1 + X (sinh x) 1 (kπ)2   = + x2 sinh x x k=1 1 + (kπ)2 0

∞

1 X 2x ∴ coth x = + x k 2 π 2 + x2 k=1

Here term by term differentiation is justified since the series on the RHS is absolutely convergent. Substituting x → πx then gives: ∞

coth πx =

X 2πx 1 + 2 2 πx k π + π 2 x2 k=1

∞ 1 1 X 2x = + πx π k 2 + x2

(10.11)

k=1

Manipulating the above expression, ∞ X k=1

k2

1 πx coth πx − 1 = 2 +x 2x2

We can extend the sum above by adding the term when k = 0, which is x12 : ∞ X 1 1 + πx coth πx = 2 2 k +x 2x2 k=0

294

CHAPTER 10. POLYGAMMA FUNCTIONS

Substituting x2 = 1 then gives ∞ X k=0

10.5

1 1 + π coth π = +1 2

Exercise Problems

1. Evaluate lim ln

√ x

x→0

2. Find

k2

Z

∞

0

x!

x2 ln x dx ex

3. Derive the reflection identity ψ(1 − z) − ψ(z) = π cot πz

4. Find

∞ X ψ (1) (k) k=1

k

5. Challenge Problem Prove that

Z

∞

ψ(x) = 0

e−xt e−t − dt t 1 − e−t

6. Challenge Problem Evaluate

Z 0

π/2

x3 csc x dx

Chapter 11

Beta Function

295

296

11.1

CHAPTER 11. BETA FUNCTION

Definition

Definition The beta function, or the Euler integral of the first kind, is most commonly defined as: Z 1 B(x, y) = tx−1 (1 − t)y−1 dt (11.1) 0

For R(x), R(y) > 0. A key property of the beta function is B(x, y) =

Γ(x)Γ(y) Γ(x + y)

Figure 11.1: Graph of z = B(x, y)

(11.2)

297

11.2. SPECIAL VALUES

11.2

Special Values  1 1 =π B , 2 2   1 2 2π B , =√ 3 3 3   √ 1 3 , =π 2 B 4 4 

B (x, 1) =

11.3

1 x

Properties and Representations

We will begin this section by deriving a trigonometric integral for the beta function. Theorem The beta function can be given by: B(x, y) = 2

Z

π 2

cos2x−1 (θ) sin2y−1 (θ)dθ, R(x), R(y) > 0

0

(11.3)

Here, R(·) denotes the real part.

Proof. Consider the definition of the gamma function, (1.7) Z Γ(n + 1) = n! = 0

We can then write

∞

tn e−t dt

298

CHAPTER 11. BETA FUNCTION ∞

Z Γ(n + 1)Γ(k + 1) =

tn e−t dt

∞

Z

uk e−u du

0

0

Substituting t = x2 and u = y 2 , Z

∞

Γ(n + 1)Γ(k + 1) = 4

2

x2n+1 e−x dx

Z

∞

2

y 2k+1 e−y dy

0

0 2

Notice that e−x is symmetric around x = 0. Thus, ∞

Z Γ(n + 1)Γ(k + 1) =

2

|x|2n+1 e−x dx

Z

−∞

Z

∞

Z

∞

= −∞

∞

2

|y|2k+1 e−y dy

(11.4)

−∞

2

2

|x|2n+1 e−x |y|2k+1 e−y dx

−∞

Transforming into polar coordinates with x = r cos θ, y = r sin θ Z 2π Z ∞ 2 Γ(n + 1)Γ(k + 1) = r · e−r |r cos θ|2n+1 |r sin θ|2k+1 drdθ 0

Z

2π

Z

= 0

Z =

∞

0

∞

0 2

r2n+2k+3 e−r |cos θ|2n+1 |sin θ|2k+1 drdθ #  "Z 2π

2

r2n+2k+3 e−r dr

0

|cos θ|2n+1 |sin θ|2k+1 dθ

0

Notice that the underlined integrand is periodic with a period of Thus,

Γ(n + 1)Γ(k + 1) Z ∞  "Z 2n+2k+3 −r2 =4 r e dr 0

0

π 2

π 2.

# 2n+1

|cos θ|

2k+1

|sin θ|

dθ

299

11.3. PROPERTIES AND REPRESENTATIONS

  Since cos θ and sin θ are non-negative for all θ ∈ 0, π2 , we can get rid of the absolute value sign. Z  "Z π ∞

Γ(n+1)Γ(k+1) = 4

r

2n+2k+3 −r2

e

2

dr

0

cos

2n+1

2k+1

(θ) sin

#

(θ)dθ

0

Substituting u = r2 , du = 2rdr gives

Z Γ(n+1)Γ(k+1) = 2

∞

 "Z un+k+1 e−u du

π 2

# cos2n+1 (θ) sin2k+1 (θ)dθ

0

0

We will now use the definition of the gamma function to get

Z Γ(n + 1)Γ(k + 1) = 2Γ(n + k + 2)

π 2

cos2n+1 (θ) sin2k+1 (θ)dθ

0

Γ(n + 1)Γ(k + 1) =2 Γ(n + k + 2)

π 2

Z

cos2n+1 (θ) sin2k+1 (θ)dθ

0

By (11.2), we can express the LHS as:

B(n + 1, k + 1) = 2

Z

π 2

cos2n+1 (θ) sin2k+1 (θ)dθ

0

Or alternatively,

B(x, y) = 2

Z 0

For R(x), R(y) > 0.

π 2

cos2x−1 (θ) sin2y−1 (θ)dθ

300

CHAPTER 11. BETA FUNCTION

A shorter method to derive (11.3) is through the definition given in (11.1). The substitution t = cos2 u, dt = −2 cos u sin u du into (11.1) gives

B(x, y) =

Z

1 x−1

t

y−1

(1 − t)

Z

0

cos2x−1 (u) sin2y−1 (u) du

dt = −2 π 2

0

Z =2

π 2

cos2x−1 (u) sin2y−1 (u) du

0

We will now prove an interesting identity. Theorem The beta function has a recursive property given by: B(x, y) = B(x, y + 1) + B(x + 1, y)

(11.5)

Proof. Notice that by (11.2) we can write B(x + 1, y) = =

Γ(x + 1)Γ(y) Γ(x + y + 1)

xΓ(x)Γ(y) x Γ(x)Γ(y) = · (x + y)Γ(x + y) x + y Γ(x + y)

We now have the recurrence relation B(x + 1, y) =

x · B (x, y) x+y

B(x, y + 1) =

y · B(x, y) x+y

And, by symmetry,

(11.6)

11.3. PROPERTIES AND REPRESENTATIONS

301

Hence,

B(x, y + 1) + B(x + 1, y) =

x y · B (x, y) + · B(x, y) x+y x+y

∴ B(x, y) = B(x, y + 1) + B(x + 1, y)

Now, for one more! Theorem An alternative integral representation of the beta function is given by Z ∞ ux−1 (11.7) B(x, y) = du , R(x), R(y) > 0 (1 + u)x+y 0

Proof. The substitution t =

u u+1 ,

B(x, y) =

dt =

1 du (u+1)2

into (11.1) gives

1

Z

tx−1 (1 − t)y−1 dt

0

Z

∞

= 0

ux−1 du (1 + u)x−1 (1 + u)y−1 (1 + u)2 Z = 0

For R(x), R(y) > 0.

