Preface
Notations
Series Transformations and Special Functions
Generating Functions and Powerful Identities
Logarithmic Integrals
Harmonic Series
Table of Mathematica Commands
References
Index

##### Citation preview

An Introduction to The Harmonic Series And Logarithmic Integrals For High School Students Up To Researchers

Copyright © 2021 by Ali Shadhar Olaikhan. All rights reserved. No part of this book may be reproduced, distributed, or transmitted in any form or by any means, including photocopying, recording, or other electronic or mechanical methods, without the prior written permission of the publisher, except as provided by United States of America copyright law. For permission requests, e-mail the publisher at [email protected]. First edition published April 2021 Book cover designed by Islam Farid and Aqil Almosawi LATEX class prepared by Elio A. Farina ISBN 978-1-7367360-1-2 (eBook)

To my parents

Preface The harmonic series and logarithmic integrals, which are strongly interrelated, are not commonly found in the standard textbooks. Evaluating them can be challenging to new learners, as it requires specific approaches and a good knowledge of special functions such as the gamma function, the polygamma function, the beta function, the polylogarithm function, and various other special functions and constants. It also requires a lot of experience and patience, since it involves plenty of tricks and time-consuming calculations. The purpose of this book is to introduce the harmonic series in a way suitable for all readers with a good knowledge of calculus, from high school students to researchers. The book is the result of over five years of working on the harmonic series. As I taught myself this topic, I struggled to find the proofs for most of the identities required for evaluating the harmonic series. With the experience gained over years, I managed to prove these identities in detail using only basic definitions and well– known techniques, and without using contour integration or the residue theorem, which require a deep understanding of complex analysis. I would like to inform the reader that I borrowed a few proofs from some sites, mainly from the Mathematics Stack Exchange site, adding more details and modifying them my own way. Also, most of the text is written in equations, so the reader won’t find much unnecessary verbiage in this book. The book consists of four chapters. Chapter 1 presents some essential series transformations and special functions and shows how these functions are related to each other. It explains the definition and properties of each function and also derives many special values needed for the calculations in chapters 3 and 4. In chapter 2, the reader will find the derivations of plenty of useful identities: generating functions involving the harmonic number and series expansion of powers of arcsin(x). Other identities are derived using the beta function, the Cauchy product, Abel’s summation, and Fourier series. Chapter 3 prepares all the integral results required to calculate the harmonic series in chapter 4, including some new results. These were derived using algebraic identities, integral manipulations and the beta function. Chapter 4 shows how to calculate many types of harmonic series: non-alternating series, alternating series, series with powers of 2 in the denominator, series with powers of 2n + 1 in the denominator, series with rational argument, series with skew iv

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harmonic number, series with central binomial coefficient, and many others. Several solutions are presented using two different methods. At the end of the book, I have provided a table of Mathematica commands for approximating or evaluating limits, derivatives, integrals, and series, so that the reader can verify any result of interest throughout the book. More advanced and challenging problems about the harmonic series may be found on my Mathematics Stack Exchange page, https://math.stackexchange.com /users/432085/ali-shadhar. I decided not to include them in the book for the sake of simplicity. To keep up to date with any new identities or results, you can follow my Facebook group, Harmonic Series, https://www.facebook.com/gro ups/178723409566339. Finally, I would like to express my gratitude to my friend Cornel Ioan Vălean for being a big motivation for me to explore the realm of the harmonic series through his amazing problems and solutions, many of which are included in his book, (Almost) Impossible Integrals, Sums, and Series, and for his valuable tips for writing this book. I would also like to thank Elio Arturo and my brother Hasan Shadhar for their help in using LaTeX. I extend my gratitude to my friends, Khalaf Ruhemi, Shivam Sharma, and Hasan Hussein for all the support and encouragement they offered me while writing this book. I also want to thank my parents, to whom I am dedicating this book, for all their support. Phoenix, Arizona, USA April 2021

Contents 1

Series Transformations and Special Functions 1.1 Shifting the Sum Index . . . . . . . . . . . . . . 1.2 Reversing the Order of the Sum Terms . . . . . . 1.3 Splitting a Sum Into its Odd and Even Parts . . . 1.4 Converting the Summand a2n to an . . . . . . . 1.5 Converting the Summand a2n+1 to an . . . . . . 1.6 Converting the Summand (−1)n a2n to in an . . . 1.7 Converting the Summand (−1)n a2n+1 to in an . 1.8 Converting a Sum to a Product . . . . . . . . . . 1.9 Double Summation . . . . . . . . . . . . . . . . 1.10 The Logarithm of a Complex Number . . . . . . 1.11 Gamma Function . . . . . . . . . . . . . . . . . 1.11.1 Definition . . . . . . . . . . . . . . . . . 1.11.2 Functional Equation . . . . . . . . . . . 1.11.3 Stirling’s Approximation . . . . . . . . . 1.11.4 Expressing Gamma Function as a Product 1.11.5 Euler’s Definition as an Infinite Product . 1.11.6 Euler’s Reflection Formula . . . . . . . . 1.11.7 Legendre Duplication Formula . . . . . . 1.12 Beta Function . . . . . . . . . . . . . . . . . . . 1.12.1 Definition . . . . . . . . . . . . . . . . . 1.12.2 Trigonometric Integral Representation . . 1.12.3 Improper Integral Representation . . . . 1.12.4 Powerful Integral Representation . . . . . 1.13 Riemann Zeta Function . . . . . . . . . . . . . . 1.13.1 Definition . . . . . . . . . . . . . . . . . 1.13.2 Integral Representation . . . . . . . . . . 1.13.3 Evaluation of ζ(0) . . . . . . . . . . . . 1.13.4 Evaluation of ζ(2) . . . . . . . . . . . . 1.13.5 Evaluation of ζ(2n) . . . . . . . . . . . 1.14 Dirichlet Eta Function . . . . . . . . . . . . . . . 1.14.1 Definition . . . . . . . . . . . . . . . . .

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1.14.2 Integral Representation . . . . . . . . . . . . . . . 1.15 Dirichlet Beta Function . . . . . . . . . . . . . . . . . . . 1.15.1 Definition . . . . . . . . . . . . . . . . . . . . . . 1.15.2 Integral Representation . . . . . . . . . . . . . . . 1.15.3 Evaluation of β(2a) . . . . . . . . . . . . . . . . 1.15.4 Evaluation of β(2a + 1) . . . . . . . . . . . . . . 1.16 Polylogarithm Function . . . . . . . . . . . . . . . . . . . 1.16.1 Definition . . . . . . . . . . . . . . . . . . . . . . 1.16.2 Dilogarithm Reflection Formula . . . . . . . . . . 1.16.3 Landen’s Dilogarithm Identity . . . . . . . . . . . 1.16.4 Dilogarithm Inversion Formula . . . . . . . . . . . 1.16.5 Relation Involving Four Dilogarithm Functions . . 1.16.6 Another Relation Involving Dilogarithm Functions 1.16.7 Landen’s Trilogarithm Identity . . . . . . . . . . . 1.16.8 Polylogarithm Inversion Formula . . . . . . . . . . 1.17 Harmonic Number . . . . . . . . . . . . . . . . . . . . . 1.17.1 Definition . . . . . . . . . . . . . . . . . . . . . . 1.17.2 Infinite Series Representation . . . . . . . . . . . 1.18 Skew Harmonic Number . . . . . . . . . . . . . . . . . . 1.18.1 Definition . . . . . . . . . . . . . . . . . . . . . . 1.18.2 Infinite Series Representation . . . . . . . . . . . 1.19 Digamma Function . . . . . . . . . . . . . . . . . . . . . 1.19.1 Definition . . . . . . . . . . . . . . . . . . . . . . 1.19.2 Digamma Reflection Formula . . . . . . . . . . . 1.19.3 Digamma–Harmonic Number Identity . . . . . . . 1.20 Polygamma Function . . . . . . . . . . . . . . . . . . . . 1.20.1 Definition . . . . . . . . . . . . . . . . . . . . . . 1.20.2 Series Representation . . . . . . . . . . . . . . . . 1.20.3 Integral Representation . . . . . . . . . . . . . . . 1.20.4 Evaluation of ψ (a) (1). . . . . . . . . . . . . . . . 1.20.5 Evaluation of ψ (a) 12 . . . . . . . . . . . . . . .  1.20.6 Evaluation of ψ (2a) 14 . . . . . . . . . . . . . .  1.20.7 Evaluation of ψ (2a) 34 . . . . . . . . . . . . . . 1.21 Catalan’s Constant . . . . . . . . . . . . . . . . . . . . . 1.22 Euler–Mascheroni Constant . . . . . . . . . . . . . . . . . 2

Generating Functions and Powerful Identities 2.1 Generating Functions . . . . . . . . . . . . P∞ (a) 2.1.1 Pn=1 Hn xn . . . . . . . . . . . ∞ Hn n 2.1.2 . . . . . . . . . . . . n=1 n x P∞ Hn n 2.1.3 . . . . . . . . . . . . n=1 n2 x P∞ Hn(2) n 2.1.4 . . . . . . . . . . . n x Pn=1 (2) n ∞ 2 2.1.5 . . . . . . . n=1 (Hn − Hn )x

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Contents

2.1.6 2.1.7 2.1.8 2.1.9 2.1.10 2.1.11 2.1.12 2.1.13 2.1.14 2.1.15

2.2

2.3

2.4

2 (2) Hn −Hn xn . . n=1 n P∞ Hn2 n . . . . . n x Pn=1 ∞ Hn n . . . . . n=1 n3 x P∞ Hn(2) n . . . . n=1 n2 x P∞ Hn(3) n . . . . n x Pn=1 ∞ 3 n H x . . . . . n 2 Pn=1 ∞ Hn n . . . . . n=1 n2 x P∞ (2) n . . Hn Hn x Pn=1 (2) ∞ 3 − 3H H (H n n n n=1 P∞ Hn Hn(2) n x . . . n=1 P∞ Hn3 n n . . . . . n=1 n x

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2.1.16 . . . . . . . . . . . . . . . . . . P∞ (2) 4 2 2.1.17 Pn=1 (Hn − 6Hn Hn + 8Hn Hn(3) + 3(Hn(2) )2 − 6Hn(4) )xn ∞ H n xn . . . . . . . . . . . . . . . . . . . . . . . 2.1.18 Pn=1 ∞ Hn n 2.1.19 . . . . . . . . . . . . . . . . . . . . . . . n x Pn=1 ∞ Hn n 2.1.20 Pn=1 n2 x . . . . . . . . . . . . . . . . . . . . . . . ∞ n n . . . . . . . . . . . . . . . . . . . . . . . 2.1.21 n=1 H 2 x P∞ Hn/2 n . . . . . . . . . . . . . . . . . . . . . . 2.1.22 n x Pn=1 ∞ Hn/2 n 2.1.23 . . . . . . . . . . . . . . . . . . . . . . n=1 n2 x P∞ (2n n) n 2.1.24 n=1 4n Hn x . . . . . . . . . . . . . . . . . . . . . P∞ (2n n ) Hn n 2.1.25 n=1 4n n x . . . . . . . . . . . . . . . . . . . . . 2n P∞ ( n ) Hn n 2.1.26 4n n2 x . . . . . . . . . . . . . . . . . . . . . Pn=1 ∞ 2H2n −Hn 2n 2.1.27 x . . . . . . . . . . . . . . . . . . . n=1 n P∞ H2n 2n+1 2.1.28 x . . . . . . . . . . . . . . . . . . . . n=1 P∞ 2n+1 (−1)n H2n 2n+1 x . . . . . . . . . . . . . . . . . . 2.1.29 2n+1  2n Pn=1 ∞ Hn −H2n 1 2.1.30 − 2n2 x . . . . . . . . . . . . . . . n=1 n Series Expansion of Powers of arcsin(z) . . . . . . . . . . . . . 2.2.1 Series Expansion of arcsin(z) . . . . . . . . . . . . . . √ 2.2.2 Series Expansion of arcsin(z) . . . . . . . . . . . . . . . 1−z 2

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2.2.3 Series Expansion of arcsin3 (z) . . . . . 2.2.4 Series Expansion of arcsin4 (z) . . . . . Identities by Beta Function . . . . . . . . . . . . 2.3.1 Expressing Beta Function as a Product . . 2.3.2 Evaluation of Four Logarithmic Integrals Identities by Cauchy Product . . . . . . . . . . . 2.4.1 Cauchy Product of Two Power Series . . 2.4.2 Cauchy Product of − ln(1 − x) Li2 (x) . . 2.4.3 Cauchy Product of Li22 (x) . . . . . . . . 2.4.4 Cauchy Product of − ln(1 − x) Li3 (x) . .

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2.5

2.6

3

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2.4.5 Cauchy Product of Li2 (x) Li3 (x) . . . 2.4.6 Cauchy Product of Li23 (x) . . . . . . . 2.4.7 Cauchy Product of − ln(1 − x) Li4 (x) . Identities by Abel’s Summation . . . . . . . . . 2.5.1 Abel’s Summation . . . . . . . . . . . 2.5.2 First Application . . . . . . . . . . . . 2.5.3 Second Application . . . . . . . . . . . 2.5.4 Third Application . . . . . . . . . . . . Identities By Fourier Series . . . . . . . . . . . 2.6.1 Fourier Series . . . . . . . . . . . . . . 2.6.2 Fourier Series of Even Function . . . . 2.6.3 Fourier Series of Odd Function . . . . 2.6.4 Fourier Series of cos(zx) . . . . . . . . 2.6.5 Fourier Series of sin(zx) . . . . . . . . 2.6.6 Fourier Series of ln(sin x) . . . . . . . 2.6.7 Fourier Series of ln(cos x) . . . . . . . 2.6.8 Fourier Series of ln(tan x) . . . . . . . π 2.6.9 Series Representation of sin(πz) . . . . 2.6.10 Series Representation of cot(πz) . . . . 2.6.11 Euler’s Product Formula of sin(πz)  . . 2.6.12 Series Representation of sec π2 z . . . 2.6.13 Series Representation of sin(x) . . . . 2.6.14 Series Representation of tan x ln(sin x) 2.6.15 Series Representation of ln2 (2 cos x) .

Logarithmic Integrals 3.1 Generalized Logarithmic Integrals R 1 lna (x) 3.1.1 dx . . . . . . . . R01 ln1−x a (x) 3.1.2 dx . . . . . . . . 0 1+x R 1 lna ( 1−x 1+x ) 3.1.3 dx . . . . . . x 0 a−1 R 1 ln( 1−x ln (x) ) 1+x 3.1.4 dx . . R01 lna (1−x)x dx . . . . . . 3.1.5 1+x 0 R 12 lna (x) 3.1.6 dx . . . . . . . R01 ln1−x a (1+x) 3.1.7 dx . . . . . . x 0 x R 1 ln2a−1 ( 1−x ) 3.1.8 dx . . . . 1+x R0∞ lna (1+x) 3.1.9 dx . . . . . . 2 R01 lna1+x (1−x) 3.1.10 0 1+x2 dx . . . . . . R ∞ 2a 3.1.11 0 ln1+x(x) 2 dx . . . . . . . R ∞ Lia (−x) 3.1.12 0 1+x2 dx . . . . . . R1 (−x) 3.1.13 0 Li2a+1 dx . . . . . 1+x2

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3.1.14 3.1.15 3.1.16 3.1.17 3.1.18 3.1.19 3.1.20 3.1.21 3.1.22 3.1.23 3.1.24 3.1.25 3.1.26 3.1.27 3.1.28 3.1.29 3.1.30 3.1.31

3.2

ln2a (x) ln(1+x) dx . . 1+x2 0 R ∞ Lia (−x2 ) 1+x2 dx . . . . . 0 R 1 Li2a+1 (−x2 ) dx . . . . 1+x2 R01 ln2a (x) arctan(x) dx . . 1−x2 R0∞ ln2a (x) ln(1+x) √ dx . . 0 x(1+x) R 1 ln2a (x) ln(1+x2 ) dx . . 2 R01 lna (x)1+x lna (1−x) dx . . x(1−x) 0 R 12 lna (x) lna (1−x) dx . . x(1−x) R01 lna (x) ln(1−x) dx . . . R01 lna (x)1−x ln(1−x) dx . . . x(1−x) R01 lna (x) ln(1+x) dx . . . R01 lna (x)1+x ln(1+x) dx . . . x(1+x) R01 lna (1−x) ln(1+x) dx . x 0 R 1 lna (x) ln( 1+x 2 ) dx . . . 1−x R01 lna (1−x) Li2 (x) dx . . R0∞ ln2a−1x(x) ln(1+x) dx . x(1+x) R01 x2n dx . . . . . . . . R01 1+x xn dx . . . . . . . . R01 1+x 2n−1 x arctanh(x)dx R01 n−1 x Li a (x)dx . . . 0

R1

3.1.32 3.1.33 Results of Logarithmic Integrals R π2 3.2.1 sin(2nx) cot(x)dx . R0π2 ln(sin x)dx . . . . . 3.2.2 R0π2 2 3.2.3 ln (sin x)dx . . . . R0π2 3.2.4 ln(sin x) ln(cos x)dx R01 ln(x) ln(1−x) √ 3.2.5 dx . . . 0 x 1−x R ∞ ln2 (x) ln(1+x2 ) 3.2.6 dx . . 1+x2 R01 ln(1−x) ln(1+x) 3.2.7 dx . . x R01 ln(x) ln(1−x) ln(1+x) 3.2.8 dx R01 ln(1−x) lnx2 (1+x) 3.2.9 dx . . R01 ln2 (1−x)x ln(1+x) 3.2.10 0 dx . . R 1 ln3 (1−x)x ln(1+x) 3.2.11 0 dx . . R 1 ln(1−x) xln3 (1+x) 3.2.12 0 dx . . R 1 ln3 (1+x)x ln(x) 3.2.13 0 dx . . . x

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3.2.14 3.2.15 3.2.16

ln(x) ln(1+x) dx 1−x ln(x) ln(1−x) dx 1+x R01 ln(x) ln2 (1−x) dx 1+x 0

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R01

3.2.22

R 21

3.2.23

1

R01

3.2.24

R01

3.2.25

R1

3.2.26 3.2.27 3.2.28 3.2.29 4

R1

0

R012 R01 R01 R01 0

. . . . . . . . . . . . . . . . . . . . . . 174 . . . . . . . . . . . . . . . . . . . . . . 175 R 1 2 ln(1−x) & 0 ln (x)1+x dx . . . . . . . . . . 175

R1 R1 ln(1+x) ln(1+x) ln2 (1+x) dx, 0 ln(1−x) dx, & 0 ln(x)1+x dx 2 1+x2 1+x2 ln(x) ln2 (1−x) √ dx . x(1−x) ln2 (x) ln(1−x) dx . 1−x ln2 (x) ln(1+x) dx . 1+x ln3 (1−x) ln(x) dx . x Li2 (−x) 1+x2 dx . . . . ln(x) arctan x dx . 1+x ln2 (x) arctan x dx . x(1+x2 ) Li22 (−x) dx . . . . x Li22 (−x) dx . . . . x ln2 (1−x) Li2 (x) dx x ln3 (1−x) Li2 (x) dx x ln4 (1−x) Li2 (x) dx x