∞

ux−1 du (1 + u)x+y

302

11.4

CHAPTER 11. BETA FUNCTION

Some Problems

Example 1: Find

∞

Z 0

ln x dx 1 + xn

Figure 11.2: Graph of y =

ln x 1+x2

Solution Substituting u = xn , du = nxn−1 dx gives

I=

1 n2

Z

∞

0

1

u n −1 ln u du 1+u

From the definition of the beta function given in (11.7),

B(x, y) =

Z 0

∞

tx−1 dt (1 + t)x+y

303

11.4. SOME PROBLEMS Our integral is simply 1 ∂ I= 2 B(x, 1 − x) n ∂x x= 1

n

Since B(x, y) =

Γ(x)Γ(y) Γ(x + y)

We can express I as

I=

 1 ∂  Γ(x)Γ(1 − x) 1 x= n n2 ∂x

Recall Euler’s reflection formula, Γ(x)Γ(1 − x) = π csc(πx) Hence,  1 ∂  π csc(πx) 1 x= n n2 ∂x     π2 π π = − 2 cot csc n n n

I=

Example 2: Evaluate

Z 0

∞

√

dx 1 + x4 (1 + xπ )

Solution We will solve a generalized form of the integral here. Define Z ∞ dx √ f (α) = 1 + x4 (1 + xα ) 0

304

CHAPTER 11. BETA FUNCTION

Figure 11.3: Graph of y =

√

1 1+x4 (1+xπ )

We then proceed to apply the u-substitution u = ∞

Z

√

f (α) = 0

Z

0

= ∞

q 1+ ∞

Z =

√ 0

Z = 0

1 u4

√

, dx = −

du u2

dx 1 + x4 (1 + xα )

1 

1+

1 uα

 ·−

du  u4 + 1 1 +

∞

1 x

1 uα

du u2



du uα · u4 + 1 1 + uα

Therefore, Z 2f (α) = 0

∞

√

du uα · + u4 + 1 1 + uα

Z 0

∞

√

dx 1 · 4 1 + xα 1+x

305

11.4. SOME PROBLEMS After combining these integrals we obtain   Z ∞ 1 dx xα √ + 2f (α) = · 1 + xα 1 + xα 1 + x4 0 | {z }=1 Z ∞ dx √ 2f (α) = 1 + x4 0

Turns out that π was just a distractor! We will now substitute x = √ sec2 u tan u, dx = 2√ du tan u 2f (α) =

=

Z

1 2

Z

1 2

π 2

Z

√

0

= =

π 2

1 2

π 2

0

1 2

Z

sec2 u ·√ tan u 1 + tan2 u du

du sec2 u ·√ sec u tan u π 2

0

sec u √ du tan u

1

1

cos− 2 (u) sin− 2 (u)du

0

Using (11.3) gives   2   2 1 Γ 14 Γ 4   = √ 8f (α) = π Γ 21 Therefore, Z

∞

0

Example 3: Find

Z 0

  2 Γ 14 dx √ √ = 8 π 1 + x4 (1 + xπ ) π 2

√

cos2 x dx 1 + cos2 x

(11.8)

306

CHAPTER 11. BETA FUNCTION

Figure 11.4: Graph of y =

2

√ cos x 1+cos2 x

Solution We first substitute u = cos4 x =⇒ du = −4 cos3 x sin x dx. We proceed to solve for dx in terms of u. −du 4 cos3 x sin x

dx =

−du

=

3

4u 4 sin x Note that sin x =

√

1 − cos2 u = dx =

Thus, Z I= 1

0

p

1−

√

u. Therefore,

−du p √ 4u 1− u

√

3 4

u du p √ · √ 3p 1 + u 4u 4 1 − u

307

11.4. SOME PROBLEMS 1 = 4

1

Z

− 14

u

− 12

(1 − u)

0

We recall that B(x, y) = Therefore,

1 du = B 4



3 1 , 4 2



Γ(x)Γ(y) Γ(x + y)

    3 1 4 Γ 2   I= 4Γ 54     Γ 43 Γ 21   = Γ 14 Γ

Upon simplifying we obtain:  2 Γ 43 √ dx = √ 2π 1 + cos2 x

π 2

Z ∴

cos2 x

0

Example 4: Evaluate

Z 0

∞

dx 1 + xn

Solution Substituting x = Z I= 0

∞

√ n

tan2 u, dx =

dx 2 = 1 + xn n

Z

2 n

csc u sec u tan2/n u du gives

π/2

cos1−2/n u sin−1+2/n du

0

Using (11.3),   1 1 1 I = B 1− , n n n     1 1 1 = Γ 1− Γ n n n

(11.9)

308

CHAPTER 11. BETA FUNCTION

  π n

π I = csc n

Where in the last step we used Euler’s reflection formula. We will now prove the following result. Theorem The Legendre duplication formula is given by:   √ 1 Γ(x)Γ x + = 21−2x Γ(2x) π 2

Proof. Recall that sin 2t = 2 sin t cos t sin (2t) =⇒ = cos2x−1 (t) sin2x−1 (t) 2x−1 2 2x−1

And,

B(x, x) = 2

Z

π 2

sin2x−1 (2t) dt 22x−1

0

By (11.3). The substitution u = 2t then gives B(x, x) = 2

1−2x

Z

π

sin2x−1 (u)du

0

Since the integrand is symmetric around u = π2 , we can write

B(x, x) = 2 · 2

1−2x

Z 0

π 2

sin2x−1 (u)du

309

11.5. EXERCISE PROBLEMS   1 =2 · B x, 2   Γ(x)Γ 21 1−2x  =2 ·  Γ x + 12 1−2x

Moreover, B(x, x) =

Γ(x)Γ(x) Γ(2x)

Putting it all together,   √ 1 Γ(x) = 21−2x · Γ(2x) π ∴Γ x+ 2 Since Γ

11.5

  1 2

=

√

(11.10)

π.

Exercise Problems

1. Prove that B(x, y) = B(y, x) using the integral definition given in (11.1). 2. Evaluate

Z

π/2 √

sin x dx

0

3. Find a closed form for

Z

π/2

ln sin x ln tan x dx 0

4. Use differentiation under the integral sign and the beta function Z 1 π2 to show that ln x ln(1 − x) dx = 2 − 6 0

310

CHAPTER 11. BETA FUNCTION

5. Prove that the value of an integral of the form Z ∞ dx 0

(1 + xa )b (1 + xc )

Is independent of c if ab = 2 (See (11.8)).

6. Evaluate

Z

∞

p 4 1 + x4 − x dx

0

7. Challenge Problem Evaluate Watson’s integral Z πZ πZ π 0

0

0

dx dy dz 1 − cos x cos y cos z

Hint: Use the substitutions x = 12 tan α, y = 12 tan β, and z = 21 tan γ and then use spherical coordinates.

Chapter 12

Zeta Function

311

312

12.1

CHAPTER 12. ZETA FUNCTION

Definition

Introduced by Euler in the first half of the 18th century, and then expanded on by Riemann’s seminal 1859 paper On the Number of Primes Less Than a Given Magnitude, the zeta function continues to be vital not only in analysis but other domains of mathematics such as number theory. Take the Riemann hypothesis, a centuries old unsolved problem, which arose from investigating the nontrivial zeroes of the zeta function. This problem has intimate ties to number theory and the distribution of primes, as shown by Riemann in his 1859 paper1 . Definition The Riemann zeta function, usually denoted as ζ(·), is most commonly defined as: ζ(s) =

∞ X 1 ns

(12.1)

n=1

12.2

Special Values

Some notable values include: 1 Via analytic continuation ζ(−1) = − 12 1 ζ(0) = − Via analytic continuation 2 π2 ζ(2) = 6 ζ(3) ≈ 1.202 This is known as Apery’s constant ζ(4) = 1

π4 90

B. Riemann, Über die Anzahl der Primzahlen unter einer gegebenen Grösse, Monatsber. Berlin. Akad. (1859).

313

12.2. SPECIAL VALUES

Figure 12.1: Graph of y = ζ(x). Notice that the function ζ(x) diverges at x = 1. Moreover, limx→∞ ζ(x) = 1 We can easily prove some of these values. The most popular result here is the evaluation of ζ(2), which is also known as the Basel problem. We will present Euler’s original approach to the problem below. Proposition. The summation of the reciprocals of the squares of the natural numbers is given by:

ζ(2) =

π2 6

Proof. Recall the Taylor series expansion of sin x:

sin x =

∞ X (−1)n 2n+1 x x3 x5 x = − + + ··· (2n + 1)! 1! 3! 5!

n=0

(12.2)

314

CHAPTER 12. ZETA FUNCTION

Dividing both sides by x, ∞

sin x X (−1)n 2n 1 x2 x4 = x = − + + ··· x (2n + 1)! 1! 3! 5! n=0

Now, consider the infinite product definition of

sin x = x

x2 1− 2 π

!

x2 1− 2 4π

!

sin x x

x2 1− 2 9π

given in (9.8). ! ···

We will now take the sum of the coefficients of x2 (Which is allowed by Newton’s Identities2 ):

Sum of the coefficients of x = − 2

=−



1 1 1 + 2 + 2 + ··· 2 π 4π 9π



∞ 1 X 1 π2 n2 n=1

The above expression must be equal to the sum of the coefficients of x2 in the Taylor series expansion of sinx x . Therefore, ∞ 1 1 X 1 − =− 2 3! π n2 n=1

∴

∞ X π2 1 = ζ(2) = 2 n 6

n=1

2

See Mead, D. G. (1992). Newtons Identities. The American Mathematical Monthly, 99(8), 749. doi:10.2307/2324242

315

12.2. SPECIAL VALUES Now, using similar logic, we will attempt to evaluate ζ(4). Proposition. ζ(4) can be written as: π4 90

ζ(4) =

Proof. Since ζ(2) =

π2 6 ,

(12.3)

we can write: ∞ X n=1

1

=

π 2 n2

1 6

Squaring the expression above gives 

∞ X

 n=1

2 1 1  = π 2 n2 36

The LHS can be expressed as:



∞ X

 n=1

2 1  1 = 4 π 2 n2 π

=



1 1 1 + + ··· 12 22 32

∞ X n=1

1 π 4 n4

+2

X n 1, the RHS is equivalent to ζ(z). Therefore, Z ζ(z)Γ(z) = 0

∞

uz−1 du eu − 1

(16.2)

Which is a beautiful formula. It gives rise to this interesting result,

16.5. FERMI-DIRAC (F-D) STATISTICS

∞

Z 0

393

u π2 du = eu − 1 6

We can then use (16.2) to determine that Z 0

∞

x3 π4 π4 dx = ζ(4)Γ(4) = 6 · = ex − 1 90 15

Back to our original problem, we have that P π5 = A 15

=



2h c2



2π 5 k 4 15h3 c2

kT h

4

! T4

This is the Stefan–Boltzmann law, which is used in a variety of applications. Markedly, it is used in astrophysics to calculate the temperature of stars!

16.5

Fermi-Dirac (F-D) Statistics

Fermi–Dirac statistics describe a distribution of particles over energy states in systems consisting of many identical particles that obey the "Pauli exclusion principle". It is named after the Italian physicist Enrico Fermi and English physicist Paul Dirac5,6 . 5

Fermi, Enrico (1926). Sulla quantizzazione del gas perfetto monoatomico. Rendiconti Lincei (in Italian). 3: 145–9., translated as Zannoni, Alberto (1999-12-14). On the Quantization of the Monoatomic Ideal Gas. arXiv:condmat/9912229. 6 Dirac, Paul A. M. (1926). On the Theory of Quantum Mechanics. Proceedings of the Royal Society A. 112 (762): 661–77. Bibcode:1926RSPSA.112..661D. doi:10.1098/rspa.1926.0133. JSTOR 94692.

394

CHAPTER 16. STATISTICAL MECHANICS

Scientific Model The Fermi-Dirac (F-D) distribution characterizes a system of identical fermions (particles with half integer spin) in thermodynamic equilibrium. In that scenario, the average number of fermions in a single-particle state i is given by the sigmoid functiona : ni =

1 e(Ei −µ)/kT

+1

(16.3)

a

Reif, F. (1965). Fundamentals of Statistical and Thermal Physics. McGraw–Hill. ISBN 978-0-07-051800-1.

Where

• k is Boltzmann’s constant • T is the absolute temperature • Ei is the energy of a single-particle state i • µ is the total chemical potential

Fermions include all quarks and leptons, as well as all composite particles made of an odd number of these. They can be either elementary particles (electrons) or composite particles (protons). The F-D distribution is most frequently applied to electrons, which have spin 21 . The F–D distribution is only valid if the number of fermions in the system is large enough so that adding fermions has a negligible effect on µ. Derivation. We begin by noting the Pauli Exclusion Principle:

16.5. FERMI-DIRAC (F-D) STATISTICS

395

Scientific Law The Pauli exclusion principle was formulated by Austrian physicist Wolfgang Pauli in 1925 for electrons, and later extended to all fermions with his spin–statistics theorem of 1940a . It states that two or more identical fermions cannot occupy the same quantum state within a quantum system simultaneously. a Pauli, Wolfgang (1980). General principles of quantum mechanics. Springer-Verlag. ISBN 9783540098423.