. . . 177

. . . . . . . . . . . . . . . . . . . . . 179 . . . . . . . . . . . . . . . . . . . . . 180 . . . . . . . . . . . . . . . . . . . . . 181 . . . . . . . . . . . . . . . . . . . . . 181 . . . . . . . . . . . . . . . . . . . . . 183 . . . . . . . . . . . . . . . . . . . . . 183 . . . . . . . . . . . . . . . . . . . . . 184 . . . . . . . . . . . . . . . . . . . . . 186 . . . . . . . . . . . . . . . . . . . . . 187 . . . . . . . . . . . . . . . . . . . . . 188 . . . . . . . . . . . . . . . . . . . . . 188 . . . . . . . . . . . . . . . . . . . . . 188

Harmonic Series 4.1 Generalized Series . . . P∞Harmonic Hn/p 4.1.1 . . . . . . . . q n Pn=1 ∞ Hn . . . . . . . . . 4.1.2 n=1 nq P∞ Hn 4.1.3 nq . . . . . . . . . Pn=1 ∞ (−1)n Hn . . . . . . 4.1.4 n=1 n2q P∞ (−1)n H n 4.1.5 . . . . . . n2q Pn=1 ∞ Hn/2 4.1.6 . . . . . . . . n=1 n2q n P∞ (−1) Hn/2 4.1.7 . . . . . n=1 n2q P∞ ζ(q)−Hn(q) 4.1.8 . . . . . n=1 n P∞ Hn(2) 4.1.9 . . . . . . . n=1 n2q+1 P∞ Hn(2q+1) 4.1.10 . . . . . . n=1 n2 2 P∞ Hn 4.1.11 . . . . . . . n2q+1 Pn=1 ∞ Hn 4.1.12 . . . . . . q n=1 (2n+1) P∞ (−1)n Hn(q) . . . . . 4.1.13 n=1 n P∞ (−1)n Hn(2q+1) 4.1.14 . . . n=1 2n+1 n P∞ (−1) Hn 4.1.15 n=1 (2n+1)2q+1 . . . . . 4.2 Non–Alternating Harmonic Series

189 . . . . . . . . . . . . . . . . . . 189 . . . . . . . . . . . . . . . . . . 189 . . . . . . . . . . . . . . . . . . 190 . . . . . . . . . . . . . . . . . . 194 . . . . . . . . . . . . . . . . . . 198 . . . . . . . . . . . . . . . . . . 201 . . . . . . . . . . . . . . . . . . 202 . . . . . . . . . . . . . . . . . . 202 . . . . . . . . . . . . . . . . . . 203 . . . . . . . . . . . . . . . . . . 205 . . . . . . . . . . . . . . . . . . 207 . . . . . . . . . . . . . . . . . . 208 . . . . . . . . . . . . . . . . . . 210 . . . . . . . . . . . . . . . . . . 212 . . . . . . . . . . . . . . . . . . 213 . . . . . . . . . . . . . . . . . . 214 . . . . . . . . . . . . . . . . . . 216

xii

Contents

4.2.1

P∞

. . . . . . . . . . . . . . . . . . . . . . . . . 216

4.2.2

P∞

. . . . . . . . . . . . . . . . . . . . . . . . . 216

4.2.3 4.2.4 4.2.5 4.2.6 4.2.7 4.2.8 4.2.9 4.2.10 4.2.11 4.2.12 4.2.13 4.2.14 4.2.15 4.2.16 4.2.17 4.2.18 4.2.19 4.2.20 4.2.21 4.2.22 4.2.23 4.3

Hn n=1 n2 . . (2) Hn . n=1 n2 P∞ Hn2 n2 . . Pn=1 ∞ Hn H2n n=1 n2 P∞ Hn(2) . n=1 n3 P∞ Hn(3) . n=1 n2 P∞ Hn2 n=1 n3 . . P∞ Hn Hn(2) n=1 n2 P∞ Hn3 n=1 n2 . . P∞ Hn(2) . n=1 n4 P∞ Hn2 n=1 n4 . . P∞ Hn(4) . n=1 n2 P∞ (Hn(2) )2 n=1 n2 P∞ Hn Hn(3) n=1 n2 P∞ Hn2 Hn(2) n=1 n2 P∞ Hn4 n=1 n2 . . P∞ Hn Hn(2) n=1 n3 P∞ Hn3 n=1 n3 . . P∞ Hn(2) . n=1 n5 P∞ Hn2 n=1 n5 . . P∞ Hn(3) . n=1 n4 P∞ Hn(4) . n=1 n3 P∞ Hn2 Hn(2) n=1 n3 P∞ Hn(2) . n=1 n7

. . . . . . . . . . . . . . . . . . . . . . . . . 217 . . . . . . . . . . . . . . . . . . . . . . . . . 218 . . . . . . . . . . . . . . . . . . . . . . . . . 220 . . . . . . . . . . . . . . . . . . . . . . . . . 220 . . . . . . . . . . . . . . . . . . . . . . . . . 220 . . . . . . . . . . . . . . . . . . . . . . . . 222 . . . . . . . . . . . . . . . . . . . . . . . . . 225 . . . . . . . . . . . . . . . . . . . . . . . . . 225 . . . . . . . . . . . . . . . . . . . . . . . . . 226 . . . . . . . . . . . . . . . . . . . . . . . . . 226 . . . . . . . . . . . . . . . . . . . . . . . . . 227 . . . . . . . . . . . . . . . . . . . . . . . . 227 . . . . . . . . . . . . . . . . . . . . . . . . 228 . . . . . . . . . . . . . . . . . . . . . . . . . 230 . . . . . . . . . . . . . . . . . . . . . . . . 230 . . . . . . . . . . . . . . . . . . . . . . . . . 231 . . . . . . . . . . . . . . . . . . . . . . . . . 232 . . . . . . . . . . . . . . . . . . . . . . . . . 233 . . . . . . . . . . . . . . . . . . . . . . . . . 234 . . . . . . . . . . . . . . . . . . . . . . . . . 234 . . . . . . . . . . . . . . . . . . . . . . . . 235

4.2.24 . . . . Alternating Harmonic Series P∞ (−1)n Hn 4.3.1 . . . n Pn=1 ∞ (−1)n H2n 4.3.2 . . n=1 n P∞ (−1)n Hn 4.3.3 . . . n2 Pn=1 ∞ (−1)n H2n 4.3.4 . . n=1 n2 P∞ (−1)n Hn(2) 4.3.5 . . n=1 n P∞ (−1)n Hn(3) 4.3.6 . . n Pn=1 ∞ (−1)n Hn 4.3.7 . . . n=1 n3 P∞ (−1)n Hn(2) 4.3.8 . . n=1 n2 P∞ (−1)n Hn2 4.3.9 . . . n=1 n2

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236 238 238 238 239 240

. . . . . . . . . . . . . . . . . . . . . 240 . . . . . . . . . . . . . . . . . . . . . 241 . . . . . . . . . . . . . . . . . . . . . 241 . . . . . . . . . . . . . . . . . . . . . 243 . . . . . . . . . . . . . . . . . . . . . 243

Contents

xiii

4.3.10 4.3.11 4.3.12 4.3.13 4.3.14 4.3.15 4.3.16 4.3.17 4.4

4.5

P∞

n=1 P∞ Pn=1 ∞ n=1 P∞ n=1 P∞ n=1 P∞ n=1 P∞ n=1 P∞ n=1 P∞ n=1

(2) (−1)n Hn Hn n 3 (−1)n Hn . . n (−1)n Hn . . n4 (2) (−1)n Hn . n3 2 (−1)n Hn . . n3 (4) (−1)n Hn . n (3) (−1)n Hn . n2 (2) (−1)n Hn Hn n2 3 (−1)n Hn . . n2

. . . . . . . . . . . . . . . . . . . . . . 244 . . . . . . . . . . . . . . . . . . . . . . 245 . . . . . . . . . . . . . . . . . . . . . . 245 . . . . . . . . . . . . . . . . . . . . . . 247 . . . . . . . . . . . . . . . . . . . . . . 247 . . . . . . . . . . . . . . . . . . . . . . 248 . . . . . . . . . . . . . . . . . . . . . . 249 . . . . . . . . . . . . . . . . . . . . . . 249

4.3.18 . . . . . . . . . . . . . . . . . Harmonic Series with Powers of 2 in the Denominator . . P∞ Hn 4.4.1 . . . . . . . . . . . . . . . . . . . . . n n=1 n2 P∞ Hn 4.4.2 . . . . . . . . . . . . . . . . . . . . . n=1 n2 2n P∞ Hn(2) 4.4.3 n=1 n2n . . . . . . . . . . . . . . . . . . . . . P∞ Hn2 4.4.4 n2n . . . . . . . . . . . . . . . . . . . . . Pn=1 ∞ Hn 4.4.5 n=1 n3 2n . . . . . . . . . . . . . . . . . . . . . P∞ Hn(2) 4.4.6 n=1 n2 2n . . . . . . . . . . . . . . . . . . . . . P∞ Hn2 4.4.7 n=1 n2 2n . . . . . . . . . . . . . . . . . . . . . P∞ Hn(3) 4.4.8 n2n . . . . . . . . . . . . . . . . . . . . . Pn=1 ∞ Hn 4.4.9 n=1 n4 2n . . . . . . . . . . . . . . . . . . . . . P∞ Hn(4) 4.4.10 n=1 n2n . . . . . . . . . . . . . . . . . . . . . P∞ Hn(2) 4.4.11 n=1 n3 2n . . . . . . . . . . . . . . . . . . . . . P∞ Hn(3) 4.4.12 n=1 n2 2n . . . . . . . . . . . . . . . . . . . . . P∞ Hn2 4.4.13 n=1 n3 2n . . . . . . . . . . . . . . . . . . . . . P∞ Hn Hn(2) 4.4.14 . . . . . . . . . . . . . . . . . . . n=1 n2 2n P∞ Hn3 4.4.15 n=1 n2 2n . . . . . . . . . . . . . . . . . . . . . Harmonic Series with Powers of 2n + 1 in the denominator P∞ (−1)n H2n+1 . . . . . . . . . . . . . . . . . 4.5.1 n=0 n P∞ (−1)2n+1 H2n+1 . . . . . . . . . . . . . . . . . 4.5.2 n=0 (2n+1)2 (2) P∞ (−1)n H2n+1 4.5.3 . . . . . . . . . . . . . . . . . 2n+1 Pn=0 ∞ Hn 4.5.4 n=1 (2n+1)2 . . . . . . . . . . . . . . . . . . . P∞ (−1)n Hn 4.5.5 n=0 (2n+1)2 . . . . . . . . . . . . . . . . . . . P∞ (−1)n Hn(2) . . . . . . . . . . . . . . . . . . 4.5.6 n=0 2n+1 n P∞ (−1) H2n+1 4.5.7 . . . . . . . . . . . . . . . . . n=0 (2n+1)3 (2) n P∞ (−1) H2n+1 4.5.8 . . . . . . . . . . . . . . . . . n=0 (2n+1)2

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251 252 252 252

. . . . . 253 . . . . . 254 . . . . . 254 . . . . . 255 . . . . . 256 . . . . . 256 . . . . . 257 . . . . . 258 . . . . . 258 . . . . . 259 . . . . . 260 . . . . . 262 . . . . . 265 . . . . . 266 . . . . . 266 . . . . . 266 . . . . . 267 . . . . . 267 . . . . . 268 . . . . . 269 . . . . . 271 . . . . . 272

xiv

Contents

4.5.9 4.5.10 4.5.11

4.7

4.8

4.9

n=1 P∞ n=1 P∞ n=1 P∞ n=1 P∞ n=1

(2) Hn (2n+1)2 2 Hn (2n+1)2 (2) Hn (2n+1)3 (3) Hn (2n+1)2 (3) Hn (2n+1)3

. . . . . . . . . . . . . . . . . . . . . . . . 272 . . . . . . . . . . . . . . . . . . . . . . . . 275 . . . . . . . . . . . . . . . . . . . . . . . . 275

. . . . . . . . . . . . . . . . . . . . . P∞ (−1)n Hn(3) + 4 n=1 . . . . . . . . . . 4.5.13 n3 Skew Harmonic Series . . . . . . . . . . . . . . . . . . . . . P∞ (−1)n H n 4.6.1 . . . . . . . . . . . . . . . . . . . . . n=1 n P∞ Hn 4.6.2 . . . . . . . . . . . . . . . . . . . . . . . . n=1 n3 P∞ (−1)n H n 4.6.3 . . . . . . . . . . . . . . . . . . . . . n=1 n3 P∞ (−1)n H n Hn 4.6.4 . . . . . . . . . . . . . . . . . . . n Pn=1 ∞ H n Hn 4.6.5 . . . . . . . . . . . . . . . . . . . . . . n=1 n2 P∞ (−1)n H n Hn . . . . . . . . . . . . . . . . . . . 4.6.6 n2 Pn=1 ∞ H 2n H2n 4.6.7 . . . . . . . . . . . . . . . . . . . . . n=1 n2 Harmonic Series with Rational Argument . . . . . . . . . . . P∞ (−1)n Hn/2 4.7.1 . . . . . . . . . . . . . . . . . . . . n Pn=1 ∞ Hn/2 4.7.2 n=1 n2 . . . . . . . . . . . . . . . . . . . . . . . n P∞ (−1) Hn/2 4.7.3 . . . . . . . . . . . . . . . . . . . . n=1 n2 P∞ Hn/2 4.7.4 n=1 n3 . . . . . . . . . . . . . . . . . . . . . . . n P∞ (−1) Hn/2 4.7.5 . . . . . . . . . . . . . . . . . . . . n=1 n3 P∞ Hn Hn/2 . . . . . . . . . . . . . . . . . . . . . 4.7.6 n=1 n2n P∞ (−1) Hn Hn/2 4.7.7 . . . . . . . . . . . . . . . . . . 2 n=1 P∞ (−1)n nHn/2 4.7.8 . . . . . . . . . . . . . . . . . . . . n=1 n4 Harmonic Series2nwith Binomial Coefficient in the Numerator . P∞ ( n ) Hn 4.8.1 . . . . . . . . . . . . . . . . . . . . . n=1 4n n n P∞ (2n n ) (−1) Hn 4.8.2 . . . . . . . . . . . . . . . . . . n=1 4n n P∞ (2n ) Hn n . . . . . . . . . . . . . . . . . . . . . 4.8.3 n=1 4n n2 (2) P∞ (2n n ) Hn 4.8.4 n=1 4n n . . . . . . . . . . . . . . . . . . . . . (2) P∞ (2n ) H n 2n 4.8.5 n=1 4n n . . . . . . . . . . . . . . . . . . . . . 2 P∞ (2n n ) Hn 4.8.6 . . . . . . . . . . . . . . . . . . . . . n=1 4n n 2 P∞ (2n ) H n n 4.8.7 . . . . . . . . . . . . . . . . . . . . . n=1 4n n2 Harmonic Series with Binomial Coefficient in the Denominator P∞ 4n Hn 4.9.1 . . . . . . . . . . . . . . . . . . . . . n=1 (2n) n2 P∞ 4nn H2n 4.9.2 n=1 (2n) n2 . . . . . . . . . . . . . . . . . . . . . P∞ 4nn Hn 4.9.3 . . . . . . . . . . . . . . . . . . . . . n=1 (2n) n3 n 4.5.12

4.6

P∞

. . . 276 . . . 276 . . . 277 . . . 277 . . . 277 . . . 278 . . . 279 . . . 280 . . . 281 . . . 284 . . . 285 . . . 285 . . . 286 . . . 286 . . . 287 . . . 288 . . . 288 . . . 289 . . . 290 . . . 291 . . . 291 . . . 292 . . . 292 . . . 293 . . . 294 . . . 295 . . . 297 . . . 299 . . . 299 . . . 301 . . . 302

Contents

xv

4.9.4 4.9.5 4.9.6

P∞

n=1

P∞

(2) 4n Hn 2n n ( ) 2 n n

2 Hn n2

4 n=1 (2n) P∞ 4nn H2n n=1 (2n) n3

. . . . . . . . . . . . . . . . . . . . . . . . 304 . . . . . . . . . . . . . . . . . . . . . . . . 305 . . . . . . . . . . . . . . . . . . . . . . . . 307

n

Table of Mathematica Commands

312

References

314

Index

316

Notations C R Z Z≥0 Z≤0 Z+ Z− R(z) J(z) n! a b



Γ B ζ η β Lin Hn (a)

Hn

Hn

The set of complex numbers The set of real numbers The set of integers (Z = {. . . , −3, −2, −1, 0, 1, 2, 3, . . . }) The set of non-negative integers (Z≥0 = {0, 1, 2, . . . }) The set of non-positive integers (Z≤0 = {. . . , −2, −1, 0}) The set of positive integers (Z+ = {1, 2, 3, . . . }) The set of negative integers (Z− = {. . . , −3, −2, −1}) The real part of a complex number z The imaginary part of a complex number z n factorial Qn n! = 1 · 2 · 3 · · · (n − 1) · n = k=1 k, n ∈ Z+ The central binomial coefficient  Γ(a+1) a b = Γ(b+1)Γ(a−b+1) The Gamma R ∞ function Γ(z) = 0 tz−1 e−t dt, R(z) > 0 The Beta function R1 B(a, b) = 0 xa−1 (1 − x)b−1 dx, R(a) > 0, R(b) > 0 The Riemann zeta function P ∞ ζ(z) = 1 + 21z + 31z + · · · = k=1 k1z , R(z) > 1 The Dirichlet eta function k−1 P∞ η(z) = 1 − 21z + 31z − · · · = k=1 (−1) , R(z) > 0 kz The Dirichlet beta function P∞ (−1)k β(z) = 1 − 31z + 51z − · · · = k=0 (2k+1) R(z) > 0 z, The Polylogarithm function P∞ k 2 3 Lin (z) = z + 2zn + 3zn + · · · = k=1 kzn , |z| ≤ 1 The n-th harmonic number Pn Hn = 1 + 21 + 13 + · · · + n1 = k=1 k1 , n ∈ Z+ The n-th harmonic number of order a Pn (a) Hn = 1 + 21a + 31a + · · · + n1a = k=1 k1a , n ∈ Z+ The skew harmonic number n−1 k−1 Pn H n = 1 − 12 + 13 − · · · + (−1)n = k=1 (−1)k , n ∈ Z+ xvi

Notations

ψ ψ (a) ln e γ G

xvii

The Digamma function d ψ(n) = dn ln(Γ(n)) =

ln0 (Γ(n)) Γ(n)

The Polygamma function da da+1 ψ (a) (n) = dn a ψ(n) = dna+1 ln(Γ(n)) The natural logarithm (loge ) The base of the natural logarithm e = limn→∞ (1 + 1/n)n = 2.7182818284590 . . . The Euler–Mascheroni constant γ = limn→∞ (Hn − ln(n)) = 0.5772156649015 . . . The Catalan’s constant P∞ (−1)k G = 1 − 312 + 512 − · · · = k=0 (2k+1) 2 = 0.9159655941772 . . .

Chapter 1

Series Transformations and Special Functions 1.1

Shifting the Sum Index n X k=m

Proof. The index k in

Pn

k=m

ak =

n+c X

ak−c .

(1.1)

k=m+c

ak ranges from m to n: m ≤ k ≤ n.

Replace k by j − c, m ≤ j − c ≤ n. On solving this compound inequality, we get m + c ≤ j ≤ n + c. This indicates that if we replace the index k by j − c, the index j will range from m + c to n + c: n n+c X X ak = aj−c . k=m

j=m+c

Replace j by k in the latter equality to finish the proof. 1

2

Chapter 1. Series Transformations and Special Functions

Example 1: Let ak =

Hk k+1

and m = 0 then shift the index by −1, n n+1 X X Hk−1 Hk = . k+1 k

k=0

k=1

Example 2: Let ak = Hk xk−1 and m = 3 then shift the index by +2, n X

Hk xk−1 =

Hk+2 xk+1 .

k=1

k=3

1.2

n−2 X

Reversing the Order of the Sum Terms n X

ak =

n X

an−k+m .

(1.2)

k=m

k=m

Proof. Following the previous proof, we have m ≤ k ≤ n. Replace k by n − j + m, m≤n+m−j ≤n or m ≤ j ≤ n. This shows that if we replace the index k by n − j + m, the index j will range from m to n as well: n n X X ak = an−j+m . k=m

j=m

The proof completes on replacing j by k in the latter equality. This type of transformation reverses the order of the sum terms. To see that, let m = 1 and n = 4 in (1.2), the LHS sum gives 4 X

ak = a1 + a2 + a3 + a4 ,

k=1

which is equivalent to the RHS sum: 4 X k=1

a5−k = a4 + a3 + a2 + a1 ,

1.3. Splitting a Sum Into its Odd and Even Parts

3

but in reversed order. Example 1: Put ak =

1 k

and m = 1, n n X 1 X 1 = . k n−k+1

k=1

(1.3)

k=1

Example 2: Put ak = k 2 and m = 3, n X

k2 =

k=3

1.3

n X

(n − k + 3)2 .

k=3

Splitting a Sum Into its Odd and Even Parts ∞ X

an =

∞ X

a2n+1 +

a2n .

(1.4)

n=1

n=0

n=1

∞ X

Proof. ∞ X

an = a1 + a2 + a3 + · · ·

n=1

= (a1 + a3 + a5 + · · · ) + (a2 + a4 + a6 + · · · ) ∞ ∞ X X a2n , a2n+1 + = n=0

n=1

and the proof is complete Example 1: Put an =

1 n2 ,

∞ ∞ ∞ ∞ ∞ X X X X 1 1 1 1 1X 1 = + = + n2 (2n + 1)2 n=1 (2n)2 (2n + 1)2 4 n=1 n2 n=1 n=0 n=0

or

∞ ∞ X 1 1 4X = . 2 n 3 n=0 (2n + 1)2 n=1

Example 2: Put an =

Hn n3 ,

∞ ∞ ∞ X X X Hn H2n+1 H2n = + . 3 3 3 n (2n + 1) (2n) n=1 n=0 n=1

4

Chapter 1. Series Transformations and Special Functions

Converting the Summand a2n to an

1.4

∞ X

a2n =

n=1

∞ ∞ 1X 1X an + (−1)n an . 2 n=1 2 n=1

(1.5)

Proof. Starting with the RHS, ∞ X

an +

∞ X

(−1)n an = a1 + a2 + a3 + · · · + (−a1 + a2 − a3 + · · · )

n=1

n=1

= 2a2 + 2a4 + 2a6 + · · · ∞ X a2n . = 2(a2 + a4 + a6 + · · · ) = 2 n=1

The proof finalizes on dividing both sides by 2. Following the same approach, we also find ∞ ∞ ∞ X 1X 1X a2n = an + (−1)n an . (1.6) 2 n=0 2 n=0 n=0 Example 1: Let an =

1 (n+1)4

in (1.5),

∞ X

∞ ∞ 1 1 1X 1 X (−1)n = + . (2n + 1)4 2 n=1 (n + 1)4 2 n=1 (n + 1)4 n=1

Example 2: Let an =

Hn+1 (n+3)3

in (1.6),

∞ ∞ ∞ X H2n+1 1 X Hn+1 1X Hn+1 = + (−1)n . 3 3 (2n + 3) 2 n=0 (n + 3) 2 n=0 (n + 3)3 n=0

Converting the Summand a2n+1 to an

1.5

∞ X n=0

a2n+1 =

∞ ∞ 1X 1X an − (−1)n an . 2 n=1 2 n=1

(1.7)

Proof. ∞ X n=1

an −

∞ X

(−1)n an = a1 + a2 + a3 + · · · − (−a1 + a2 − a3 + · · · )

n=1

= 2a1 + 2a3 + 2a5 + · · ·

1.6. Converting the Summand (−1)n a2n to in an

5 ∞ X

= 2(a1 + a3 + a5 + · · · ) = 2

a2n+1 .

n=0

Divide both sides by 2 to complete the proof. Let’s shift the index of the LHS sum in (1.8) by −1, ∞ X

a2n−1 =

n=1

Example 1: Set an =

Hn n3

∞ ∞ 1X 1X an − (−1)n an . 2 n=1 2 n=1

(1.8)

in (1.7),

∞ ∞ ∞ X H2n+1 1 X Hn 1X Hn (−1)n 3 . = − 3 3 (2n + 1) 2 n=1 n 2 n=1 n n=0

Example 2: Set an =

1 n4

in (1.8),

∞ ∞ 1 1X 1 1 X (−1)n = − . (2n − 1)4 2 n=1 n4 2 n=1 n4 n=1 ∞ X

1.6

Converting the Summand (−1)n a2n to in an ∞ X

n

(−1) a2n = R

n=1

∞ X

in an .

(1.9)

n=1

Proof. ∞ X

in an = ia1 + i2 a2 + i3 a3 + i4 a4 + i5 a5 + i6 a6 + · · ·

n=1

= ia1 − a2 − ia3 + a4 + ia5 − a6 + · · · = i(a1 − a3 + a5 − · · · ) + (−a2 + a4 − a6 + · · · ) ∞ ∞ X X =i (−1)n a2n+1 + (−1)n a2n , n=0

n=1

and the proof follows on comparing the real parts of both sides. Example 1: Put an =

xn n3 , ∞ X n=1

(−1)n

∞ n X x2n nx = R i . (2n)3 n3 n=1

(1.10)

6

Chapter 1. Series Transformations and Special Functions

Example 2: Put an =

Hn+1 n2 , ∞ X

H2n+1 (−1)n (2n)2 n=1

1.7

=R

∞ X

in

n=1

Hn+1 . n2

Converting the Summand (−1)n a2n+1 to in an ∞ X

(−1)n a2n+1 = J

∞ X

in an .

(1.11)

n=1

n=0

Proof. Compare the imaginary parts of both sides of (1.10). Example 1: Let an =

1 n3 , ∞ X

∞ X (−1)n in = J . (2n + 1)3 n3 n=0 n=1

Example 2: Let an =

Hn (n+1)2 ,

∞ X

(−1)n

n=0

1.8

∞ X H2n+1 Hn in = J . (2n + 2)2 (n + 1)2 n=1

Converting a Sum to a Product r X n=m

ln(an ) = ln

r Y

an .

(1.12)

n=m

Proof. r X

ln(an ) = ln(am ) + ln(am+1 ) + · · · + ln(ar )

n=m

= ln(am × am+1 × · · · × ar ) = ln

r Y

an .

n=m

Example 1: Let an = n, r X n=1

ln(n) = ln

r Y n=1

n = ln(1 × 2 × 3 × · · · × r) = ln(r!).

1.9. Double Summation

7

Example 2: Let an = en , ln

r Y n=1

1.9

r X

en =

ln(en ) =

n=1

r X

n=

n=1

r(r + 1) . 2

Double Summation ∞ X m X

a m bn =

∞ X ∞ X

am bn .

(1.13)

n=1 m=n

m=1 n=1

Proof. m ∞ X X

1 X

am bn = a1

m=1 n=1

bn + a2

n=1

2 X

3 X

bn + a3

n=1

bn + · · ·

n=1

= a1 (b1 ) + a2 (b1 + b2 ) + a3 (b1 + b2 + b3 ) + · · · = b1 (a1 + a2 + · · · ) + b2 (a2 + a3 + · · · ) + b3 (a3 + a4 + · · · ) + · · · ∞ ∞ ∞ X X X = b1 am + b2 am + b3 am + · · · m=1 ∞ X

=

m=2 ∞ X

bn

am =

am bn ,

n=1 m=n

m=n

n=1

m=3 ∞ ∞ X X

and the proof is complete. If we follow the same steps above, we also find ∞ m−1 X X

am bn =

m=1 n=1

∞ ∞ X X

am bn .

n=1 m=n+1

Example 1: Let am = pm and bn = pn , ∞ X m X m=1 n=1

m+n

p

=

∞ X n=1

n

p

∞ X

! m

p

m=n

{use the geometric series formula for the inner sum asuming |p| < 1}  n  ∞ ∞ X p 1 X 2 n = pn = (p ) 1−p 1 − p n=1 n=1 {use the geometric series formula again}  2  1 p p2 = = . 2 1−p 1−p (1 − p)(1 − p2 )

(1.14)

8

Chapter 1. Series Transformations and Special Functions

Example 2: Let am = xm and bn = H n , ∞ X m X

m

x Hn =

m=1 n=1

∞ X

Hn

n=1

∞ X

! x

m

m=n

{use the geometric series formula for the inner sum asuming |x| < 1}  n  ∞ ∞ X x 1 X = Hn = H n xn 1 − x 1 − x n=1 n=1 {substitute the result from (2.28)}   1 ln(1 + x) ln(1 + x) = . = 1−x 1−x (1 − x)2

1.10

The Logarithm of a Complex Number

The logarithm of a complex number, z = x + iy, is given by ln(x + iy) =

y 1 ln(x2 + y 2 ) + i arctan , 2 x

x > 0.

(1.15)

Proof. We begin with converting to polar coordinates, x + iy = r cos(θ) + ir sin(θ) = r(cos(θ) + i sin(θ)). On using Euler’s formula: eix = cos x + i sin x,

(1.16)

which can be proved by expanding cos x, sin x, and ex in Taylor series, we have x + iy = reiθ . Take the logarithm of both sides,   ln(x + iy) = ln reiθ = ln(r) + ln eiθ = ln(r) + iθ. p The proof completes on substituting r = x2 + y 2 and θ = arctan(y/x). The constraint x > 0 in (1.15) shows that this rule is valid only when the complex number is in the first or fourth quadrant of the complex plane. In general, for positive x and y, where x = y 6= 0 and y 6= 0, we have y 1 ln(x2 + y 2 ) + i arctan ; 2 x h  y i 1 ln(−x + iy) = ln(x2 + y 2 ) + i π − arctan ; 2 x ln(x + iy) =

(1.17) (1.18)

1.10. The Logarithm of a Complex Number

9

h  y i 1 ; ln(x2 + y 2 ) − i π − arctan 2 x   1 y ln(x − iy) = ln(x2 + y 2 ) − i arctan . 2 x

ln(−x − iy) =

(1.19) (1.20)

Note that (1.19) and (1.20) follows from replacing i by −i in (1.18) and (1.17) respectively. To prove (1.18), we sum it with its conjugate, ln(−x + iy) + ln(x + iy) = ln(−x2 − y 2 ) = ln(−1) + ln(x2 + y 2 ) or ln(−x + iy) = ln(−1) + ln(x2 + y 2 ) − ln(x + iy) {substitute ln(−1) = iπ and recall the result from (1.17)} y 1 = iπ + ln(x2 + y 2 ) − ln(x2 + y 2 ) − arctan 2 x h  y i 1 2 2 = ln(x + y ) + i π − arctan , 2 x which matches (1.18). The value ln(−1) = iπ follows from using the identity: ln(−x) = ln(x) + iπ,

x > 0.

(1.21)

To show that, set x = π in Euler’s formula in (1.16), eiπ = −1, take the log of both sides, iπ = ln(−1), add ln(x) to both sides, ln(x) + iπ = ln(−1) + ln(x) = ln(−x). Using the rules from (1.17) to (1.20) we find: π 1 ln(2) + i ; 2 4 1 3π ln(−1 + i) = ln(2) + i ; 2 4 1 3π ln(−1 − i) = ln(2) − i ; 2 4 1 π ln(1 − i) = ln(2) − i ; 2 4 π ln(i) = i ; 2 ln(1 + i) =

(1.22) (1.23) (1.24) (1.25) (1.26)

10

Chapter 1. Series Transformations and Special Functions

π ln(−i) = −i . 2

1.11

Gamma Function

1.11.1

Definition

The gamma function is defined by Z ∞ Γ(z) = tz−1 e−t dt,

(1.27)

R(z) > 0.

(1.28)

0

For a different integral form, set t = −n ln(x) in (1.28), 1

Z

Γ(z) = (−1)z−1 nz

xn−1 lnz−1 (x)dx,

R(z) > 0.

(1.29)

0

Divide both sides of (1.29) by nz Γ(z), 1 (−1)z−1 = z n Γ(z)

1

Z

xn−1 lnz−1 (x)dx,

R(z) > 0.

(1.30)

z ∈ Z+ .

(1.31)

0

On using Γ(z) = (z − 1)! given in (1.33), we obtain (−1)z−1 1 = nz (z − 1)!

1.11.2

Z

1

xn−1 lnz−1 (x)dx,

0

Functional Equation

One of the key properties of the gamma function is Γ(z + 1) = zΓ(z),

z∈ / Z≤0 .

Proof. Replace z by z + 1 in (1.28), Z

Γ(z + 1) =

tz e−t dt

0

{apply integration by parts (IBP)} Z ∞ ∞ = −tz e−t 0 +z tz−1 e−t dt | {z } 0 0

= zΓ(z).

(1.32)

1.11. Gamma Function

11

Note that

Z

Γ(1) = 0

∞ e−t dt = −e−t 0 = −0 + 1 = 1,

and by using Γ(z + 1) = zΓ(z), we see that: Γ(2) = 1 · Γ(1) = 1 · 1 = 1! Γ(3) = 2 · Γ(2) = 2 · 1! = 2! Γ(4) = 3 · Γ(3) = 3 · 2! = 3! Γ(5) = 4 · Γ(4) = 4 · 3! = 4!. So in general we have Γ(z) = (z − 1)!,

1.11.3

z ∈ Z+ .

(1.33)

Stirling’s Approximation

Stirling’s Approximation is an approximation for factorials: √ 1 n! ∼ 2πnn+ 2 e−n , where the sign ∼ means lim √ n→∞

n! 1

2πnn+ 2 e−n

= 1.

The following proof may be found in [7, p. 277–278]: Proof. Begin with the definition of the gamma function: Z ∞ n! = Γ(n + 1) = tn e−t dt 0 n √ o let t = n + x 2n r !n √ √ n+ 1 −n Z ∞ 2 −x 2n = 2n 2 e 1+x dx √n e n − 2 ( r !n  √ 2 ) 2 n ln 1+x n write 1 + x =e n  √2 √ √ n+ 1 −n Z ∞ n ln 1+x n −x 2n 2 = 2n e dx. √ e −

1

n 2

2nn+ 2 e−n then let n → ∞, we get Z ∞  √2 √ n! n ln 1+x n −x 2n lim √ n+ 1 = lim dx √n e −n n→∞ n→∞ 2 2n e − 2

Divide both sides by

(1.34)

12

Chapter 1. Series Transformations and Special Functions

Z

=

e

h  √2 √ i −x 2n limn→∞ n ln 1+x n

dx.

−∞

To find the remaining limit, let x

q

2 n

= y and so n =

2x2 y2 ,

"

# r ! √ 2 ln(1 + y) − y n ln 1 + x − x 2n = 2x2 lim y→0 n y2

lim

n→∞

{apply L’Hopital’s rule, since we have the case 0/0} 1 1+y

= 2x2 lim

−1

2y

y→0

It follows that lim √

n→∞

−1 = −x2 . y→0 1 + y

= x2 lim

Z

n! 2n

n+ 12

e−n

2

e−x dx.

= −∞

This integral is called the Gaussian Integral, which evaluates to polar coordinates (see[39]). Thus, lim √

n→∞

n! 1 2nn+ 2 e−n

=

π by using the

π.

√ Divide both sides by π to finish the proof. To evaluate the Gaussian Integral in a different way, split the integral at x = 0, Z 0 Z ∞ Z ∞ 2 −x2 e dx = + e−x dx −∞

Z =

0

2

−∞

e−x dx + −∞ {z } | Z

=2

x→−x ∞ −x2

e

Z

2

e−x dx =

Z

0

√ x= y

0 ∞

2

e−x dx +

0

Z

dx =

1

y − 2 e−y dy = Γ

0

0

1.11.4

Z

2

e−x dx

0

  1 use (1.41) √ = π. 2

Expressing Gamma Function as a Product

The gamma function is also expressed as Γ(z) =

n Γ(z + n + 1) Y 1 , z z+k k=1

Proof. By the functional equation in (1.32), we have: Γ(z + 1) = zΓ(z);

z∈ / Z≤0 .

(1.35)

1.11. Gamma Function

13

Γ(z + 2) = (z + 1) · Γ(z + 1) = (z + 1) · zΓ(z); Γ(z + 3) = (z + 2) · Γ(z + 2) = (z + 2) · (z + 1) · zΓ(z). This can be generalized to Γ(z + n + 1) = (z + n) · (z + n − 1) · · · · (z + 1) · zΓ(z) or Γ(z) =

n Γ(z + n + 1) Γ(z + n + 1) Γ(z + n + 1) Y 1 = Qn . = z(z + 1) · · · (z + n) z z+k z k=1 (z + k) k=1

1.11.5

Euler’s Definition as an Infinite Product

Another form of the gamma function is z ∞ 1 Y 1 + k1 Γ(z) = , z 1 + kz

z∈ / Z≤0 .

(1.36)

k=1

Proof. Multiply both sides of (1.35): Γ(z) =

n Γ(z + n + 1) Y 1 z z+k k=1

by

n!nz Γ(z+n+1) ,

we obtain n 1 Y n!nz Γ(z)n!nz = Γ(z + n + 1) z z+k k=1

( write n! =

n Y

k then use

k=1

=

n Y k=1

ak

n Y

bk =

k=1

n Y

) ak bk

k=1

n n 1 Y knz 1 Y nz = . z k+z z 1 + kz k=1

k=1

Take the limit on both sides letting n 7→ ∞, n

Y nz 1 n!nz = lim . n→∞ Γ(z + n + 1) z n→∞ 1 + kz

Γ(z) lim

k=1

For the LHS limit, use Stirling’s approximation for n! and Γ(z + n + 1), √ 1 n!nz 2πnn+ 2 e−n nz lim = lim √ 1 n→∞ Γ(z + n + 1) n→∞ 2π(n + z)n+z+ 2 e−n−z

(1.37)

14

Chapter 1. Series Transformations and Special Functions 1

nn+z+ 2

= ez lim

1

(n + z)n+z+ 2  z+ 12  n n n z = e lim n→∞ n + z n+z  z+ 12 n lim z+ 21 n→∞ n + z z (1)   = ez = e = 1. n z ez lim 1 + n→∞ n n→∞

Since n−1 Y k=1

1 1+ k

z =

n−1 Y k=1

(k + 1)z = kz

Qn−1

k=1 (k + 1) Qn−1 z k=1 k

z

2z · 3z · · · (n − 1)z · nz = nz = 1z · 2z · · · (n − 1)z and since

n−1 Y

Qn ak =

k=1

ak

an

k=1

z

n =

, we have

n−1 Y k=1

1 1+ k

Qn

z

1 + k1 z 1 + n1

z

k=1

=

.