So we have the restriction that in our system, each allowed energy state si can accommodate one and only one fermion. We can also set the restrictions that • N , the number of fermions in the system, is a fixed number P • i Ei Ni , the total energy of the system, is constant Now, consider the problem of placing Ni indistinguishable fermions in the ith level into Si states. This is simply:  Wi =

Si Ni



Si ! (Si − Ni ) ! Ni !

=

Therefore, the total number of ways to arrange indistinguishable fermions in a multi-level system is

W =

Y i

Wi =

Y i

Si ! (Si − Ni ) ! Ni !

Where the product is taken over all i. We now seek to set Ni values such that we maximize W using our basic optimization tools. Remember from the fundamental laws of thermodynamics that

396

CHAPTER 16. STATISTICAL MECHANICS

the most likely distribution is that with the largest amount of microstates. But, before we proceed we need to take the natural logarithm of W . Simply put, solving d (ln W ) = 0 is much simpler than solving d(W ) = 0. Since d (ln W ) =

d(W ) W

And W 6= 0, we are justified in this step. Taking the natural logarithm of W , ln W =

X


ln (Si ! ) − ln (Si − Ni )! − ln (Ni ! )

i

Since we are dealing with a large number of fermions, we can use Stirling’s formula from (1.5). ln x! = x ln x − x For large x. Applying this approximation gives:

ln W =

X

Si ln (Si ) − Si − (Si − Ni ) ln (Si − Ni )

i

+ (Si − Ni ) − Ni ln (Ni ) + Ni

=

X

Si ln (Si ) − (Si − Ni ) ln (Si − Ni ) − Ni ln (Ni )

i

Now, we are in a comfortable place to perform our optimization. X ∂W d ln W = dNi ∂Ni i

397

16.5. FERMI-DIRAC (F-D) STATISTICS X ∂W

d (ln W ) =

∂Ni

i

Since

dSi dNi

(16.4)

dNi

= 0 (Si is a constant with respect to Ni ), we can write: ∂W = ln (Si − Ni ) + 1 − ln (Ni ) − 1 ∂Ni 

Si −1 Ni

"



= ln



(16.5)

Plugging this value into (16.4),

X

d (ln W ) =

ln

i

Si −1 Ni

# dNi

Now, our optimization step. " X

 ln

i

Si −1 Ni

# dNi = 0

But, we have the restrictions: • There is a fixed number of particles • The energy of the system is constant Although many readers are familiar with how to deal with such constrained problems (Using Lagrange multipliers), a brief introduction to Lagrange multipliers will be provided.

398

CHAPTER 16. STATISTICAL MECHANICS

Theorem The method of Lagrange multipliers is a strategy for finding the local maxima or minima of a function subject to equality constraintsa . The method for one constraint can be summarized as follows: in order to find the stationary points of f (x) under the constraint g(x) = 0, find the stationary points of: L(x, λ) = f (x) + λg(x) Where λ is the Langrage multiplier. a

Hoffmann, Laurence D.; Bradley, Gerald L. (2004). Calculus for Business, Economics, and the Social and Life Sciences (8th ed.). pp. 575–588. ISBN 0-07-242432-X.

This method is advantageous since it allows optimization problems to be solved without explicit parameterization in terms of the constraints. Therefore, the method of Lagrange multipliers is very popular in solving challenging constrained optimization problems. Using this method gives the two constraints:

N−

X

Ni = 0

i

E−

X

Ei Ni = 0

i

Therefore,

 L(x, λ1 , λ2 ) = ln W + λ1 N −

 X

Ni  + λ2 E −

i

In general, we can maximize W using



 X i

dL dNi

= 0,

Ei Ni 

399

16.5. FERMI-DIRAC (F-D) STATISTICS

∂W = λ1 + λ2 Ei ∂Ni Using (16.5) and exponentiating both sides„  ln

Si −1 Ni

 = λ1 + λ2 Ei

Si − 1 = eλ1 +λ2 Ei Ni

∴ Ni =

1+

Si λ e 1 +λ2 Ei

For some constants λ1 , λ2 . Now, our focus shifts to trying to evaluate λ1 , λ2 . Recall Boltzmann’s law, which states that:

S = k ln W = kL(x, λ1 , λ2 ) Where S denotes entropy. We can then see that ∂S ∂S = kλ1 , = kλ2 ∂N ∂E We can now utilize the thermodynamic equation dE = T dS − P dV + µdN =⇒ dS =

 =⇒

1 (dE + P dV − µdN ) T

∂S ∂N

 = kλ1 = − E,V

µ T

400

CHAPTER 16. STATISTICAL MECHANICS 

∂S ∂E

 = kλ2 = N,V

1 T

Solving for λ1 , λ2 λ1 = −

λ2 =

µ kT

1 kT

Therefore,

ni =

1 e(Ei −µ)/kT

+1

Chapter 17

Miscellaneous

401

402

17.1

CHAPTER 17. MISCELLANEOUS

Volume of a Hypersphere of Dimension N

The N deminsional hypersphere, or N −ball, is a generalization of the notion of a sphere to higher dimensions. A a circle of radius R (which can be regarded as a 2−ball) has the equation:

x2 + y 2 ≤ R 2

Similarly, for 3 dimensions, the equation of a sphere of radius R is given by:

x2 + y 2 + z 2 ≤ R 2

In general, for an N −ball, its volume is the volume enclosed by

x21 + x22 + · · · + x2N ≤ R2

17.1.1

Spherical Coordinates

To the reader that is familiar with multiple integrals and the use of spherical coordinates, this section will be rather easy. However, a brief introduction is given below.

17.1. VOLUME OF A HYPERSPHERE OF DIMENSION N

403

Definition The spherical coordinates of a point are given by (r, θ, ϕ), where p • r = x2 + y 2 + z 2
• ϕ = arctan xy
• θ = arccos zr And r, θ, ϕ denote radius, inclination, and azimuth, respectively. Conversely, one can write: • x = r sin θ cos ϕ • y = r sin θ sin ϕ • z = r cos θ We can extend this spherical coordinate system to an arbitrary dimension. By using the notion of the Jacobian, ∂(x1 , x2 , · · · , xn ) d V = ∂(r, ϕ1 , · · · , ϕn−1 ) n



=

 cos(ϕ1 ) −r sin(ϕ1 ) 0 0 ··· 0   sin(ϕ1 ) cos(ϕ2 ) r cos(ϕ1 ) cos(ϕ2 ) −r sin(ϕ1 ) sin(ϕ2 ) 0 · · · 0     .. .. .. ..   . . . . . . .   det   0     ··· ··· −r sin(ϕ1 ) · · · sin(ϕn−2 ) sin(ϕn−1 ) sin(ϕ1 ) · · · sin(ϕn−2 ) cos(ϕn−1 ) sin(ϕ1 ) · · · sin(ϕn−2 ) sin(ϕn−1 ) r cos(ϕ1 ) · · · sin(ϕn−1 ) ··· r sin(ϕ1 ) · · · sin(ϕn−2 ) cos(ϕn−1 )

= rn−1 sinn−2 (ϕ1 ) sinn−3 (ϕ2 ) · · · sin(ϕn−2 ) dr dϕ1 dϕ2 · · · dϕn−1

404

CHAPTER 17. MISCELLANEOUS

Where |·| denotes the determinant. In this coordinate system, there is one radial coordinate, r, and n − 1 angular coordinates. The domain of each angular coordinate is [0, π], except φn−1 which has domain [0, 2π). A more detailed derivation can be found in the Blumenson (1960)1 .