Substitute this result in the RHS limit,  z Qn ∞ n 1 z Y Y 1 + k1 nz k=1 1 + k z = lim = lim . n→∞ n→∞ 1 + kz 1 + kz 1 + n1 1 + kz k=1 k=1 k=1 n Y

The proof follows on plugging the two limits in (1.37).

1.11.6

Euler’s Reflection Formula

A well-known relationship is Euler’s reflection formula: Γ(z)Γ(1 − z) =

π , sin(πz)

z∈ / Z.

Proof. Set a = z and b = 1 − z in (1.51): B(a, b) =

Γ(a)Γ(b) = Γ(a + b)

Z 0

xa−1 dx, (1 + x)a+b

(1.38)

1.11. Gamma Function

15

we obtain Z ∞ z−1 x Γ(z)Γ(1 − z) = dx 1 +x 0 Z 1 Z ∞  z−1 Z 1 z−1 Z ∞ z−1 x x x + dx = dx + dx = 1 + x 1 + x 1 +x 0 1 0 } | 1 {z

(1.39)

x→1/x

Z

1

z−1

1

Z

−z

x x dx + dx 0 1+x 0 1+x   1 in series in both integrals expand 1+x ! ! Z 1 Z 1 ∞ ∞ X X z−1 n n −z n n = x (−1) x dx + x (−1) x dx =

0

0

n=0

n=0

{interchange integration and summation} Z 1 Z 1 ∞ ∞ X X n n n+z−1 (−1) (−1) = x dx + xn−z dx 0

n=0

=

∞ X

n

0

n=0 ∞ X

n

(−1) (−1) + n+z n−z+1 n=0 n=0

{seperate the first term of the first sum and shift the index of the second} ∞ ∞ ∞ 1 X (−1)n X (−1)n 1 X 2z(−1)n = + − = − z n=1 n + z n−z z n=1 n2 − z 2 n=1 {recall the identity in (2.130)} π . = sin(πz)

(1.40)

Moreover, setting z = 1/2 in (1.38) yields   π 1 = =π Γ2 2 sin( π2 ) or Γ

  √ 1 = π. 2

(1.41)

Remark: The interchange of integration and summation used in the proof above is justified by Lebesgue’s dominated convergence theorem (see [13]): Z X ∞ X n=1

fn (x)dx =

∞ Z X n=1

X

fn (x)dx,

(1.42)

16

Chapter 1. Series Transformations and Special Functions

where X is the integration interval. This theorem will be frequently used throughout the book where it’s applicable. So we are not going to reexplain why the interchange is justified if used again. More results of the gamma function may be found in [38].

1.11.7

Legendre Duplication Formula

The Legendre duplication formula is given by   √ 2 π Γ(2n) 1 +n = . Γ 2 4n Γ(n)

(1.43)

The following proof may be found in [40]: Proof. Letting a = b = n in the beta function in (1.47): 1

Z

Γ(a)Γ(b) = B(a, b) = Γ(a + b)

xa−1 (1 − x)b−1 dx

0

yields Γ2 (n) = Γ(2n)

Z

1

x

n−1

n−1

(1 − x)

dx

x= 1+u 2

=

2

1−2n

Z

= 21−2n

Z

0

Z 1

Z

0

0 2 n−1

(1 − u ) −1 {z | u→−u

1−2n

(1 − u2 )n−1 du

+ −1

=2

(1 − u2 )n−1 du

−1

0

1−2n

1

1−2n

du +2 }

2 n−1

(1 − u )

=2

du + 2

(1 − u2 )n−1 du

0

1

Z

1

Z

1−2n

0

Z

1

(1 − u2 )n−1 du

0 1−2n

1

Z

2(1 − u2 )n−1 du

=2

0 √ u= x

= 21−2n

Z

1

1

x− 2 (1 − x)n−1 dx

0

{use the defnition of the beta function in (1.47)}   1 1−2n =2 B ,n 2 {recall the property of the beta function in (1.48)} =

2 Γ( 12 )Γ(n) . 4n Γ( 12 + n)

The proof completes on substituting Γ(1/2) =

π given in (1.41).

(1.44)

1.12. Beta Function

17

Further, let’s set u = cos(x) in (1.44), π 2

Z

sin2n−1 (x)dx =

0

4n Γ2 (n) . 4Γ(2n)

(1.45)

By the definition of the central binomial coefficient:   a Γ(a + 1) , = Γ(b + 1)Γ(a − b + 1) b we have 

2n n

 =

Γ(2n + 1) Γ2 (n + 1)

{use Γ(z + 1) = zΓ(z) given in (1.32)} = or

2nΓ(2n) 2Γ(2n) = 2 (nΓ(n)) nΓ2 (n) Γ2 (n) = Γ(2n)

2  . n

2n n

Plug this result in (1.45), Z

π 2

sin2n−1 (x)dx =

0

1.12

Beta Function

1.12.1

Definition

4n 1  2n 2n . n

(1.46)

The beta function is defined by Z B(a, b) =

1

xa−1 (1 − x)b−1 dx,

R(a) > 0, R(b) > 0.

(1.47)

0

One of the key identities of the beta function is the Beta–Gamma identity: B(a, b) =

Γ(a)Γ(b) . Γ(a + b)

The following proof may be found in [42]: Proof. Using the definition of the gamma function in (1.28), we have  Z ∞  Z ∞ e−x xa dx e−y y b dy Γ(a)Γ(b) = 0

0

(1.48)

18

Chapter 1. Series Transformations and Special Functions

{let x = zt and y = z(1 − t)} Z 1 e−z (zt)a−1 (z(1 − t))b−1 z dt dz = 0 0 Z ∞   Z 1 a−1 b−1 −z a+b−1 t (1 − t) dt = e z dz Z

0

0

= (Γ(a + b)) (B(a, b)) , and the proof follows on dividing both sides by Γ(a + b). Another identity of the beta function is the symmetry identity: B(a, b) = B(b, a).

(1.49)

To prove it, let 1 − x = y in (1.47), 1

Z

xa−1 (1 − x)b−1 dx =

B(a, b) = 0

1.12.2

1

Z

y b−1 (1 − y)a−1 dy = B(b, a).

0

Trigonometric Integral Representation

A different integral form of the beta function is Z

π 2

B(a, b) = 2

cos2a−1 (x) sin2b−1 (x)dx,

R(a) > 0, R(b) > 0. (1.50)

0

Proof. Make the change of variable x = cos2 (y) in (1.47).

1.12.3

Improper Integral Representation

Another form of the beta function is Z ∞ xa−1 B(a, b) = dx, (1 + x)a+b 0 Proof : Make the substitution x =

1.12.4

y 1+y

R(a) > 0, R(b) > 0.

(1.51)

in (1.47).

Powerful Integral Representation

We also have Z B(a, b) = 0

1

xa−1 + xb−1 dx, (1 + x)a+b

R(a) > 0, R(b) > 0.

(1.52)

1.13. Riemann Zeta Function

19

The following proof may be found in [28, p. 72]: Proof. Let x → 1/x, Z

1

0 1

xa−1 + xb−1 dx = (1 + x)a+b

Z

1

xa−1 + xb−1 dx (1 + x)a+b

 xa−1 + xb−1 add dx to both sides then divide by 2 (1 + x)a+b 0 Z Z 1 1 xa−1 + xb−1 1 ∞ xa−1 + xb−1 = dx + dx 2 0 (1 + x)a+b 2 1 (1 + x)a+b Z 1 Z ∞  a−1 1 x + xb−1 = dx + 2 (1 + x)a+b 0 1 Z 1 ∞ xa−1 + xb−1 = dx 2 0 (1 + x)a+b Z ∞ Z 1 xa−1 1 ∞ xb−1 = dx + dx 2 0 (1 + x)a+b 2 0 (1 + x)a+b {use the definition of the beta function given in (1.51)} 1 1 = B(a, b) + B(b, a) 2 2 {use B(b, a) = B(a, b) given in (1.49)}



Z

= B(a, b).

1.13

Riemann Zeta Function

1.13.1

Definition

The zeta function is defined by ζ(z) = 1 +

1.13.2

∞ X 1 1 1 + + · · · = , z z 2 3 nz n=1

R(z) > 1.

(1.53)

Integral Representation

The integral form of the zeta function is given by ζ(z) =

(−1)z−1 Γ(z)

Z 0

1

lnz−1 (x) dx, 1−x

R(z) > 1.

(1.54)

20

Chapter 1. Series Transformations and Special Functions

Proof : Take the summation for both sides of (1.31): (−1)z−1 1 = nz Γ(z)

Z

1

xn−1 lnz−1 (x)dx

0

over n ≥ 1, we get ∞ ∞ Z X 1 (−1)z−1 X 1 n−1 z−1 := ζ(z) = x ln (x)dx nz Γ(z) n=1 0 n=1

{interchange integration and summation } ! Z ∞ X (−1)z−1 1 z−1 = ln (x) xn−1 dx Γ(z) 0 n=1 {use the geometric series formula}   Z 1 (−1)z−1 1 z−1 = ln (x) dx, Γ(z) 1−x 0 and the proof is finalized.

1.13.3

Evaluation of ζ(0) 1 ζ(0) = − . 2

The Following solution may be found in [9]: Solution Let’s consider (1.79): Z Z ∞ z−1 (−1)z 1 lnz (x) 1 t x=e−t η(z) = dx = dt. Γ(z) 0 1 + x Γ(z) 0 1 + et Substitute η(z) = (1 − 21−z )ζ(z) given in (1.75), Z ∞ z−1 t 1 ζ(z) = dt (1 − 21−z )Γ(z) 0 1 + et     ∞ 1 Z ∞ tz et   1 tz IBP  = + dt  t ) t )2 (1 − 21−z )Γ(z)  z(1 + e z (1 + e 0   | {z 0} 0

{write zΓ(z) = Γ(z + 1)} Z ∞ 1 tz et dt. = 1−z (1 − 2 )Γ(z + 1) 0 (1 + et )2

(1.55)

1.13. Riemann Zeta Function

21

Next, take the limit on both sides and let z → 0, ∞

Z lim ζ(z) = ζ(0) = −

z→0

1.13.4

0

∞ 1 et 1 dt = =− . (1 + et )2 1 + et 2 0

Evaluation of ζ(2) π2 . 6

ζ(2) =

(1.56)

The following solution may be found in [3]: Solution Squaring both sides of π = 4

Z

1

0

dx , 1 + x2

we have π2 = 16

Z Z

1

dx 1 + x2

0 1Z 1

= Z

0 0 1Z x

 Z

1

0

dy 1 + y2



dy dx (1 + x2 )(1 + y 2 )

dt dx + t2 /x2 ) 0 0 x(1 + {change the order of integration} Z Z 1 1 1 dx dt = 2 0 t x(1 + x2 )(1 + t2 /x2 )  Z 1 Z 1 √ du x= u 1 = dt 2 2 0 t2 (1 + u)(u + t )    1 1 1 1 write = − (1 + u)(u + t2 ) 1 − t2 u + t2 1+u  Z 1 Z 1 Z 1 1 1 du du = − dt 2 0 1 − t2 t2 u + t2 t2 1 + u Z u=1 1 1 1  = ln(u + t2 ) − ln(1 + u) u=t2 dt 2 2 0 1−t      Z 1 1 1 1 + t2 2t2 = ln − ln dt 2 0 1 − t2 2 1 + t2   Z 1 1 1 (1 + t2 )2 = ln dt 2 0 1 − t2 4t2 t=xy

=

x2 )(1

22

Chapter 1. Series Transformations and Special Functions

Z

1

ln



1+t2 2t

 t= 1−x 1+x

Z

1

ln



1+x2 1−x2



1 dt = dx 1− 2 0 x 0     1−y 2 Z 1 ln 1+y Z √ x= y 1 1−y 1 1 ln (1−y)2 = dy = dy 4 0 y 4 0 y Z 1 Z 1 1 ln(1 − y) 1 ln(1 − y 2 ) = dy − dy 4 0 y 2 0 y {z } | =

t2

y 2 →y

=

1 8

Z ln(1 − y) 1 1 ln(1 − y) dy − dy y 2 0 y 0 Z 3 1 ln(1 − y) =− dy 8 0 y {expand ln(1 − y) in series} ! Z ∞ X 3 11 yn =− − dy 8 0 y n n=1

Z

1

{interchange intergation and summation} Z ∞ ∞ 3 X 1 1 n−1 3X 1 3 = y dy = = ζ(2). 2 8 n=1 n 0 8 n=1 n 8 So we have

1.13.5

π2 16

= 38 ζ(2) or ζ(2) =

π2 6 .

Evaluation of ζ(2n)

The following generalization is derived by Rob Johnson (see [14]): n

ζ(2n) = −

(−1)n 2n X ζ(2n − 2k)π 2k π − (−1)k , 2(2n)! (2k + 1)!

n ∈ Z+ .

(1.57)

k=1

Proof. His proof starts with letting fn = −2

ζ(2n) . π 2n

Multiply both series by x2n then consider the summation over n ≥ 0, ∞ X n=0

fn x2n =

∞ X n=0

−2

ζ(2n) 2n x π 2n

{separate the first term then expand ζ(2n) in series}

(1.58)

1.13. Riemann Zeta Function

23

= −2ζ(0) − 2

∞ ∞ X x2n X 1 π 2n k 2n n=1 k=1

{write ζ(0) = −1/2 given in (1.55) and change the order of summations} n ! ∞ ∞  X X x2 =1−2 π2 k2 n=1 k=1

{use the geometric series formula} x2 π 2 k2 2 − πx2 k2

∞ X

∞ X

x2 π 2 k 2 − x2 1 k=1 k=1 n o x to get this sum, set z = in (2.131) π   1 x =1−2 − cot x . 2 2

=1−2

As a result, we have

∞ X

=1−2

fn x2n = x cot x.

n=0

Multiply both sides of (1.59) by

sin x x ,

cos x =

∞ sin x X fn x2n . x n=0

Expanding cos x and sin x in Taylor series: cos x =

∞ X (−1)n n=0

(2n)!

x2n ,

sin x =

∞ X (−1)n 2n+1 x (2n + 1)! n=0

yields ∞ X (−1)n n=0

(2n)!

x2n =

∞ ∞ X (−1)n X fn x2n (2n + 1)! n=0 n=0

{make use of the Cauchy product in (2.81)}   (−1)n where = an and fn = bn (2n + 1)! ! ∞ n X X (−1)k = fn−k x2n . (2k + 1)! n=0 k=0

(1.59)

24

Chapter 1. Series Transformations and Special Functions

Separate the first term of the inner sum, ∞ X (−1)n n=0

(2n)!

x

2n

=

∞ X

n X

fn +

n=0

fn−k (−1) (2k + 1)! k

! x2n .

(1.60)

k=1

Compare the coefficients of x2n in both sides of (1.60), n

X fn−k (−1)n (−1)k = fn + (2n)! (2k + 1)! k=1

or

n

fn =

(−1)n X fn−k − (−1)k . (2n)! (2k + 1)!

(1.61)

k=1

Substitute (1.61) in (1.58), n

−2

X (−1)n ζ(2n − 2k) ζ(2n) = +2 (−1)k , 2n π (2n)! (2k + 1)!π 2n−2k k=1

2n

and the proof finishes on multiplying both sides by − π2 . Examples π2 ; 6 π4 ζ(4) = ; 90 π6 ; ζ(6) = 945 π8 ζ(8) = ; 9450 π 10 ζ(10) = ; 93555 691π 12 ζ(12) = . 638512875 ζ(2) =

Also note the following relations: π4 = 36 π6 ζ 3 (2) = = 216 π6 ζ(2)ζ(4) = 4500 ζ 2 (2) =

5 ζ(4); 2 35 ζ(6); 8 7 = ζ(6); 4

(1.62) (1.63) (1.64)

1.14. Dirichlet Eta Function

25

π8 7 = ζ(8); 8100 6 π8 5 ζ(2)ζ(6) = = ζ(8); 5670 3 33 π 10 = ζ(10); ζ(2)ζ(8) = 56700 20 π 10 11 ζ(4)ζ(6) = = ζ(10); 85050 10 π 12 875875 ζ 6 (2) = = ζ(12); 46656 44224 7007 π 12 = ζ(12); ζ 3 (4) = 729000 5528 π 12 715 ζ 2 (6) = = ζ(12); 893025 691 12 π 3003 ζ(4)ζ(8) = = ζ(12); 850500 2764 2275 π 12 = ζ(12). ζ(2)ζ(10) = 561330 1382 ζ 2 (4) =

1.14

Dirichlet Eta Function

1.14.1

Definition

(1.65) (1.66) (1.67) (1.68) (1.69) (1.70) (1.71) (1.72) (1.73)

The Dirichlet eta function is defined by η(z) =

∞ X (−1)n−1 , nz n=1

R(z) > 0.

This function is related to the zeta function. To show that, let an =

(1.74)

1 nz

in (1.5),

∞ X

∞ ∞ X X 1 1 (−1)n 2 = + (2n)z nz n=1 nz n=1 n=1

21−z

∞ ∞ ∞ X X X 1 1 (−1)n−1 = − nz nz n=1 nz n=1 n=1

21−z ζ(z) = ζ(z) − η(z). Rearrange the terms, η(z) = (1 − 21−z )ζ(z).

(1.75)

26

Chapter 1. Series Transformations and Special Functions

Examples 1 ζ(2); 2 3 η(3) = ζ(3); 4 7 η(4) = ζ(4). 8 η(2) =

1.14.2

(1.76) (1.77) (1.78)

Integral Representation

The integral representation of eta function is (−1)z−1 η(z) = Γ(z)

Z

1

0

lnz−1 (x) dx, 1+x

R(z) > 0.

(1.79)

Proof. Multiply both sides of (1.30): (−1)z−1 1 = z n Γ(z)

Z

1

xn−1 lnz−1 (x)dx.

0

by (−1)n−1 then consider the summation over n ≥ 1, Z 1 ∞ ∞ X (−1)n−1 (−1)z−1 X n−1 (−1) := η(z) = xn−1 lnz−1 (x)dx z n Γ(z) 0 n=1 n=1 {interchange integration and summation} ! Z ∞ X (−1)z−1 1 z−1 n−1 = ln (x) (−x) dx Γ(z) 0 n=1 {use the geometric series formula} Z (−1)z−1 1 lnz−1 (x) = dx. Γ(z) 1+x 0

1.15

Dirichlet Beta Function

1.15.1

Definition

The Dirichlet beta function is defined by β(z) =

∞ X

(−1)n 1 1 = 1 − z + z − ··· , z (2n + 1) 3 5 n=0

R(z) > 0.

(1.80)

1.15. Dirichlet Beta Function

1.15.2

27

Integral Representation

The integral form of the Dirichlet beta function is (−1)z−1 β(z) = Γ(z)

1

Z 0

lnz−1 (x) dx, 1 + x2

R(z) > 0.

(1.81)

Proof. By (1.30), we have 1 (−1)z−1 = (2n + 1)z Γ(z)

1

Z

x2n lnz−1 (x)dx.

0

Multiply both sides by (−1)n then consider the summation over n ≥ 0, Z 1 ∞ (−1)z−1 X (−1)n n (−1) := β(z) = x2n lnz−1 (x)dx z (2n + 1) Γ(z) 0 n=0 n=0 ∞ X

{interchange integration and summation} ! Z ∞ X (−1)z−1 1 z−1 = ln (x) (−x2 )n dx Γ(z) 0 n=0 {use the geometric series formula} Z (−1)z−1 1 lnz−1 (x) dx. = Γ(z) 1 + x2 0

1.15.3

Evaluation of β(2a)

For a ∈ Z+ , we have 21−4a β(2a) = − ψ (2a−1) (2a − 1)!

  1 − (1 − 2−2a )ζ(2a). 4

(1.82)

Proof. Replace z by 2a in (1.80),

β(2a) = ( use

∞ X

n

(−1) an = 2

n=0

=2

∞ X

(−1)n (2n + 1)2a n=0

∞ X n=0

∞ X

a2n −

∞ X

n=0 ∞ X

) an given in (1.6)

1 1 − . 2a (4n + 1) (2n + 1)2a n=0 n=0

(1.83)

28

Chapter 1. Series Transformations and Special Functions

Evaluation of the first sum: ∞ 1 X 1 1 = 2a 2a 2a (4n + 1) 4 (n + 1/4) n=0 n=0 ∞ X

{make use of the definition in (1.177)}   2−4a 1 (2a−1) =− ψ . (2a − 1)! 4 Evaluation of the second sum: Set an =

1 (n+1)a

(1.84)

in (1.6),

∞ X

∞ ∞ 1 1X 1 1 X (−1)n = + (2n + 1)a 2 n=0 (n + 1)a 2 n=0 (n + 1)a n=0

{shift the index n by −1 in both series on the RHS} ∞ ∞ 1 X (−1)n−1 1X 1 + = 2 n=1 na 2 n=1 na 1 1 ζ(a) + η(a) 2 2 1−a {substitute η(a) = (1 − 2 )ζ(a) given in (1.75)} =

= (1 − 2−a )ζ(a). Thus, we have

∞ X

1 = (1 − 2−a )ζ(a). a (2n + 1) n=0

(1.85)

Replace a by 2a in (1.85), ∞ X

1 = (1 − 2−2a )ζ(2a). 2a (2n + 1) n=0

(1.86)

The proof completes on plugging (1.84) and (1.86) in (1.83).

Examples ∞ X

 ψ (1) 14 3 ln(x) dx = − ζ(2); 2 48 4 0 1+x  Z ∞ X (−1)n ψ (3) 14 1 1 ln3 (x) 15 β(4) = =− dx = − ζ(4); 4 2 (2n + 1) 6 1 + x 768 16 0 n=0  Z 1 5 ∞ n X ψ (5) 14 (−1) 1 ln (x) 63 β(6) = =− dx = − ζ(6); 6 2 (2n + 1) 120 1 + x 245760 64 0 n=0 (−1)n β(2) = =− (2n + 1)2 n=0

Z

1

(1.87) (1.88) (1.89)

1.15. Dirichlet Beta Function

β(8) =

29

∞ X

1 (−1)n =− 8 (2n + 1) 5040 n=0

Z 0

1

 ψ (7) 41 255 ln7 (x) dx = − ζ(8); 1 + x2 165150720 256 (1.90)

β(10) =

1.15.4

∞ X

n

Z

9

1

1 ln (x) (−1) =− dx 10 (2n + 1) 362880 0 1 + x2 n=0  ψ (9) 14 1023 = − ζ(10). 190253629440 1024

(1.91)

Evaluation of β(2a + 1)

For a ∈ Z≥0 , we have β(2a + 1) =

4−a−1 π d2a lim1 2a csc(πs). (2a)! s→ 2 ds

(1.92)

Proof. Replace z by 2a + 1 in (1.81) then write Γ(2a + 1) = (2a)!, β(2a + 1) = =

1 (2a)!

Z

1 (2a)!

Z

1

0 ∞

ln2a (x) 1 dx = 2 1+x (2a)!

0

2a

ln (x) 1 dx − 1 + x2 (2a)!

Z

Z

Z

0 ∞

|1

1

ln2a (x) dx 1 + x2

2a

ln (x) dx 1 + x2 {z }

x→1/x

Z

=

4−a−1 d2a lim1 2a [π csc(πs)]. (2a)! s→ 2 ds

ln (x) 1 ln2a (x) dx − dx 1 + x2 (2a)! 0 1 + x2 0   Z 1 2a 1 ln (x) add β(2a + 1) := dx to both sides then divide by 2 (2a)! 0 1 + x2 Z ∞ 2a 1 ln (x) = dx 2(2a)! 0 1 + x2 {recall the result from (3.53)} =

1 (2a)!

2a

∞

Z

1

Examples Z 1 ∞ X (−1)n 1 π = dx = ; 2 2n + 1 1 + x 4 0 n=0 Z ∞ X (−1)n 1 1 ln2 (x) π3 β(3) = = dx = ; (2n + 1)3 2 0 1 + x2 32 n=0 β(1) =

(1.93) (1.94)

30

Chapter 1. Series Transformations and Special Functions

β(5) =

∞ X

1 (−1)n = 5 (2n + 1) 24 n=0

Z 0

1

5π 5 ln4 (x) dx = ; 2 1+x 1536

∞ X

Z 1 6 1 ln (x) 16π 7 (−1)n = dx = ; β(7) = (2n + 1)7 720 0 1 + x2 184320 n=0 Z 1 8 ∞ X 1 277π 9 ln (x) (−1)n = dx = β(9) = . (2n + 1)9 40320 0 1 + x2 8257536 n=0

1.16

Polylogarithm Function

1.16.1

Definition

(1.95) (1.96) (1.97)

The polylogarithm function is defined by Lia (z) = where |z| =

p

∞ X zn z2 z3 = z + a + a + ··· , a n 2 3 n=1

|z| ≤ 1,

(1.98)

x2 + y 2 is the modulus of the complex number, z = x + iy.

Let’s discuss the case a = 1: Li1 (z) =

∞ X zn = − ln(1 − z). n n=1

(1.99)

This function diverges when |z| = 1 and converges to − ln(2) when z = −1, and so the case a = 1 is valid when z = −1, but invalid when |z| = 1. Actually the range of a can be extended to the whole complex plane. To keep it simple, we will consider only the case a ∈ Z+ , since that is all we need for later calculations. Note that ∞ ∞ X X (1)n 1 Lia (1) = = = ζ(a) (1.100) a n na n=1 n=1 and Lia (−1) =

∞ X (−1)n = −η(a) = (21−a − 1)ζ(a). a n n=1

(1.101)

Examples 1 Li2 (−1) = − ζ(2); 2 3 Li3 (−1) = − ζ(3); 4

(1.102) (1.103)

1.16. Polylogarithm Function

31

7 Li4 (−1) = − ζ(4); 8 15 Li5 (−1) = − ζ(5); 16 31 Li6 (−1) = − ζ(6). 32

(1.104) (1.105) (1.106)

It’s also valid to set z = i in (1.98), since |i| = 1 assuming a = 2, 3, 4, · · · , ∞ X in Lia (i) = na n=1

( use

∞ X

f (n) =

n=1

=

∞ X

f (2n + 1) +

∞ X

) f (2n) given in (1.4)

n=1 ∞ X

n=0 ∞ X

i2n+1 i2n + a (2n + 1) (2n)a n=0 n=1

∞ X (−1)n (−1)n −a + 2 =i (2n + 1)a na n=1 n=0 ∞ X

{use the definitions in (1.80) and (1.74)} = iβ(a) − 2−a η(a) {write η(a) = (1 − 21−a )ζ(a) given in (1.75)} = iβ(a) + (21−2a − 2−a )ζ(a). Thus, Lia (i) = (21−2a − 2−a )ζ(a) + iβ(a).

(1.107)

Examples 1 1 Li2 (i) = − ζ(2) + iβ(2) = − ζ(2) + iG; 8 8 3 π3 3 Li3 (i) = − ζ(3) + iβ(3) = − ζ(3) + i ; 32 32 32 7 Li4 (i) = − ζ(4) + iβ(4). 128 In the calculations above, we used β(2) = G and β(3) = (1.94) respectively.

π3 32

(1.108) (1.109) (1.110)

given in (1.206) and

To write the polylogarithm function in integral form, multiply both sides of (1.31): 1 (−1)a−1 = a n (a − 1)!

Z 0

1

tn−1 lna−1 (t)dt

32

Chapter 1. Series Transformations and Special Functions

by z n then take the summation over ≥ 1, ∞ Z ∞ X (−1)a−1 X 1 n n−1 a−1 zn := Li (z) = z t ln (t)dt a na (a − 1)! n=1 0 n=1

{switch integration and summation} ! Z ∞ (−1)a−1 1 lna−1 (t) X n = (zt) dt (a − 1)! 0 t n=1 a−1

=

(−1) (a − 1)!

1

Z 0

{use the geometric series formula}   Z lna−1 (t) zt (−1)a−1 1 z lna−1 (t) dt. dt = t 1 − zt (a − 1)! 0 1 − zt

Therefore, Lia (z) =

(−1)a−1 (a − 1)!

Z 0

1

z lna−1 (t) dt. 1 − zt

(1.111)

This integral form extends the range of z to the whole complex plane except the real line for x > 1. In other words, z ∈ / (1, ∞). Note that a 6= 1 when z = 1, since we R1 1 will have 0 1−t dt, which is a divergent integral. Let’s replace z by

z z−1

in (1.111), 

Lia

z z−1

 =

(−1)a (a − 1)!

Z 0

1

z lna−1 (t) dt. 1 − z + zt

(1.112)

Like the integral in (1.111), this integral also extends the range of z to thewhole  z complex plane except the real line for x > 1 but notice here z 6= 1, since Lia z−1 is undefined for this value deducing that z ∈ / [1, ∞). Both of (1.111) and (1.112) can be found in [28, p. 4]. For a different integral form, we begin with Z z zn tn−1 dt. = n 0 Divide both sides by na−1 then consider the summation over n ≥ 1, Z z ∞ ∞ X X zn 1 := Lia (z) = tn−1 dt a a−1 n n 0 n=1 n=1 {interchange integration and summation} ! Z z ∞ 1 X tn = dt na−1 0 t n=1 {recall the definition in (1.98)}

1.16. Polylogarithm Function

33 z

Z = 0

Lia−1 (t) dt. t

Thus,

z

Z Lia (z) = 0

Lia−1 (t) dt. t

(1.113)

This integral form also extends the range of z to the whole complex plane. Note that R1 a 6= 1 when z = 1, since we will have 0 1t dt, which is a divergent integral. If we start with the integral representation in (1.111) for both Lia (z) and Lia (−z), we find   Z z z (−1)a−1 1 a−1 ln (t) − Lia (z) + Lia (−z) = dt (a − 1)! 0 1 − zt 1 + zt   Z (−1)a−1 1 a−1 2tz 2 = ln (t) dt (a − 1)! 0 1 − z 2 t2 Z 1 2 a−1 √ t= y 1−a (−1)a−1 z ln (y) dy = 2 2 (a − 1)! 0 1−z y {to get this integral, replace z by z 2 in (1.111)} = 21−a Lia (z 2 ). Then, we have Lia (z) + Lia (−z) = 21−a Lia (z 2 ).

(1.114)

This relation extends the range of z to the whole complex plane. n For a different approach assuming |z| ≤ 1, put an = nz a in (1.5): 2

∞ X n=1

a2n =

∞ X n=1

an +

∞ X

(−1)n an ,

n=1

we get 2

∞ ∞ ∞ X X z 2n z n X (−z)n = + (2n)a na n=1 na n=1 n=1

21−a

∞ ∞ ∞ X X (z 2 )n z n X (−z)n = + na na n=1 na n=1 n=1

21−a Lia (z 2 ) = Lia (z) + Lia (−z). Examples 1 Li2 (z 2 ); 2 1 Li3 (z) + Li3 (−z) = Li3 (z 2 ); 4 Li2 (z) + Li2 (−z) =

(1.115) (1.116)

34

Chapter 1. Series Transformations and Special Functions

Li4 (z) + Li4 (−z) =

1 Li4 (z 2 ). 8

(1.117)

Let’s differentiate both sides of (1.98) with respect to z, ∞ ∞ ∞ X 1 X zn Lia−1 (z) ∂ ∂ X zn nz n−1 = = = Lia (z) = . a a−1 ∂z ∂z n=1 na n z n z n=1 n=1

Therefore, ∂ Lia−1 (z) Lia (z) = . ∂z z

1.16.2

(1.118)

Dilogarithm Reflection Formula

For z ∈ C, the following formula holds: Li2 (z) + Li2 (1 − z) = ζ(2) − ln(z) ln(1 − z).

(1.119)

Proof. Differentiate ln(z) ln(1 − z) then integrate, Z ln(z) ln(1 − z) = d(ln(z) ln(1 − z))  Z  ln(z) ln(1 − z) = − + dz 1−z z = − Li2 (1 − z) − Li2 (z) + C. To find C, take the limit on both sides and let z → 0, lim ln(z) ln(1 − z) = − Li2 (1) − Li2 (0) + C.

z→0

Since lim ln(z) ln(1 − z) = 0, we have C = Li2 (1) = ζ(2) and the proof is z→0 complete. To show the limit is 0, we write    ln(1 − z) , lim ln(z) ln(1 − z) = lim z ln(z) lim z→0 z→0 z→0 z where, by using L’Hopital’s rule, the first limit is 0 and the second limit is −1. If we put z = 1/2 in (1.119), we obtain     1 1 Li2 + Li2 = ζ(2) − ln2 (2) 2 2 or Li2

  1 1 1 = ζ(2) − ln2 (2). 2 2 2

(1.120)

1.16. Polylogarithm Function

35

Furthermore, set z = i in (1.119), Li2 (i) + Li2 (1 − i) = ζ(2) − ln(i) ln(1 − i). Substitute the values from (1.26), (1.25), and (1.108) in (1.121), π  3 ln(2) + G i. Li2 (1 − i) = ζ(2) − 8 4

(1.121)

(1.122)

Replacing i by −i in (1.122) yields Li2 (1 + i) =

1.16.3

π  3 ζ(2) + ln(2) + G i. 8 4

(1.123)

Landen’s Dilogarithm Identity

For z ∈ C, z 6= 0, the following identity holds:   1 z−1 Li2 (1 − z) + Li2 = − ln2 (z). z 2 Proof. Differentiate Li2

z−1 z

(1.124)



then integrate,  Z    z−1 z−1 Li2 = d Li2 z z   ∂ Lia−1 (z) employ Lia (z) = given in (1.118) ∂z z Z Z Z ln(z) ln(z) ln(z) = dz = − dz − dz z(1 − z) z 1−z 1 = − ln2 (z) − Li2 (1 − z) + C. 2 

The proof finishes on extracting C = 0 by putting z = 1. Let’s set z = 1 − i in (1.124),   1−i 1 Li2 (i) + Li2 = − ln2 (1 − i). 2 2 Substitute the values from (1.108), (1.26), and (1.25),     1−i 5 1 π Li2 = ζ(2) − ln2 (2) − G − ln(2) i. 2 16 8 8 Replace i by −i,     1+i 5 1 π Li2 = ζ(2) − ln2 (2) + G − ln(2) i. 2 16 8 8

(1.125)

(1.126)

36

Chapter 1. Series Transformations and Special Functions

This result can again be found by setting z = −1 + i in (1.127).

1.16.4

Dilogarithm Inversion Formula

For z ∈ C, z 6= 0, the following identity holds:   1 ln2 (z) Li2 (−z) + Li2 − + 2 Li2 (−1). =− z 2

(1.127)

Proof.   Z   Z  ln 1+z 1 1 z Li2 − = d Li2 − = dz z z z Z Z ln(z) 1 ln(1 + z) = dz − dz = − Li2 (−z) − ln2 (z) + C. z z 2 

To find the constant C, set z = 1, Li2 (−1) = − Li2 (−1) −

1 2 ln (1) + C 2

or C = 2 Li2 (−1), and we are done with the proof.

1.16.5

Relation Involving Four Dilogarithm Functions

For z ∈ C, z 6= −1, the following identity holds:       2z z 1+z Li2 − Li2 + Li2 1+z 1+z 2   1 = Li2 (z) + Li2 + ln(2) ln(1 + z). 2

(1.128)

Proof. Let  f (z) = Li2

2z 1+z



 − Li2

z 1+z



 + Li2

1+z 2

we have Z 

Z f (z) =

d f (z) =

ln(2) ln(1 − z) − 1+z z

= ln(2) ln(1 + z) + Li2 (z) + C.

 dz

 ,

1.16. Polylogarithm Function

37

Setting z = 0 gives C = Li2 ( 12 ) and the proof is finalized.

1.16.6

Another Relation Involving Dilogarithm Functions

For z ∈ C, z 6= −1, the following identity holds:         2z z 1−z 1 − Li2 − Li2 − Li2 (z) + Li2 Li2 1+z 1+z 2 2   1+z = ln(1 − z) ln . (1.129) 2 Proof. Let  f (z) = Li2

2z 1+z



 − Li2

z 1+z



 − Li2

1−z 2



  1 − Li2 (z) + Li2 , 2

we have Z 

ln(1 − z) ln(1 + z) ln(2) f (z) = d f (z) = − + 1+z 1−z 1−z  Z Z  ln(1 − z) ln(1 + z) ln(2) − dz + dz = 1+z 1−z 1−z Z = d(ln(1 − z) ln(1 + z)) − ln(2) ln(1 − z) Z

 dz

= ln(1 − z) ln(1 + z) − ln(2) ln(1 − z)   1+z = ln(1 − z) ln + C. 2 The proof completes on extracting C = 0 by setting z = 0.