17.1.2

Calculation

Using spherical coordinates, the volume of an N −ball is RZ π

Z

Z

π

Z

2π

···

VN (R) = 0

0

0

rN −1 sinN −2 (ϕ1 ) · · ·

0

sin(ϕN −2 ) dϕN −1 · · · dϕ2 dϕ1 dr We can write the iterated integral above as a product of integrals

Z VN (R) =

R

r

N −1

! Z dr

0

! Z

2π

π

dϕN −1

N −2

sin

0

Z

 (ϕ1 )dϕ1

0 π

···

 sin(ϕN −2 )dϕN −2

0

Evaluating the first integral,

RN VN (R) = N

Z

! Z

2π

dϕN −1 0

π

N −2

sin

 (ϕ1 )dϕ1

0

Z ···

π

 sin(ϕN −2 )dϕN −2

0 1

Blumenson, L. E. (1960). A Derivation of n-Dimensional Spherical Coordinates. The American Mathematical Monthly. 67 (1): 63–66. doi:10.2307/2308932. JSTOR 2308932.

17.1. VOLUME OF A HYPERSPHERE OF DIMENSION N

405

Notice that this is very close to the trigonometric integral form of beta function,

B(x, y) = 2

Z

π 2

sin2x−1 (t) cos2y−1 (t)dt

0

See (11.3) for derivation. However,  we need to find a way to change the interval of our integrals to 0, π2 . The first integral’s change in domain is trivial to compute. But, for the integrals with a sin term, we need to notice that sinN −1 (x) is symmetric around π2 in the interval [0, π].

Figure 17.1: Graph of y = sinN x for different values of N on x ∈ [0, π] Therefore,

406

CHAPTER 17. MISCELLANEOUS

RN VN (R) = N

Z 2

π 2

! N −2

sin

π 2

Z ··· 2

(ϕ1 )dϕ1

0

! sin(ϕN −2 )dϕN −2

0

Z ·

4

π 2

! dϕN −1

0

Now, we have a form that is equivalent to (11.3)! We can then write VN (R) as

VN (R) =

=

RN N

RN ·B N

Γ ·





       N −1 1 N −2 1 1 1 1 , ·B , · · · B 1, ·2B , 2 2 2 2 2 2 2

           Γ 21 Γ N 2−2 Γ 12 Γ N 2−3 Γ 12       · Γ N2 Γ N 2−1 Γ N 2−2       Γ (1) Γ 12 Γ 21 Γ 21   ·2 ··· Γ (1) Γ 32

N −1 2

Notice that the finite product telescopes. Thus, using nΓ(n) = Γ(n + 1), we have N

2π 2 RN   VN (R) = N Γ N2 N

∴ VN (R) =

π 2 RN   Γ N2 + 1

(17.1)

17.1. VOLUME OF A HYPERSPHERE OF DIMENSION N

17.1.3

407

Discussion

Using (17.1), the unit ball in N dimensions has volume N

π2

VN (1) = Γ



N 2

+1



An interesting question might be: When does the volume of a unit N −ball reach its maximum? We can plot VN (1) to get an idea:

Figure 17.2: Graph of y = Vx (1) The volume demonstrably peaks at N = 5. But, what is so special about 5? Nothing particular. If the radius were to be changed, the peak in (17.2) would shift. Consider, for example, an N −ball of radius 2 (figure (17.3)). Which demonstrably peaks much later. This trend of larger peaks

408

CHAPTER 17. MISCELLANEOUS

Figure 17.3: For the case of R = 2, the volume peaks around N = 25

should be intuitive from the RN term in the numerator. Challenge Problem Let n ∈ N be the value at which VN (R) has a maximum. Prove that: b2πR2 − 2e−γ c ≤ n ≤ b2πR2 − 1c Hint: Use the result:a,b     1 log x − < ψ(x) < ln x + e−γ − 1 2 a

F. Qi and B.-N. Guo. Sharp inequalities for the psi function and harmonic numbers. arXiv:0902.2524. b N. Elezovic, C. Giordano and J. Pecaric. The best bounds in Gautschi’s inequality. Math. Inequal. Appl. 3 (2000). 239–252.

However, one feature for all radii holds true, the volume of the hy-

17.1. VOLUME OF A HYPERSPHERE OF DIMENSION N

409

persphere always tends to 0 as N → ∞. This is easy to show by taking the limit of VN (R):

N

π 2 RN  lim VN (R) = lim  N →∞ N →∞ Γ N2 + 1 Which can be easily shown to equal 0 by using Stirling’s formula. But, what is an intuitive explanation for this? From the definition of an N −ball, we know that the region defined by

x21 + x22 + · · · x2N ≤ R2 Gives the volume of a N −ball of radius R. For the sake of simplicity, let us consider R = 1. This argument can, however, easily be extended to any R > 0. As N grows, most xn must be close to 0. Consider the line given by:

x1 = x2 = · · · = xN   This line intersects our N −ball at ± √1N , · · · , √1N . As N → ∞, this bounding region becomes smaller and smaller.

17.1.4

Applications

This brings us to the importance of this rather abstract problem. In fact, the result we obtained about the limiting behavior of VN (R) is a part of the larger curse of dimensionality:

410

CHAPTER 17. MISCELLANEOUS

Definition The curse of dimensionality refers to various phenomena that arise when analyzing data in high-dimensional spaces, that do not occur in low-dimensinal space such as the 3−dimensional space we experience everyday. This expression was coined by the American mathematician Richard E. Bellman. Bellman used this expression to characterize some phenomena in dynamic programming, a method developed by Bellmana,b . a Richard Ernest Bellman (1961). Adaptive control processes: a guided tour. Princeton University Press. b Richard Ernest Bellman (2003). Dynamic Programming. Courier Dover Publications. ISBN 978-0-486-42809-3.

This phenomenon arises often in statistics and machine learning. Interestingly, with a fixed amount of training data, the predictive power of a machine learning classifier or regressor first increases but then decreases. This phenomenon is named the peaking phenomenon and is similar to the trend in volume of a N −ball (See figure 17.2 and 17.3)2 . Indeed, it is because the notion of N −balls is vital to data analysis in higher dimensions. Our analysis of N −balls is especially relevant to k−nearest neighbor classification3 .

17.1.5

Mathematical Connections

The notion of the volume of a N −ball appears in many fields of mathematics, especially in measure theory. For example, for any Borel set S, the following equality holds: 2 Koutroumbas, Sergios Theodoridis, Konstantinos (2008). Pattern Recognition - 4th Edition. Burlington. 3 Radovanović, Miloš; Nanopoulos, Alexandros; Ivanović, Mirjana (2010). Hubs in space: Popular nearest neighbors in high-dimensional data. Journal of Machine Learning Research. 11: 2487–2531.

17.1. VOLUME OF A HYPERSPHERE OF DIMENSION N

411

λN (S) = 2−N VN (1)H N (S) Where • λN denotes the N −dimensional Lebesgue measure • H N (S) denotes the N −dimensional Hausdorff measure The reader should be familiar with the Lebesgue measure, which was discussed briefly earlier. The Hausdorff measure is fundamental in geometric measure theory, and appears in harmonic analysis4,5 .

4

Morgan, F. (2016). Geometric measure theory: a beginners guide. Amsterdam: Academic Press. 5 Toro, T. (2019). Geometric Measure Theory–Recent Applications. Notices of the American Mathematical Society, 66(04), 1. doi: 10.1090/noti1853

412

CHAPTER 17. MISCELLANEOUS

Part V

Appendices

413

Appendix A It √ is standard knowledge that the imaginary unit, i, is equal to −1. The term "imaginary" was coined by the French polymath René Descartes in an effort to devalue imaginary numbers’ validity6 . In the third book of his La Géométrie, Descartes states: Moreover, the true roots as well as the false [roots] are not always real; but sometimes only imaginary [quantities]; that is to say, one can always imagine as many of them in each equation as I said; but there is sometimes no quantity that corresponds to what one imagines, just as although one can imagine three of them in this [equation], x3 −6x2 +13x−10 = 0, only one of them however is real, which is 2, and regarding the other two, although one increase, or decrease, or multiply them in the manner that I just explained, one would not be able to make them other than imaginary [quantities]. However, after the pioneering work of Euler and Gauss, both acceptance and interest in imaginary numbers rose. Interest in e and the imaginary unit first arose from Bernoulli’s ob-

6

Giaquinta, Mariano; Modica, Giuseppe (2004). Mathematical Analysis: Approximation and Discrete Processes. Springer Science Business Media. p. 121. ISBN 978-0-8176-4337-9. Extract of page 121

415

416 servation that:7 1 1 = 1 + x2 2



1 1 + 1 + ix 1 − ix



Which, when combined with the standard integral Z ln(1 + αx) dx = +C 1 + αx α Can give us an idea of what the complex natural logarithm is. Another notable formula is due to the English mathematician Roger Cotes:8 ln(cos x + i sin x) = ix Which was discovered in 1714. In 1748, Euler published his famous formula which states that eix = cos x + i sin x Proof. We will present a standard proof due to Euler. Recall the Taylor series for sin x and cos x:

sin x =

∞ X (−1)n x2n+1 n=0

cos x =

(2n + 1)!

∞ X (−1)n x2n n=0

(2n)!

=x−

=1−

x3 x5 + + ··· 3! 5!

x2 x4 + + ··· 2! 4!

Also, recall the Taylor series of the exponential function 7

Bernoulli, Johann (1702). Solution d’un problème concernant le calcul intégral, avec quelques abrégés par rapport à ce calcul [Solution of a problem in integral calculus with some notes relating to this calculation]. Mémoires de l’Académie Royale des Sciences de Paris. 1702: 197–289. 8 Grattan-Guinness, I. (2000). The rainbow of mathematics: a history of the mathematical sciences. New york: W.W. Norton Company.

417

ex =

∞ X xn

n!

n=0

=1+x+

x2 + ··· 2!

Substituting x → ix in the series expansion of ex gives

e

ix

=

∞ X (ix)n n=0

= 1 + ix +

n!

i2 x2 i3 x3 i4 x4 + + + ··· 2! 3! 4!

= 1 + ix −

x2 ix3 x4 − + + ··· 2! 3! 4!

Notice that we can separate even and odd powers of x,

eix =

x2 x4 1− + + ··· 2! 4!

!

x3 x5 +i x− + + ··· 3! 5!

!

The first sum is cos x and the second sum is sin x. Therefore, eix = cos x + i sin x

(17.2)

Substituting x → −x in (17.2), e−ix = cos x − i sin x

(17.3)

Since cos(·) and sin(·) are even and odd functions, respectively. Adding (17.2) and (17.3) gives

418

cos x =

eix + e−ix 2

While subtracting (17.2) and (17.3) gives

sin x =

eix − e−ix 2i

We can also derive the beautiful result eiπ + 1 = 0 Using (17.2), which is famously known as Euler’s identity. The physicist Richard Feynman called this equation "our jewel" and "the most remarkable formula in mathematics"9 . The author agrees wholeheartedly with Feynman’s assessment! It is perhaps useful to introduce another approach to deriving Euler’s formula. Proof. Let

f (x) = cos x + i sin x We will begin by differentiating f (x): df = − sin x + i cos x dx = if (x) We now have the differential equation 9

Feynman, Richard P. (1977). The Feynman Lectures on Physics, vol. I. Addison-Wesley. p. 22-10. ISBN 0-201-02010-6.

419

df = if (x) dx df = idx f (x) Integrating,

Z

df = f (x)

Z i dx

ln f (x) = ix + C =⇒ f (x) = eix+C = C1 eix Now, f (0) = 1, so C1 = 1. Therefore, eix = cos x + i sin x

We now move our discussion to the complex logarithm. Definition The complex logarithm is defined similarly to the real logarithma . Let ex = z Then ln z = x a

Sarason, Donald (2007). Complex Function Theory (2nd ed.). American Mathematical Society.

Proposition. For any complex number, there are infinitely many complex logarithms.

420 Proof. Let z = reiθ for r, θ ∈ R and r > 0. Then one value of the complex logarithm is: ln z = ln r + iθ However, note that z can also be represented as z = reiθ+2ikπ for any k ∈ N since e2ikπ = cos(2πk) + i sin(2πk) = 1 Taking the complex logarithm of both sides of z = reiθ+2ikπ gives, ln z = ln r + (θ + 2kπ)i Since we can choose any k ∈ N, there are infinitely many values of the complex logarithm. The complex logarithm is defined as a multi-valued function. Hence, a branch is required to define the complex logarithm as a function. In most cases, we will deal with the principal value of the complex logarithm. For each complex number z ∈ C \ 0, the principal value of ln z is the logarithm whose imaginary part lies in the interval (−π, π]. We can then write the principal value of the complex logarithm as  ln z = ln r +

θ 2π



(17.4)

For some z = reiθ . Here {·} denotes the fractional part. The interested reader should refer to chapter 8 for a survey of problems involving fractional parts.