1.16.7

Landen’s Trilogarithm Identity

For z ∈ C, z 6= 0, the following identity holds:   1 z−1 Li3 (z) + Li3 (1 − z) + Li3 = ζ(3) + ln3 (z) + ζ(2) ln(z) z 6 1 − ln2 (z) ln(1 − z). (1.130) 2 Proof.  Li3

z−1 z



Z =

    Z Li2 z−1 z−1 z d Li3 = − dz z z(1 − z)

38

Chapter 1. Series Transformations and Special Functions

{make use of Landen’s dilogarithm identity in (1.124)}  Z  ln2 (z) Li2 (1 − z) = + dz z(1 − z) 2z(1 − z)   Z Li2 (1 − z) Li2 (1 − z) 1 ln2 (z) 1 ln2 (z) = + + + dz z 1−z 2 z 2 1−z Z Z 1 1 ln2 (z) Li2 (1 − z) dz − Li3 (1 − z) + ln3 (z) + dz = x 6 2 1−z | {z } IBP

= ln(z) Li2 (1 − z) − Li3 (1 − z) +

1 3 1 ln (z) − 6 2

Z

ln2 (z) dz. 1−z

For the remaining integral, force integration by parts twice, Z

ln2 (z) dz = − ln(1 − z) ln2 (z) + 2 1−z

ln(1 − z) ln(z) dz z Z Li2 (z) dz = − ln(1 − z) ln2 (z) − 2 Li2 (z) ln(z) + 2 z = − ln(1 − z) ln2 (z) − 2 Li2 (z) ln(z) + 2 Li3 (z). Z

(1.131)

Plug this integral back in,   1 1 z−1 Li3 = ln(z)[Li2 (1 − z) + Li2 (z)] + ln(1 − z) ln2 (z) + ln3 (z) z 2 6 − Li3 (1 − z) − Li3 (z) {make use of (1.119) for the first term} 1 1 = ζ(2) ln(z) − ln(1 − z) ln2 (z) + ln3 (z) − Li3 (1 − z) − Li3 (z) + C, 2 6 and the proof follows on extracting C = ζ(3) by setting z = 1. Setting z = 1/2 in (1.130) yields     1 1 1 1 Li3 + Li3 + Li3 (−1) = ζ(3) − ln3 (2) − ζ(2) ln(2) + ln3 (2). 2 2 6 2 Substitute Li3 (−1) = − 43 ζ(3) given in (1.103),   7 1 1 1 = ζ(3) − ln(2)ζ(2) + ln3 (2). Li3 2 8 2 6

(1.132)

For another result, set z = i in (1.130) then consider the real parts of both sides,    i−1 R Li3 (i) + Li3 (1 − i) + Li3 i

1.16. Polylogarithm Function

39

 = ζ(3) + R Since Li3

i−1 i



 1 3 1 2 ln (i) + ζ(2) ln(i) − ln (i) ln(1 − i) . 6 2

= Li3 (1 + i) and R Li3 (1 − i) = R Li3 (1 + i), we have R {Li3 (i) + 2 Li3 (1 + i)} 

= ζ(3) + R

 1 1 3 ln (i) + ζ(2) ln(i) − ln2 (i) ln(1 − i) . 6 2

Collect the results from (1.109), (1.26), and (1.25), R Li3 (1 + i) =

1.16.8

35 3 ln(2)ζ(2) + ζ(3) = R Li3 (1 − i). 16 64

(1.133)

Polylogarithm Inversion Formula

For z ∈ C, z 6= 0, the following two identities hold:   a X ln2a−2n (z) ln2a (z) 1 Li2n (−1) Li2a (−z) + Li2a − =2 − ; (1.134) z (2a − 2n)! (2a)! n=1   a X 1 ln2a+1 (z) ln2a−2n+1 (z) Li2a+1 (−z) − Li2a+1 − − . =2 Li2n (−1) z (2a − 2n + 1)! (2a + 1)! n=1 (1.135) Proof. Divide both sides of (1.127) by z then integrate from z = 1 to z,   1 ln3 (z) Li3 (−z) − Li3 − =− + 2 Li2 (−1) ln(z). z 2·3 Repeat the same process,   ln4 (z) ln2 (z) 1 =− + 2 Li2 (−1) + 2 Li4 (−1), Li4 (−z) + Li4 − z 2·3·4 2

(1.136)

(1.137)

again,   1 ln5 (z) ln3 (z) =− + 2 Li2 (−1) + 2 Li4 (−1) ln(z). Li5 (−z) − Li5 − z 2·3·4·5 2·3 (1.138) This can be generalized to (1.134) and (1.135) and the proof is finished. Let’s put z = −1 + i in (1.136) then consider the real parts of both sides,      1+i 1 R Li3 (1 − i) − Li3 = R −ζ(2) ln(−1 + i) − ln3 (−1 + i) . 2 6

40

Chapter 1. Series Transformations and Special Functions

Substitute the values from (1.23) and (1.133),     1+i ln3 (2) ln(2) 35 1−i = . R Li3 − ζ(2) + ζ(3) = R Li3 2 48 32 64 2

(1.139)

For more values, let z = −2 in (1.127), (1.136), (1.137), and (1.138) using ln(−2) = ln(2) + iπ, which follows from (1.21), we obtain: 3 ζ(2) − π ln(2)i ; (1.140) 2 7 3 π Li3 (2) = ζ(3) + ln(2)ζ(2) − ln2 (2)i ; (1.141) 8 2 2   1 4 π 1 + 2ζ(4) + ln2 (2)ζ(2) − ln (2) − ln3 (2)i ; (1.142) Li4 (2) = − Li4 2 24 6   1 π 4 1 1 3 Li5 (2) = Li5 ln5 (2) − ln (2)i. + 2 ln(2)ζ(4) + ln (2)ζ(2) − 2 3 120 24 (1.143) Li2 (2) =

The values of Li2 ( 12 ) and Li3 ( 21 ), given in (1.120) and (1.132), were used in (1.140) and (1.141). For more results of polylogarithm functions, check [16].

1.17

Harmonic Number

1.17.1

Definition

The n-th harmonic number of order a is defined by Hn(a) =

n X 1 1 1 1 = 1 + a + a + ··· + a, ka 2 3 n

n, a ∈ Z+ .

(1.144)

k=1

One of the properties of the harmonic number is (a)

Hn(a) − Hn−1 =

1 . na

To show that, subtract the following two harmonic numbers: 1 1 1 1 + a + ··· + + a, 2a 3 (n − 1)a n 1 1 1 = 1 + a + a + ··· + . 2 3 (n − 1)a

Hn(a) = 1 + (a)

Hn−1

(1.145)

1.17. Harmonic Number

41

With a = 1 in (1.144) and (1.145), we have Hn(1) = Hn =

n X 1 1 1 1 = 1 + + + ··· + , k 2 3 n

(1.146)

1 . n

(1.147)

1 . (2k − 1)a

(1.148)

k=1

Hn − Hn−1 = Another property is (a)

H2n − 2−a Hn(a) =

n X k=1

To show that, begin with the RHS: n X

1 1 1 1 = 1 + a + a + ··· + (2k − 1)a 3 5 (2n − 1)a k=1   1 1 1 1 add and subtract a + a + a + · · · + 2 4 6 (2n)a   1 1 1 1 1 1 1 − + + + · · · + = 1 + a + a + ··· + 2 3 (2n)a 2a 4a 6a (2n)a   1 1 1 1 1 1 1 = 1 + a + a + ··· + − + · · · + 1 + 2 3 (2n)a 2a 2a 3a na =

2n n X 1 1 X 1 (a) − = H2n − 2−a Hn(a) . ka 2a ka

k=1

k=1

Further, if we let (a)

fn = H2n − 2−a Hn(a) =

n X k=1

and so (a)

(a)

fn+1 = H2n+2 − 2−a Hn+1 =

1 (2k − 1)a

n+1 X k=1

1 , (2k − 1)a

we have fn+1 − fn =

n+1 X k=1

( use

n+1 X k=1

f (k) =

n X k=1

n

X 1 1 − a (2k − 1) (2k − 1)a k=1

) f (k) + f (n + 1) for the first sum

42

Chapter 1. Series Transformations and Special Functions

=

n

n X

X 1 1 1 1 + − = . a a (2k − 1) (2n + 1) (2k − 1)a (2n + 1)a k=1

k=1

(a)

(a)

Thus, with fn = H2n − 2−a Hn , we have fn+1 − fn =

1 . (2n + 1)a

(1.149)

(a)

To write Hn in integral form, take the summation for both sides of (1.31): 1 (−1)a−1 = ka (a − 1)!

Z

1

xk−1 lna−1 (x)dx

0

from k = 1 to n, ∞ Z n X (−1)a−1 X 1 lna−1 (x) 1 (a) dx := Hn = ka (a − 1)! n=1 0 1 − x

k=1

{switch integration and summation} ! Z n X (−1)a−1 1 a−1 k−1 = x dx ln (x) (a − 1)! 0 k=1

{use the geometric series}   Z (−1)a−1 1 a−1 1 − xn = ln (x) dx. (a − 1)! 0 1−x Therefore, Hn(a)

(−1)a−1 = (a − 1)!

Z 0

1

lna−1 (x)(1 − xn ) dx. 1−x

(1.150)

This integral representation extends the range of n to the whole complex plane where R(n) > −1. Note that setting a = 1 in (1.150) gives Z Hn = 0

1

1 − xn dx, 1−x

R(n) > −1.

(1.151)

Let’s break up the integrand in (1.150), Hn(a)

Z Z (−1)a−1 1 lna−1 (x) (−1)a−1 1 xn lna−1 (x) = dx − dx (a − 1)! 0 1 − x (a − 1)! 0 1−x {substitute the result from (3.1) for the first integral} Z (−1)a−1 1 xn lna−1 (x) = ζ(a) − dx. (1.152) (a − 1)! 0 1−x

1.17. Harmonic Number

43

Reorder the terms, Z 1 n a−1   x ln (x) dx = (−1)a−1 (a − 1)! ζ(a) − Hn(a) . 1−x 0

1.17.2

(1.153)

Infinite Series Representation (a)

Another form of Hn is Hn(a) =

∞ X 1 1 − , ka (k + n)a

n∈ / Z− .

(1.154)

k=1

=

(−1)a−1 (a − 1)!

lna−1 (x)(1 − xn ) dx 1−x 0  1 in series 1−x ! Z 1 ∞ X a−1 n k−1 ln (x)(1 − x ) x dx

(−1)a−1 (a − 1)!  expand

Z

1

0

k=1

{switch integration and summation} ∞ Z (−1)a−1 X 1 a−1 = ln (x)(xk−1 − xk+n−1 )dx (a − 1)! 0 k=1

{make use of the result in (1.31)} ∞ X 1 1 = − . ka (k + n)a k=1

Setting a = 1 in (1.154) gives ∞ ∞ X X 1 1 n Hn = − = . k k+n k(k + n) k=1

(1.155)

k=1

To find the derivative of the harmonic number, rewrite (1.154) as ∞ X k=1

1 = ζ(a) − Hn(a) . (k + n)a

(1.156)

44

Chapter 1. Series Transformations and Special Functions

Next, differentiate both sides of (1.154) with respect to n, ∞

X ∂ (a) a . Hn = 0 + ∂n (k + n)a+1 k=1

Employ the result in (1.156) for the latter sum,   ∂ (a) Hn = a ζ(a + 1) − Hn(a+1) . ∂n

(1.157)

Moreover, bring back (1.155) Hn =

∞ r X X 1 1 1 1 − = lim − k k + n r→∞ k k+n

k=1

k=1

and so

r X 1 1 − . r→∞ k k+m

Hm = lim

k=1

Subtracting the two harmonic numbers gives Hn − Hm = lim

r→∞

r X k=1

∞ X 1 1 1 1 − = − . k + m k + n n=1 k + m k + n

(1.158)

ThePreason we inserted the limit in the calculations above is to avoid the divergence ∞ of k=1 k1 . Using the definition in (1.154), we also see that (a) Hn(a) − Hm =

∞ X k=1

1 1 − . (k + m)a (k + n)a

1.18

Skew Harmonic Number

1.18.1

Definition

(1.159)

The n-th skew harmonic number is defined by Hn =

n X (−1)k−1 k=1

k

=1−

1 1 (−1)n−1 + − ··· + , 2 3 n

n ∈ Z+ .

(1.160)

One of the key properties of the skew harmonic number is H n − H n−1 =

(−1)n−1 . n

(1.161)

1.18. Skew Harmonic Number

45

To prove that, write (1.160) as Hn = 1 −

1 1 (−1)n (−1)n−1 + − ··· + + . 2 3 n−1 n

Replace n by n − 1, H n−1 = 1 −

1 1 (−1)n + − ··· + . 2 3 n−1

Subtracting them yields (1.161).

Another property is H 2n = H2n − Hn .

(1.162)

To show that, we begin with the definition of H 2n : H 2n =

2n X (−1)k−1

k

k=1

=1−

1 1 1 1 + − ··· + − 2 3 2n − 1 2n

{split the terms into odd and even}   1 1 1 1 1 1 1 − + + + ··· + = 1 + + + ··· + 3 5 2n − 1 2 4 6 2n   1 1 1 1 add and subtract + + + · · · + 2 4 6 2n   1 1 1 1 1 1 1 1 = 1 + + + ··· + + −2 + + + ··· + 2 3 2n − 1 2n 2 4 6 2n   1 1 1 1 1 1 1 + − 1 + + + ··· + = 1 + + + ··· + 2 3 2n − 1 2n 2 3 n =

2n n X 1 X1 − = H2n − Hn . k k

k=1

k=1

A similar property is H 2n+1 = H2n+1 − Hn ,

(1.163)

which can be proved by using the definition of H 2n+1 : H 2n+1 =

2n+1 X k=1

( use

n+1 X k=1

f (k) =

n X k=1

(−1)k−1 k )

f (k) + f (n + 1)

46

Chapter 1. Series Transformations and Special Functions

=

2n X (−1)k−1

k

k=1

+

1 1 = H 2n + 2n + 1 2n + 1

{substitute (1.162)} 1 = H2n − Hn + 2n + 1   1 = H2n+1 use H2n + 2n + 1 = H2n+1 − Hn . The skew harmonic number can be written in integral form: Hn =

n X (−1)k−1 k=1

k

=

n X

(−1)k−1

Z

1

xk−1 dx

0

k=1

{switch integration and summation} ! Z 1 X n k−1 = (−x) dx 0

Z = 0

1

k=1

{use the geometric series formula} Z 1 Z 1 1 − (−x)n dx (−x)n dx = − dx 1+x 0 1+x 0 1+x Z 1 (−x)n dx. = ln(2) − 0 1+x

Therefore,

1

Z H n = ln(2) − 0

(−x)n dx. 1+x

(1.164)

Like the integral in (1.151), this integral also extends the range of n to the whole complex plane where R(n) > −1.

1.18.2

Infinite Series Representation

Another form of the skew harmonic number is H n = ln(2) +

∞ X (−1)k+n k=1

k+n

,

n∈ / Z− .

Proof. Begin with (1.164) Z H n = ln(2) − 0

1

(−x)n dx 1+x

(1.165)

1.19. Digamma Function

47



 1 expand in series 1+x Z 1 ∞ X = ln(2) − (−x)n (−x)k−1 dx 0

k=1

{interchange integration and summation} Z 1 ∞ X k+n−1 = ln(2) − (−1) xk+n−1 dx 0

k=1

= ln(2) +

∞ X k=1

(−1)k+n , k+n

and the prof is complete. Let’s multiply both sides of (1.165) by (−1)n , ∞ X (−1)k k=1

  = (−1)n H n − ln(2) .

k+n

1.19

Digamma Function

1.19.1

Definition

(1.166)

The Digamma function is defined as ψ(n) =

1.19.2

Γ0 (n) d ln(Γ(n)) = . dn Γ(n)

(1.167)

Digamma Reflection Formula

One of the key properties of the digamma function is ψ(n) − ψ(1 − n) = −π cot(πn),

n∈ / Z.

(1.168)

Proof. Take the logarithm of both sides of the Euler’s reflection formula in (1.38): Γ(n)Γ(1 − n) = π csc(πn),

n∈ / Z,

we obtain ln(Γ(n)) + ln(Γ(1 − n)) = ln(π) + ln(csc(πn)). Next, differentiate both sides using ψ(n) − ψ(1 − n) = −

d dn

ln(Γ(n)) = ψ(n) given in (1.167),

π csc(πn) cot(πn) = −π cot(πn). csc(πn)

48

Chapter 1. Series Transformations and Special Functions

1.19.3

Digamma–Harmonic Number Identity

The relation between the digamma function and the harmonic number is n∈ / Z− .

ψ(n + 1) = Hn − γ,

(1.169)

Proof. Multiply both sides of (1.36): n ∞ 1 Y 1 + k1 , Γ(n) = n 1 + nk k=1

n∈ / Z≤0

by n then use nΓ(n) = Γ(n + 1) given in (1.32), n ∞ Y 1 + k1  Γ(n + 1) = n , 1 + k k=1

n∈ / Z− .

Take the logarithm of both sides, n ∞ Y 1 + k1  ln[Γ(n + 1)] = ln n 1 + k k=1 ( use ln

∞ Y

∞ X

ak =

k=1

=

) ln(ak ) given in (1.12)

k=1 ∞ X

n ! 1 + k1  1 + nk

ln

k=1

or ln[Γ(n + 1)] =

∞ X

 n ln

k=1

k+1 k

Differentiate both sides with respect to n using ψ(n + 1) =

∞ X

 ln

k=1

1 k



d dn

k+1 k

 − ln

k+n k

 .

ln(Γ(n + 1)) = ψ(n + 1),

 −

1 . k+n

in the RHS then insert the limit, we arrive at

ψ(n + 1) = lim

m→∞

m X k=1

 ln

k+1 k

 −

1 1 1 + − k+n k k

{rearrange the terms} "m # "m #   X m X X1 k+1 1 1 = lim ln − + lim − m→∞ m→∞ k k k k+n k=1

k=1

k=1

1.19. Digamma Function

49

" = lim

m→∞

(

m X

 ln

k=1

k+1 k

# m ∞ X X 1 1 1 − + − k k k+n



k=1

k=1

m ∞ X X 1 1 1 use = Hm and − = Hn given in (1.146) and (1.155) k k k+n k=1 k=1 # "m X k + 1 = lim ln − Hm + Hn . m→∞ k

)

k=1

The remaining sum can be evaluated using the telescoping series: m X

 ln

k=1

k+1 k

 =

m X

ln(k + 1) − ln(k)

k=1

= (ln(2) − ln(1)) + (ln(3) − ln(2)) + · · · + (ln(m) − ln(m − 1)) +(ln(m + 1) − ln(m)). Notice that all terms cancel out but ln(m + 1): m X

 ln

k=1

k+1 k

 = ln(m + 1).

Substitute this sum back, we have ψ(n + 1) = lim [ln(m + 1) − Hm ] + Hn m→∞

{let m + 1 = p } = lim [ln(p) − Hp−1 ] + Hn p→∞

 1 given in (1.147) p 1 = − lim [Hp − ln(p)] + lim + Hn p→∞ p→∞ p {the first limit is the Euler–Mascheroni constant defined in (1.209)} 

write Hp−1 = Hp −

= −γ + 0 + Hn , and the proof is finalized. Let’s set n = 0 in (1.169), ψ(1) = −γ. Now set n = −1/2 in (1.169), ψ

  Z 1 1 1 1 − x− 2 + γ = H− 12 = dx 2 1−x 0

(1.170)

50

Chapter 1. Series Transformations and Special Functions √

Z

x=y

1

= −2 0

dy = −2 ln(2). 1+y

Combining (1.170) and (1.171) yields   1 ψ − ψ(1) = −2 ln(2). 2 Moreover, we have ψ(n + 1) − ψ(n) =

1 . n

(1.171)

(1.172)

(1.173)

To show that, make use of the identity in (1.169), ψ(n + 1) − ψ(n) = Hn − γ − (Hn−1 − γ) = Hn − Hn−1 {recall the relation in (1.147)} 1 = . n

1.20

Polygamma Function

1.20.1

Definition

The Polygamma function is defined as ψ (a) (n) =

da da+1 ψ(n) = ln(Γ(n)). dna dna+1

(1.174)

Observe that ψ (0) (n) = ψ(n).