Appendix B Any introductory two-semester calculus course sequence is bound to investigate Taylor series. However, this appendix shall serve as a quick review. The Taylor series of some function, f (x), is its representation as an infinite sum of terms. These terms are calculated from the values of the derivatives of f (x) at a single point. We can present a more formal definition below: Definition Let f (x) be an infinitely differentiable complex-valued function at x = α. Its Taylor series is then given by the infinite power series:

f (x) = f (α) +

f 0 (α) f 00 (α) (x − α) + (x − α)2 1! 2! f 000 (α) + (x − α)3 + · · · 3!

Or, using summation notation, ∞ X f (n) (α) n=0

n!

(x − α)n

421

422 Where f (n) denotes the nth derivative of f . Some Taylor series include: Exponential Function

x

e =

∞ X xn n=0

n!

Which converges for all x ∈ C Logarithmic Function

ln(1 − x) = − ln(1 + x) =

∞ X xn

n=1 ∞ X n=1

n

= −x −

x2 x3 − − ···, 2 3

(−1)n−1 xn x2 x3 =x− + − ···. n 2 3

Which always converge for |x| < 1. It is worth noting that the series representations of ln(1 − x) and ln(1 + x) converge at x = −1 and x = 1, respectively. Trigonometric Functions

sin x = cos x = tan x =

∞ X (−1)n 2n+1 x (2n + 1)!

n=0 ∞ X

n=0 ∞ X n=1

=x−

x3 x5 + − ··· 3! 5!

(−1)n 2n x (2n)!

=1−

x2 x4 + − ··· 2! 4!

B2n (−4)n (1 − 4n ) 2n−1 x (2n)!

=x+

x3 2x5 + + ··· 3 15

423 arcsin x =

∞ X n=0

(2n)! x2n+1 n 4 (n! )2 (2n + 1)

=x+

π − arcsin x 2 ∞ π X (2n)! = − x2n+1 n 2 4 (n! )2 (2n + 1)

x3 3x5 + + ··· 6 40

arccos x =

=

n=0

arctan x =

∞ X n=0

(−1)n 2n+1 x 2n + 1

π x3 3x5 −x− − − ··· 2 6 40

=x−

x3 x5 + − ··· 3 5

Where Bn denotes the nth Bernoulli number, see (4.4) for more. Hyperbolic Trigonometric Functions

sinh x = cosh x = tanh x = sinh−1 x = tanh−1 x =

∞ X x2n+1 (2n + 1)!

=x+

x3 x5 + + ··· 3! 5!

x2n (2n)!

=1+

x2 x4 + + ··· 2! 4!

B2n 4n (4n − 1) 2n−1 x (2n)!

=x−

x3 2x5 17x7 + − + ··· 3 15 315

=x−

x3 3x5 + + ··· 6 40

=x+

x3 x5 + + ··· 3 5

n=0 ∞ X

n=0 ∞ X n=1 ∞ X n=0 ∞ X n=0

(−1)n (2n)! x2n+1 + 1)

4n (n! )2 (2n x2n+1 2n + 1

Again, Bn denotes the nth Bernoulli number. Binomial Series

(1 + x)α =

∞   X α n=0

n

xn

424 Where

α n




denotes the generalized binomial coefficient,

  Y n α α−k+1 α(α − 1) · · · (α − n + 1) = = n k n! k=1

Notice that when α ∈ Z− , this series is equivalent to the series for 1 1 1 , , ,··· 2 1 + x (1 + x) (1 + x)3

To derive many of the above series, one needs to employ the technique of interchanging summation and integration. A nice challenge would be trying to derive all the series listed above. Give it a try!

Acknowledgements First and foremost, I would like to thank everyone in the Daily Math community. My followers and supporters on the Instagram page @daily_math_ have been instrumental in my motivation to complete this book. At the time of publishing this book, the Daily Math community is over 40,000 members strong and is growing at an amazing rate. From this amazing community, a group of individuals with varying backgrounds were selected to give feedback on the book before it is published. I am very grateful for their feedback, as they have truly helped shape this book. These individuals, in no particular order, are • Poon Zong Wei Julian • Anton Steinfadt • Duc Van Khanh Tran • Solden Stoll • Kevin Tong • Jack Moffatt • Vasilij Shikunov • Florian Babisch • Keyvon Rashidi 425

426 • Others who did not want their name to be listed I would like to thank everyone who assisted in the publishing of this book. I am indebted to my graphic designer, Ayan Rasulova, for her great efforts in designing the book cover and for Christian Cortez for his help in the design of the book cover. Last, but certainly not least, I would like to express my gratitude to all the amazing people in my life that motivated me to start and finish this book. I truly would not have completed this journey without my parents, teachers, and friends always on my side. I would like to express my sincere gratitude to my calculus teacher, Mary Rubin, for her role in developing my mathematical interests and to my chemistry teacher, Maliha Malik, for her support throughout this journey. A special thank you is owed to Victor Del Carpio, who inspired me to start this book and Hrishik Rangaraju, who always motivated me to complete the book. Their role as friends and peers throughout this journey helped me in keeping my goals in sight through the many ups and downs. I am extremely grateful to my parents for their love, prayers, and sacrifices to give me the best education possible. Their hard work and dedication to my success has shaped my life, and with their help I navigated the many obstacles in writing this book.

Answers

427

428

Chapter 1 4. 1 5. 1

6. −1

7. −

8.

2 e

1 2019

9. 1 10.

64 27

11.

π 2

Chapter 2 √ π 3 − 3 ln 2 1. 9

429

2.

π ln 2 8

3. π √ 3− 2 4. 2 √ 5. 4 2 − 4 √

6.

π 4

π ln 2 √ 3 3   √ 8. 3 2 − 2 π

7.

9. The integrand is always positive on [0, 1], so the integral is positive. Moreover, the integral equals 22 − π, which shows 7 that 22 > π. 7 1

e4π 10. 2

Chapter 3 1. ln 3 2.

π ln 2 8

430

√ 3. π ln 4.

α+ 2

√ ! β

π 2

5. 2π

7.

π ln 2 2

8.

π 2e

π 9. √ 4 αβ



1 1 + α β



Chapter 4 2.

11 96

3.

2019 1010

4. (k + 1)(k + 1)! −1 6.

3 sin 1 4

7. ln 2 − ln 2020 − ln(2021! )

431

Chapter 6

1.

1 2e

2.

179π 360

3.

π 2

4.

(ln 2)2 ζ(2) − 2 2

5.

2k + 2 k+2

Chapter 7

1. 2 − 2. 3.

π

π2 6 

22n+1 π2 2

4. −G 5.

7 ζ(3) 4

 2n n

432

6.

7 ζ(3) 32

7. ln 2 

√

8. π ln

2+2 4



! +

√ π + 1 − 2 4

Chapter 8 n2

2n3 X √ 1. − + k 3 k=1 2.

1 2

3.

1 ζ(2021) − 2 2020 (2020)(2021)

4.

ζ(2) − 1 2

5. π

Chapter 9 2.

ln 2π 1 + 4 8

4. 5! ζ(4)

433

5.