1.20.2

Series Representation

Another series form of the polygamma function is ψ (a) (n + 1) =

∞ X (−1)a+1 a! , (k + n)a+1

n∈ / Z− .

k=1

Proof. Differentiate both sides of (1.169): ∞ X 1 1 ψ(n + 1) = Hn − γ = − −γ k k+n k=1

(1.175)

1.20. Polygamma Function

51

a times with respect to n, ∞ X 1 1 − k k+n

da da ψ(n + 1) = ψ (a) (n + 1) = a dn dna =

! −0

k=1

∞ X (−1)a+1 a! . (k + n)a+1

k=1

Replacing n by n − 1 in (1.175) gives ψ (a) (n) =

∞ X k=1

(−1)a+1 a! . (k + n − 1)a+1

(1.176)

Shift the index by +1, ψ (a) (n) =

∞ X (−1)a+1 a! , (k + n)a+1

n∈ / Z≤0 .

(1.177)

k=0

1.20.3

Integral Representation

Another form of the polygamma function is ψ

(a)

Z (n + 1) = − 0

1

xn lna (x) dx, 1−x

n∈ / Z− .

(1.178)

Proof. Differentiating both sides of (1.169): Z ψ(n + 1) = Hn − γ = 0

1

1 − xn dx − γ 1−x

a times with respect to n, Z 1 da 1 − xn dx − 0 a dn 0 1 − x {use differentiation under the integral sign theorem given in (2.78)} Z 1 a Z 1 n a ∂ 1 − xn x ln (x) = dx = − dx. a 1−x ∂n 1−x 0 0 ψ (a) (n + 1) =

Moreover, if we compare (1.178) and (1.153), we deduce that   ψ (a) (n + 1) = (−1)a a! Hn(a+1) − ζ(a + 1) .

(1.179)

52

Chapter 1. Series Transformations and Special Functions

1.20.4

Evaluation of ψ (a) (1)

For a ∈ Z+ , we have ψ (a) (1) = (−1)a−1 a!ζ(a + 1).

(1.180)

Proof. Put n = 0 in (1.175), ψ (a) (1) =

∞ X (−1)a−1 a!

k a+1

k=1

= (−1)a−1 a!ζ(a + 1).

Examples ψ (1) (1) = ζ(2);

(1.181)

(2)

(1.182)

ψ

ψ (3) (1) = 6ζ(4);

(1.183)

(4)

(1.184)

ψ

ψ

1.20.5

(1) = −2ζ(3);

(1) = −24ζ(5);

(5)

Evaluation of ψ (a)

(1) = 120ζ(6). 1 2

(1.185)



For a ∈ Z+ , we have ψ (a)

  1 = (−1)a−1 a!(2a+1 − 1)ζ(a + 1). 2

(1.186)

Proof. Put n = 1/2 in (1.177), ψ (a)

  X ∞ 1 (−1)a+1 a! = 2 (k + 1/2)a+1 k=0

a−1

= (−1)

a!2

a+1

∞ X k=0

1 . (2k + 1)a+1

This sum is given in (1.85) and the proof is finished. Examples   1 ψ (1) = 3ζ(2); 2   1 ψ (2) = −14ζ(3); 2

(1.187) (1.188)

1.20. Polygamma Function

53

  1 ψ = 90ζ(4); 2   1 ψ (4) = −744ζ(5); 2   1 ψ (5) = 7560ζ(6). 2 (3)

Evaluation of ψ (2a)

1.20.6

1 4

(1.189) (1.190) (1.191)



For a ∈ Z+ , we have   d2a 1 = (22a − 21+4a )(2a)!ζ(2a + 1) − 22a−1 π lim1 2a [π csc(πs)]. ψ (2a) 4 s→ 2 ds (1.192) Proof. Replace 2a by 2a + 1 in (1.82),   2−1−4a (2a) 1 β(2a + 1) = − ψ − (1 − 2−1−2a )ζ(2a + 1). (2a)! 4

(1.193)

Combine (1.92) and (1.193) to finish the proof. Examples   1 ψ (2) = −56ζ(3) − 2π 3 ; 4   1 ψ (4) = −11904ζ(5) − 40π 5 ; 4   1 = −5852160ζ(7) − 1952π 7 ; ψ (6) 4   1 ψ (8) = −5274501120ζ(9) − 177280π 9 ; 4   1 = −7606429286400ζ(11) − 25866752π 11 . ψ (10) 4

1.20.7

Evaluation of ψ (2a)

3 4

(1.194) (1.195) (1.196) (1.197) (1.198)



For a ∈ Z+ , we have     3 1 d2a (2a) (2a) ψ =ψ + lim1 [π cot(πn)]. 2a 4 4 n→ 4 dn

(1.199)

54

Chapter 1. Series Transformations and Special Functions

Proof. Differentiate both sides of (1.168): ψ(n) − ψ(1 − n) = −π cot(πn). 2a times with respect to n, ψ (2a) (n) − ψ (2a) (1 − n) = −

d2a [π cot(πn)]. dn2a

The proof follows on taking the limit on both sides and letting n → 1/4. Examples     3 1 ψ (2) = ψ (2) + 4π 3 ; 4 4     3 1 ψ (4) = ψ (4) + 80π 5 ; 4 4     1 3 = ψ (6) + 3904π 7 ; ψ (6) 4 4     3 1 = ψ (8) + 354560π 9 ; ψ (8) 4 4     1 3 = ψ (10) + 354560π 11 . ψ (10) 4 4

1.21

(1.200) (1.201) (1.202) (1.203) (1.204)

Catalan’s Constant

The Catalan’s constant, denoted by G, is defined as G=

∞ X

(−1)n 1 1 = 1 − 2 + 2 − ··· . 2 (2n + 1) 3 5 n=0

(1.205)

The Catalan’s constant is a special case of the Dirichlet beta function in (1.80): β(z) =

∞ X

(−1)n , (2n + 1)z n=0

G = β(2).

(1.206)

In (1.87), we also see that Z G=− 0

and

1

ln(x) dx 1 + x2

 ψ (1) 41 3 G= − ζ(2). 48 4

(1.207)

(1.208)

1.22. Euler–Mascheroni Constant

55

For more integral and series representations of the Catalan’s constant, check [36].

1.22

Euler–Mascheroni Constant

The Euler–Mascheroni Constant is defined as γ = lim (Hn − ln(n)) . n→∞

(1.209)

The Euler–Mascheroni constant, denoted by γ, is defined as the area bounded by the two functions, y = 1/x and y = 1/bxc, where bxc is the floor function, over the interval x ∈ [1, ∞). To get the form in (1.209), we calculate the bounded area over the interval x ∈ [1, n] then we let integer n → ∞: Z n  Z n dx dx γ = lim − n→∞ x 1 1 bxc   Z n 1 1 1 dx = 1 + + + ··· + = Hn−1 note that 2 3 n−1 1 bxc   Z n dx = lim Hn−1 − n→∞ x 1   Z n dx 1 = ln(n) we have Hn−1 = Hn − and n x 1   1 = lim Hn − − ln(n) n→∞ n 1 = lim (Hn − ln(n)) − lim n→∞ n→∞ n = lim (Hn − ln(n)) − 0. n→∞

For more representations of Euler–Mascheroni constant, see [37].

Chapter 2

Generating Functions and Powerful Identities Before we start deriving the generating functions, we need to prove the following series identity:

∞ X

an xn =

n=1

∞ 1 X (an − an−1 )xn , 1 − x n=1

a0 = 0.

(2.1)

Proof. We begin with the fact that 1=

1 x − . 1−x 1−x

Multiply through by an xn then take the summation over n ≥ 1, ∞ X

an xn =

n=1

∞ ∞ 1 X 1 X an xn − an xn+1 1 − x n=1 1 − x n=1

{let the index n of the second sum start from 0, since a0 = 0} ∞ ∞ 1 X 1 X = an xn − an xn+1 1 − x n=1 1 − x n=0 {shift the index n of the second sum by −1} ∞ ∞ 1 X 1 X = an xn − an−1 xn 1 − x n=1 1 − x n=1 =

∞ 1 X (an − an−1 )xn , 1 − x n=1

56

2.1. Generating Functions

57

and the proof is finalized.

2.1 2.1.1

Generating Functions P∞

n=1

Hn(a) xn

For |x| < 1, the following identity holds: ∞ X

Hn(a) xn =

n=1

(a)

Proof. Since H0

Lia (x) . 1−x

(2.2)

(a)

= 0, it’s valid to set an = Hn in (2.1), ∞ X

∞  1 X  (a) (a) Hn − Hn−1 xn 1 − x n=1 n=1   1 (a) (a) write Hn − Hn−1 = a given in (1.145) n ∞ n X x Lia (x) 1 = , = 1 − x n=1 na 1−x

Hn(a) xn =

and the proof is complete. (a) Let the index n in (2.2) start from 0, since H0 = 0, ∞ ∞ Lia (x) X (a) n X (a) n−1 = Hn x = Hn−1 x . 1−x n=0 n=1

(2.3)

The last step follows from shifting the index n by −1. For a different proof, see [28, pp. 348–349]. If we set a = 1 in (2.2) and (2.3) using Li1 (x) =

∞ X xn = − ln(1 − x), n1 n=1

we have ∞ X n=1 ∞ X

Hn xn = −

ln(1 − x) , 1−x

Hn−1 xn−1 = −

n=1

ln(1 − x) , 1−x

|x| < 1, |x| < 1.

(2.4) (2.5)

58

Chapter 2. Generating Functions and Powerful Identities

Let’s integrate both sides of (2.5) from x = 0 to x, 1 2 ln (1 − x) = 2

Z 0

∞ xX

Hn−1 xn−1 dx

n=1

{interchange integration and summation} Z x ∞ ∞ X X xn n−1 = Hn−1 x dx = Hn−1 . n 0 n=1 n=1 Therefore, ln2 (1 − x) = 2

∞ X Hn−1 n x , n n=1

|x| ≤ 1, x 6= 1.

The latter identity can be proved using the Cauchy product, as follows: ln2 (1 − x) = (− ln(1 − x))(− ln(1 − x)) {expand both logs in series} ! ∞ ! ∞ X X xn xn = n n n=1 n=1  1 1 apply (2.80) where an = and bn = n n ! ∞ n X X1 1 = xn+1 · k n − k+1 n=1 k=1    1 1 1 1 1 write · = + k n−k+1 n+1 k n−k+1 ! ∞ n n X X 1 1 X 1 n+1 = x · + n+1 k n−k+1 n=1 k=1 k=1 ( ) n n X X1 1 use = given in (1.3) n−k+1 k k=1 k=1 ! ∞ n X X 1 1 n+1 = x · 2 n+1 k n=1 k=1 ( ) n X 1 use = Hn defined in (1.146) k 

k=1

=2

∞ X Hn n+1 x n+1 n=1

{let the index n start from 0, since H0 = 0}

(2.6)

2.1. Generating Functions

59

=2

∞ X Hn n+1 x n +1 n=0

{shift the index n by −1} ∞ X Hn−1 n x . =2 n n=1

2.1.2

P∞

Hn n n=1 n x

For |x| ≤ 1, x 6= 1, the following identity holds: ∞ X Hn n 1 x = Li2 (x) + ln2 (1 − x). n 2 n=1

Proof. In (2.6), substitute Hn−1 = Hn − ln2 (1 − x) = 2

∞ X Hn − n n=1

=2

1 n

1 n

(2.7)

given in (1.147),

xn = 2

∞ ∞ X X Hn n xn x −2 n n2 n=1 n=1

∞ X Hn n x − 2 Li2 (x). n n=1

The proof follows on adding 2 Li2 (x) to both sides then dividing by 2.

2.1.3

P∞

Hn n n=1 n2 x

For |x| ≤ 1, the following identity holds: ∞ X Hn n x = Li3 (x) − Li3 (1 − x) + ln(1 − x) Li2 (1 − x) n2 n=1

1 + ln(x) ln2 (1 − x) + ζ(3). 2 Proof. Divide both sides of (2.7) by x then integrate, Z

Z Z ∞ 1 X Hn n Li2 (x) 1 ln2 (1 − x) x dx = dx + dx x n=1 n x 2 x Z 1 ln2 (1 − x) = Li3 (x) + dx. 2 x

(2.8)

60

Chapter 2. Generating Functions and Powerful Identities

Since Z

we have

Z ∞ ∞ ∞ X X 1 X Hn n Hn Hn x n x dx = xn−1 dx = , x n=1 n n n n n=1 n=1 Z ∞ X 1 Hn n ln2 (1 − x) x = Li (x) + dx. 3 n2 2 x n=1

For the remaining integral, forcing integration by parts twice yields Z ln(x) ln(1 − x) ln2 (1 − x) dx = ln(x) ln2 (1 − x) + 2 dx x 1−x   Z Li2 (1 − x) 2 = ln(x) ln (1 − x) + 2 ln(1 − x) Li2 (1 − x) + dx 1−x Z

= ln(x) ln2 (1 − x) + 2 ln(1 − x) Li2 (1 − x) − 2 Li3 (1 − x).

(2.9)

Substitute this integral back and add the constant of integration, ∞ X Hn n 1 x = Li3 (x) − Li3 (1 − x) + ln(1 − x) Li2 (1 − x) + ln(x) ln2 (1 − x) + C. 2 n 2 n=1

Set x = 0, 0 = −ζ(3) + C =⇒ C = ζ(3). Plugging in the value of the constant completes the proof.

2.1.4

P∞

(2)

Hn n=1 n

xn

For |x| ≤ 1, x 6= 1, the following identity holds: ∞ (2) X Hn n x = Li3 (x) + 2 Li3 (1 − x) − ln(1 − x) Li2 (1 − x) n n=1

−ζ(2) ln(1 − x) − 2ζ(3). Proof. Set a = 2 in (2.2), ∞ X

Hn(2) xn =

n=1

Li2 (x) . 1−x

Divide both sides by x then integrate using Z

Z ∞ ∞ ∞ X X 1 X (2) n xn (2) Hn x dx = Hn xn−1 dx = Hn(2) , x n=1 n n=1 n=1

(2.10)

2.1. Generating Functions

61

we have Z Z Z ∞ (2) X Hn n Li2 (x) Li2 (x) Li2 (x) x = dx = dx + dx n x(1 − x) x 1−x n=1 {z } | IBP

2

ln (1 − x) dx x {substitute the result from (2.9)} Z

= Li3 (x) − ln(1 − x) Li2 (x) −

= Li3 (x) + 2 Li3 (1 − x) − ln(1 − x)[Li2 (x) + ln(x) ln(1 − x)] −2 ln(1 − x) Li2 (1 − x) + C {make use of the reflection formula (1.124) for the second term} = Li3 (x) + 2 Li3 (1 − x) − ln(1 − x) Li2 (1 − x) − ζ(2) ln(1 − x) + C. Put x = 0, 0 = 2ζ(3) + C =⇒ C = −2ζ(3), and the proof is finalized. For a different method, substitute the result from (2.8) in (2.82).

2.1.5

P∞

2 n=1 (Hn

− Hn(2) )xn

For |x| < 1, the following identity holds: ∞ X

(Hn2 − Hn(2) )xn =

n=1

ln2 (1 − x) . 1−x

(2.11)

(2)

Proof. Set an = Hn2 − Hn in (2.1), ∞ X n=1

∞  1 X 2 (2) 2 (Hn − Hn(2) ) − (Hn−1 − Hn−1 ) xn 1 − x n=1 ( )  2 1 1 (2) 2 (2) write Hn−1 = Hn − and Hn−1 = Hn − 2 n n   ∞ ∞ 1 xn 2 X 2 X Hn−1 n Hn − = = x 1 − x n=1 n n 1 − x n=1 n

(Hn2 − Hn(2) )xn =

{susbstitute the result from (2.6)}   2 1 2 ln2 (1 − x) = ln (1 − x) = , 1−x 2 1−x and the proof is finished.

62

Chapter 2. Generating Functions and Powerful Identities (2)

Further, let the index in (2.11) start from 0, since Hn2 − Hn = 0, ∞ ln2 (1 − x) X 2 = (Hn − Hn(2) )xn 1−x n=0

{shift the index by −1} ∞ X (2) 2 = (Hn−1 − Hn−1 )xn−1 n=1

! 2  ∞ X 1 1 (2) − Hn + 2 xn−1 . = Hn − n n n=1 Expand and simplify, we obtain  ∞  2 ln2 (1 − x) X 2Hn = + 2 xn−1 . Hn2 − Hn(2) − 1−x n n n=1

(2.12)

Check [28, p. 355] for an alternative proof.

2.1.6

P∞

(2)

n=1

2 −H Hn n n

xn

For |x| ≤ 1, x 6= 1, the following identity holds: ∞ X

(Hn2 − Hn(2) )

n=1

xn = 2 ln(1 − x) Li2 (1 − x) − 2 Li3 (1 − x) n

+ ln(x) ln2 (1 − x) −

1 3 ln (1 − x) + 2ζ(3). 3

Proof. Divide both sides of (2.11) by x then integrate using ∞ X

(Hn2

n=1

x Hn(2) )

n

n

Z =

R

xn−1 dx =

ln2 (1 − x) dx x(1 − x)

Z ln2 (1 − x) ln2 (1 − x) = dx + dx x 1−x {substitute the result of the first integral from (2.9)} Z

= ln(x) ln2 (1 − x) + 2 ln(1 − x) Li2 (1 − x) − 2 Li3 (1 − x) 1 − ln3 (1 − x) + C. 3 On setting x = 0, we get C = 2ζ(3) and the proof is finalized.

(2.13)

xn n ,

2.1. Generating Functions

2.1.7

63

2 Hn n n=1 n x

P∞

For |x| ≤ 1, x 6= 1, the following identity holds: ∞ X Hn2 n 1 x = Li3 (x) − ln(1 − x) Li2 (x) − ln3 (1 − x). n 3 n=1

(2.14)

Proof. The proof follows on combining the results from (2.10) and (2.13). Check [28, p. 349–350] for another approach.