144 625

√ ln2 2π π 2 γ ln 2π γ 2 ζ 00 (2) ln(2π)ζ 0 (2) γζ 0 (2) + + + + 6. − − 3 48 3 12 2π 2 π2 π2

Chapter 10 1. −γ 2. 3 − 2γ 4. 2ζ(3) 6.

3π 2 G 2

−

1 128

  ψ (3) (1/4) − ψ (3) (3/4)

Chapter 11 r 2.

2 π

 !2 3 Γ 4

π3 16
√ πΓ 5/4
6. 2Γ 3/4
4 1 7. Γ(1/4) 4 3.

434

Chapter 12 1. 1 3.

π2 6

4.

π2 − ζ(3) 6

5. 1 − γ 5 6. − ζ(3) 8 7.

7 π 2 ln 2 ζ(3) − 16 8

Integral Table The integration constant has been omitted here.

Rational Functions Z

xn dx =

Z

1 xn+1 , n 6= −1 n+1

1 dx = ln|x| x

Exponential Functions Z

Z

ex dx = ex

ax dx = 435

1 x a ln a

436

Logarithmic Function Z ln x dx = x ln x − x

Trigonometric Functions Z sin x dx = − cos x

Z cos x dx = sin x

Z tan x dx = ln|sec x|

Z sec x dx = ln|sec x + tan x|

Z

sec2 x dx = tan x

Z sec x tan x dx = sec x

Miscellaneous Z

a −1 x dx = tan a2 + x 2 a

437

a 1 x + a dx = ln a2 − x 2 2 x − a

Z Z

1 x √ dx = sin−1 2 2 a a −x Z a x √ dx = sec−1 a x x 2 − a2 Z

Z

√

1 x dx = cosh−1 2 a −a   √ = ln x + x2 − a2

x2

√

1 x dx = sinh−1 a x 2 + a2   √ = ln x + x2 + a2

438

Trigonometric Identities Unit Circle Identities sin2 (θ) + cos2 (θ) = 1 Dividing sin2 (θ) + cos2 (θ) = 1 by cos2 (θ) or sin2 (θ) gives tan2 (θ) + 1 = sec2 (θ) 1 + cot2 (θ) = csc2 (θ)

Addition Formulas

sin(α + β) = sin(α) cos(β) + cos(α) sin(β) cos(α + β) = cos(α) cos(β) − sin(α) sin(β)

Double Angle Formulas

sin(2α) = 2 sin(α) cos(α) cos(2α) = cos2 (α) − sin2 (α) 439

440 The formula for cos(2α) is often rewritten by replacing cos2 (α) with 1 − sin2 (α) or replacing sin2 (α) with 1 − cos2 (α) to get cos(2α) = 1 − 2 sin2 (α) cos(2α) = 2 cos2 (α) − 1 Solving for sin2 (α) and cos2 (α) gives power reduction formulas that are very useful for integration 1 (1 − cos(2α)) 2 1 (1 + cos(2α)) 2

sin2 (α) = cos2 (α) =

An addition formula for tangent can be derived from the ones for sine and cosine. tan(α + β) = Now dividing by

sin(α) cos(β) + cos(α) sin(β) cos(α) cos(β) − sin(α) sin(β)

cos(α) cos(β) gives cos(α) cos(β) tan(α + β) =

tan(α) + tan(β) 1 − tan(α) tan(β)

tan(2α) =

2 tan(α) 1 − tan2 (α)

Triple Angle Formulas     π π sin(3θ) = 3 sin θ − 4 sin θ = 4 sin θ sin − θ sin +θ 3 3 3

cos(3θ) = 4 cos3 θ − 3 cos θ = 4 cos θ cos



   π π − θ cos +θ 3 3

441

tan(3θ) =

    3 tan θ − tan3 θ π π = tan θ tan − θ tan + θ 3 3 1 − 3 tan2 θ

Half Angle Formulas   θ 1 − cos θ sin = 2 2 2

  θ 1 + cos θ cos = 2 2 2

  θ 1 − cos θ tan = 2 sin θ 2

Try deriving other trigonometric identities using these ones as a basis!

442

Alphabetical Index A

E

Apery’s constant . . . . . . . . . 312 Augustin-Louis Cauchy . . . 30

Erwin Madelung . . . . . . . . . 365 Euler’s formula . . . . . . 200, 416 Euler’s reflection formula . 71, 90, 267, 274, 303, 308

B Basel problem . . 268, 269, 313 Bernhard Riemann . . . 60, 312 Bernoulli numbers . . . 141, 423 Binomial theorem . . . .46, 210, 423 Boltzmann’s constant . . . 378, 384, 389, 394 Brook Taylor . . . . . . . . . . . . . . 82

F Fibonacci sequence . . . 44, 154 Fractional derivative . . . . . . 51 Fundamental theorem of arithmetic . . 318, 319

G Gaussian integral . . . .116, 271 Generating function 141, 206, 319 Geometric measure theory 411 Glaisher-Kinkelin constant 253, 328 Gottfried Wilhelm Leibniz 51, 60 Guillaume de L’Hopital . . . 36

C Carl Freidrich Gauss 136, 264, 280, 415 Catalan’s constant . . 222, 275 Chain rule . . . . 50, 55, 57, 358 Crystal structure . . . 364, 368, 377 Curse of dimensionality . . 409

H Harmonic number . . . . . . . 170, 204–206, 228

D Dynamic programming . . . 410 443

444

ALPHABETICAL INDEX

Hausdorff measure . . . . . . . 411 Henri Lebesgue . . . . . . . . . . . 63

M

I

P

Integration by parts . . . . . . . 82

J James Clerk Maxwell . . . . 382 Johann Bernoulli . . . . . 36, 415 Josiah Willard Gibbs . . . . .382

K Karl Weierstrass . 30, 266, 269

L L’Hopital’s rule . 47, 112, 145, 332, 379 Lagrange multiplier . 347, 397 Lagrangian . . . . . . . . . . 346, 354 Lebesgue dominated convergence theorem 187, 188 Lebesgue integral . . . . . . 63, 64 Lebesgue measure . . . 187, 411 Legendre duplication formula 308 Leonhard Euler . 71, 241, 268, 312, 313, 317, 416, 418 Lord Rayleigh . . . . . . . . . . . .389 Ludwig Boltzmann . . . . . . . 382

Max Planck . . . . . . . . . . . . . . 390 Pafnuty Chebyshev . . . . . . 317 Pauli exclusion principle . 386, 393 Polar coordinates . . . 199, 298

R Raabe’s integral . . . . . . . . . .273 René Descartes . . . . . . . . . . 415 Richard Feynman . . . 102, 418 Riemann hypothesis . . . . . .312 Riemann series theorem . 171, 375 Riemann sums 60, 62, 63, 177 Roger Cotes . . . . . . . . . . . . . 416

S Special relativity . . . . . . . . . 362 Spherical coordinates . . . . 310, 402, 404 Stefan–Boltzmann law . . . 393 Stirling’s formula . . . . 43, 250, 396, 409

W Weierstrass factorization theorem . . . . . . . . . 269