2.1.8

P∞

Hn n n=1 n3 x

For |x| ≤ 1, x 6= 1, the following identity holds:   ∞ X Hn n x x = Li4 − Li4 (1 − x) + 2 Li4 (x) − ln(1 − x) Li3 (1 − x) n3 x−1 n=1 1 1 1 +ζ(3) ln(1−x)+ ζ(2) ln2 (1−x)− ln(x) ln3 (1−x)+ ln4 (1−x)+ζ(4). 2 6 24 (2.15) Proof. Replace z by 1 − x in (1.130),  x x−1 1 3 1 = ζ(3) + ln (1 − x) + ζ(2) ln(1 − x) − ln2 (1 − x) ln(x). 6 2 

Li3 (1 − x) + Li3 (x) + Li3

Divide both sides by x(1 − x) then integrate,   Z Z Li3 x x−1 Li3 (1 − x) Li3 (x) dx + dx + dx x(1 − x) x(1 − x) x(1 − x) {z } | {z } | {z } |

Z

I1

I2

= ζ(3)

I3

3

dx 1 ln (1 − x) ln(1 − x) + dx +ζ(2) dx x(1 − x) 6 x(1 − x) x(1 − x) | {z } | {z } | {z } Z

Z

I4

Z

I5

1 2

2

ln (1 − x) ln(x) dx. x(1 − x) | {z }

Z

I7

I6

64

Chapter 2. Generating Functions and Powerful Identities

Evaluation of I1 : Z Z Z Li3 (1 − x) Li3 (1 − x) Li3 (1 − x) I1 = dx = dx + dx x(1 − x) x 1−x {z } | IBP

ln(x) Li2 (1 − x) = ln(x) Li3 (1 − x) + dx − Li4 (1 − x) 1−x 1 = ln(x) Li3 (1 − x) + Li22 (1 − x) − Li4 (1 − x). 2 Z

Evaluation of I2 : Z I2 =

Li3 (x) dx = x(1 − x)

Z

Li3 (x) dx + x

Z

Li3 (x) dx 1−x | {z } IBP

ln(1 − x) Li2 (x) = Li4 (x) − ln(1 − x) Li3 (x) + dx x 1 = Li4 (x) − ln(1 − x) Li3 (x) − Li22 (x). 2 Z

Evaluation of I3 : Substitute Z Li3 I3 =



x x−1

= y,



x(1 − x)

Evaluation of I4 : Z I4 =

x x−1

Z dx =

dx = x(1 − x)

Z

Li3 (y) dy = Li4 (y) = Li4 y

dx + x

Z



x x−1



dx = ln(x) − ln(1 − x). 1−x

Evaluation of I5 : Z Z Z ln3 (1 − x) ln3 (1 − x) ln3 (1 − x) I5 = dx = dx + dx x(1 − x) x 1−x Z ln3 (1 − x) 1 = dx − ln4 (1 − x). x 4 Evaluation of I6 : Z ln(1 − x) ln(1 − x) dx + dx x 1−x 1 = − Li2 (x) − ln2 (1 − x) 2 {replace z by 1 − x in Landen’s formula in (1.124)} Z

I6 =

ln(1 − x) dx = x(1 − x)

Z

.

2.1. Generating Functions

65

 = Li2

x x−1

 .

Evaluation of I7 : ln2 (1 − x) ln(x) dx x(1 − x) Z Z ln2 (1 − x) ln(x) ln2 (1 − x) ln(x) dx + dx . = 1−x x | {z } | {z } Z

I7 =

I70

I700

Integrate I70 by parts, I70 =

Z

1 1 ln2 (1 − x) ln(x) dx = − ln3 (1 − x) ln(x) + 1−x 3 3

Z

ln3 (1 − x) dx. x

For I700 , expand ln2 (1 − x) in series given in (2.6), I700 =

Z

Z ∞ X ln2 (1 − x) ln(x) Hn−1 dx = 2 xn−1 ln(x)dx x n n=1  ∞ 1  X H − ln(x) n xn n IBP n x − 2 = 2 n n n n=1

∞ ∞ X X Hn n Hn n = 2 ln(x) x −2 x − 2 ln(x) Li3 (x) + 2 Li4 (x) 2 n n3 n=1 n=1

{substitute the result from (2.8)} = 2 Li4 (x) − 2 ln(x) Li3 (1 − x) + 2 ln(x) ln(1 − x) Li2 (1 − x) ∞ X Hn n 2 2 + ln (x) ln (1 − x) + 2ζ(3) ln(x) − 2 x . n3 n=1 Combining the results of I70 and I700 , I7 = 2 Li4 (x) −

1 3 ln (1 − x) ln(x) − 2 ln(x) Li3 (1 − x) + ln2 (x) ln2 (1 − x) 3 +2 ln(x) ln(1 − x) Li2 (1 − x) + 2ζ(3) ln(x) Z ∞ X Hn n 1 ln3 (1 − x) −2 x + dx. n3 3 x n=1

66

Chapter 2. Generating Functions and Powerful Identities

By collecting all integrals, the integral

R

ln3 (1−x) dx x

nicely cancels out,

  ∞ X x Hn n x = Li4 − Li4 (1 − x) + 2 Li4 (x) − ln(1 − x) Li3 (1 − x) n3 x−1 n=1 1 1 4 ln (1 − x) − ln(x) ln3 (1 − x) + ζ(3) ln(1 − x) 24 6 1 2 1 + Li2 (1 − x) + ln(x) ln(1 − x) Li2 (1 − x) + ln2 (x) ln2 (1 − x) 2 2   1 2 x − Li2 (x) − ζ(2) Li2 . 2 x−1 +

The last two lines can be further simplified: 1 1 2 Li2 (1 − x) + ln(x) ln(1 − x) Li2 (1 − x) + ln2 (x) ln2 (1 − x) 2 2  x 1 2 − Li2 (x) − ζ(2) Li2 2 x−1   2  1 x 1 2 = Li2 (1 − x) + ln(x) ln(1 − x) − Li2 (x) − ζ(2) Li2 2 2 x−1 {use the dilogarithm reflection formula (1.119) for the first term}   2  x 1 1 = ζ(2) − Li2 (x) − Li22 (x) − ζ(2) Li2 2 2 x−1   5 x = ζ(4) − ζ(2) Li2 (x) − ζ(2) Li2 4 x−1    5 x = ζ(4) − ζ(2) Li2 (x) + Li2 4 x−1 {make use of Landen’s dilogarithm identity (1.124)}   5 1 = ζ(4) − ζ(2) − ln2 (1 − x) 4 2 5 1 = ζ(4) + ζ(2) ln2 (1 − x). 4 2 Thus,   ∞ X Hn n x − Li4 (1 − x) + 2 Li4 (x) − ln(1 − x) Li3 (1 − x) x = Li 4 n3 x−1 n=1 +

1 4 1 ln (1 − x) − ln(x) ln3 (1 − x) + ζ(3) ln(1 − x) 24 6 5 1 + ζ(4) + ζ(2) ln2 (1 − x) + C. 4 2

2.1. Generating Functions

67

Set x = 0, we get C = − 14 ζ(4) and this finalizes the proof.

P∞

2.1.9

n=1

(2)

Hn n2

xn

For |x| ≤ 1, x 6= 1, the following identity holds:   ∞ (2) X Hn n x 1 x = −2 Li4 + 2 Li4 (1 − x) − Li4 (x) + Li22 (x) 2 n x−1 2 n=1 1 4 ln (1 − x) − ζ(2) ln2 (1 − x) 12 1 −2ζ(3) ln(1 − x) + ln(x) ln3 (1 − x) − 2ζ(4). 3

+2 ln(1 − x) Li3 (1 − x) −

(2.16)

Proof. Substitute the result from (2.15) in (2.83).

2.1.10

P∞

n=1

(3)

Hn n

xn

For |x| ≤ 1, x 6= 1, the following identity holds: ∞ (3) X Hn n 1 x = Li4 (x) − ln(1 − x) Li3 (x) − Li22 (x). n 2 n=1

Proof. Set a = 3 in (2.2), ∞ X

Hn(3) xn =

n=1

Divide both sides by x then integrate using

R

Li3 (x) . 1−x xn−1 dx =

xn n ,

Z ∞ (3) X Hn n Li3 (x) x = dx n x(1 − x) n=1 Z Z Li3 (x) Li3 (x) = dx + dx x 1−x | {z } IBP

ln(1 − x) Li2 (x) = Li4 (x) − ln(1 − x) Li3 (x) + dx x 1 = Li4 (x) − ln(1 − x) Li3 (x) − Li22 (x) + C. 2 Z

The proof completes on extracting C = 0 by setting x = 0.

(2.17)

68

Chapter 2. Generating Functions and Powerful Identities

2.1.11

P∞

n=1

Hn3 xn

For |x| < 1, the following identity holds: ∞ X

Hn3 xn

n=1

 3 1 Li3 (x) + 3 Li3 (1 − x) + ln(x) ln2 (1 − x) = 1−x 2  −3ζ(2) ln(1 − x) − ln3 (1 − x) − 3ζ(3) . (2.18)

Proof. Set an = Hn3 in (2.1), ∞ X

∞ 1 X 3 3 (Hn − Hn−1 )xn 1 − x n=1 n=1  3 ! ∞ 1 X 1 3 = Hn − Hn − xn 1 − x n=1 n  ∞  3Hn 1 1 X 3Hn2 − 2 + 3 xn = 1 − x n=1 n n n " ∞ # ∞ ∞ X H2 X 1 Hn n X xn n n = 3 x −3 x + . 1−x n n2 n3 n=1 n=1 n=1

Hn3 xn =

Gathering the results from (2.14) and (2.8) ends the proof. A different method may be found in [28, p. 352–354].

2.1.12

2 Hn n n=1 n2 x

P∞

For |x| ≤ 1, the following identity holds: ∞ X Hn2 n 1 x = Li4 (x) − 2 Li4 (1 − x) + 2 ln(1 − x) Li3 (1 − x) + Li22 (x) 2 n 2 n=1

− ln2 (1 − x) Li2 (1 − x) −

1 ln(x) ln3 (1 − x) + 2ζ(4). 3

Proof. Divide both sides of (2.14): ∞ X Hn2 n 1 x = Li3 (x) − ln(1 − x) Li2 (x) − ln3 (1 − x) n 3 n=1

(2.19)

2.1. Generating Functions

69

by x then integrate using

R

xn−1 dx =

xn n ,

Z ∞ X Hn2 n 1 2 1 ln3 (1 − x) x = Li (x) + Li (x) − dx. 4 2 n2 2 3 x n=1 For the remaining integral, set 1 − x = y then expand Z

1 1−y

in series,

Z ∞ Z X ln3 (y) ln3 (1 − x) dx = − dy = − y n−1 ln3 (y)dy x 1−y n=1  ∞  n n X y yn yn y IBP 3 2 − 3 ln (y) 2 + 6 ln(y) 3 − 6 4 = − ln (y) n n n n n=1

= ln3 (y) ln(1 − y) + 3 ln2 (y) Li2 (y) − 6 ln(y) Li3 (y) + 6 Li4 (y) {substitute y = 1 − x back} 3

= ln (1 − x) ln(x) + 3 ln2 (1 − x) Li2 (1 − x) − 6 ln(1 − x) Li3 (1 − x) +6 Li4 (1 − x).

(2.20)

On plugging in this integral, we arrive at ∞ X 1 Hn2 n 1 x = Li4 (x) + Li22 (x) − ln3 (1 − x) ln(x) − ln2 (1 − x) Li2 (1 − x) 2 n 2 3 n=1

+2 ln(1 − x) Li3 (1 − x) − 2 Li4 (1 − x) + C. The proof follows on extracting C = 2ζ(4) by setting x = 0.

2.1.13

P∞

n=1

Hn Hn(2) xn

For |x| < 1, the following identity holds: ∞ X

Hn Hn(2) xn =

n=1

 1 1 Li3 (x) + Li3 (1 − x) + ln(x) ln2 (1 − x) 1−x 2  −ζ(2) ln(1 − x) − ζ(3) . (2.21)

(2)

Proof. Set an = Hn Hn in (2.1), ∞ X

∞  1 X (2) Hn Hn(2) − Hn−1 Hn−1 xn 1 − x n=1 n=1     ∞ 1 X 1 1 (2) (2) = Hn Hn − Hn − Hn − 2 xn 1 − x n=1 n n

Hn Hn(2) xn =

70

Chapter 2. Generating Functions and Powerful Identities ∞ 1 X = 1 − x n=1

=

(2)

Hn Hn + n2 n

1 − 3 n

! xn

∞ ∞ (2) 1 X Hn n Li3 (x) 1 X Hn n x + x − . 1 − x n=1 n2 1 − x n=1 n 1−x

Collect the results from (2.8) and (2.10) to complete the proof. Another approach may be found in [28, pp. 350–552].

2.1.14

P∞

3 n=1 (Hn

− 3Hn Hn(2) + 2Hn(3) )xn

For |x| < 1, the following identity holds: ∞  X n=1

 ln3 (1 − x) Hn3 − 3Hn Hn(2) + 2Hn(3) xn = − . 1−x (2)

(2.22)

(3)

Proof. Set an = Hn3 − 3Hn Hn + 2Hn in (2.1), ∞  X

 Hn3 − 3Hn Hn(2) + 2Hn(3) xn

n=1 ∞ i 1 Xh 3 (2) (3) 3 = Hn − 3Hn Hn(2) + 2Hn(3) − Hn−1 + 3Hn−1 Hn−1 − 2Hn−1 xn 1 − x n=1 " ! # ∞ (2) (2) Hn2 − Hn Hn 6 1 X 3 −6 + 3 xn = 1 − x n=1 n n n

=

∞  ∞  xn X X 1 1 Hn n 6 Li3 (x) ·3 Hn2 − Hn(2) − ·6 x + , 1−x n 1 − x n2 1−x n=1 n=1

and the proof ends on collecting the results from (2.13) and (2.8).

2.1.15

P∞

n=1

(2)

Hn Hn n

xn

For |x| ≤ 1, x 6= 1, the following identity holds:   ∞ (2) X Hn Hn n x x = − Li4 + Li4 (1 − x) − ln(1 − x) Li3 (1 − x) n x−1 n=1 1 1 1 4 + ln2 (1 − x) Li2 (1 − x) + ln(x) ln3 (1 − x) − ln (1 − x) − ζ(4). 2 6 24 (2.23)

2.1. Generating Functions

71

Proof. Multiply both sides of (2.72): (2)

1

Z

xn−1 ln3 (1 − x)dx = −

0

(3)

Hn3 + 3Hn Hn + 2Hn . n

n

by −y then take the summation over n ≥ 1, Z 1 3 ∞ (2) (3) X ln (1 − x) Hn3 + 3Hn Hn + 2Hn n y =− n x 0 n=1

∞ X

! (xy)

n

dx

n=1

{use the geometric series formula}   Z 1 3 xy ln (1 − x) dx =− x 1 − xy 0 Z 1 Z 1 y ln3 (1 − x) y ln3 (t) 1−x=t =− dx = − dt 1 − xy 0 0 1 − y + yt {make use of (1.112) for the latter integral}   y = 6 Li4 . y−1 Therefore,   ∞ (3) (2) X Hn3 + 3Hn Hn + 2Hn n y y = −6 Li4 . n y−1 n=1

(2.24)

To establish another relation, divide both sides of (2.22): ∞  X n=1

 ln3 (1 − x) Hn3 − 3Hn Hn(2) + 2Hn(3) xn = − 1−x

by x then integrate from x = 0 to y using

Ry 0

xn−1 dx =

yn n ,

Z y 3 ∞ (2) (3) X Hn3 − 3Hn Hn + 2Hn n ln (1 − x) y =− dx n x(1 − x) 0 n=1 Z y 3 Z y 3 ln (1 − x) ln (1 − x) =− dx − dx. x 1−x 0 0 The first integral is calculated in (2.20) and the second integral is

1 4

ln4 (1 − y). Thus,

∞ (2) (3) X Hn3 − 3Hn Hn + 2Hn n y = − ln3 (1 − y) ln(y) − 3 ln2 (1 − y) Li2 (1 − y) n n=1

+6 ln(1 − y) Li3 (1 − y) − 6 Li4 (1 − y) +

1 4 ln (1 − y) + 6ζ(4). 4

(2.25)

72

Chapter 2. Generating Functions and Powerful Identities

The proof finalizes on combining (2.24) and (2.25) then dividing by 6. 3 Hn n n=1 n x

P∞

2.1.16

For |x| ≤ 1, x 6= 1, the following identity holds:   ∞ X Hn3 n x x = −3 Li4 − 3 Li4 (1 − x) − 2 Li4 (x) + Li22 (x) n x − 1 n=1 +3 ln(1 − x) Li3 (1 − x) + 2 ln(1 − x) Li3 (x) +

1 4 ln (1 − x) 8

3 1 − ln2 (1 − x) Li2 (1 − x) − ln(x) ln3 (1 − x) + 3ζ(4). 2 2

(2.26)

Proof. Combine (2.24) and (2.25) then divide by 2,   ∞ X Hn3 n 1 3 x x = −3 Li4 − ln3 (1 − y) ln(x) − ln2 (1 − x) Li2 (1 − x) n x − 1 2 2 n=1 +3 ln(1 − x) Li3 (1 − x) − 3 Li4 (1 − x) +

∞ (3) X Hn n 1 4 ln (1 − x) + 3ζ(4) − 2 x . 8 n n=1

Substitute the result from (2.17) to finalize the proof.

2.1.17

P∞

4 n=1 (Hn

(2)

(3)

(2)

(4)

2 − 6Hn Hn + 8Hn Hn + 3(Hn )2 − 6Hn )xn

For |x| < 1, the following identity holds: ∞  X n=1

  2 ln4 (1 − x) Hn4 − 6Hn2 Hn(2) + 8Hn Hn(3) + 3 Hn(2) − 6Hn(4) xn = . 1−x (2.27)

 2 (2) (3) (2) (4) Proof. Put an = Hn4 − 6Hn2 Hn + 8Hn Hn + 3 Hn − 6Hn in (2.1), ∞  X

Hn4

6Hn2 Hn(2)

+

8Hn Hn(3)

  2 (2) (4) + 3 Hn − 6Hn xn

n=1 ∞    1 Xh 4 (2) 4 2 Hn − Hn−1 − 6 Hn2 Hn(2) − Hn−1 Hn−1 1 − x n=1    2  2  (3) (2) (3) (2) +8 Hn Hn − Hn−1 Hn−1 + 3 Hn − Hn−1 i  (4) −6 Hn(4) − Hn−1 xn

=

2.1. Generating Functions

73

  X ∞ 1  6 = 1−x  n=1 |

(2)

2Hn 4Hn + n3 n2 {z

6 − 4 n

! xn }

S1

 +4

∞ X n=1

(2)

Hn3 3Hn Hn − n n

|

(3)

2Hn + n {z

3H 2 3 − 2n + 4 n n

S2

!

  x .  } n

The sum S1 is the Cauchy product of Li22 (x) given in (2.83). For S2 , by collecting the results from (2.14), (2.18), and (2.21), we have  ∞  X 3Hn2 3 3 (2) (3) Hn − 3Hn Hn + 2Hn − + 3 xn n n n=1 =−

x ln3 (1 − x) + 3 ln(1 − x) Li2 (x). 1−x

R Divide both sides of the latter identity by x then integrate using xn−1 dx = ! ∞ (2) (3) X 3Hn Hn 2Hn 3Hn2 Hn3 3 − + − 2 + 4 xn := S2 n n n n n n=1 Z Z 3 ln (1 − x) ln(1 − x) Li2 (x) =− dx + 3 dx 1−x x 3 1 = ln4 (1 − x) − Li22 (x) + C. 4 2

xn n ,

Setting x = 0 gives C = 0. Collect S1 and S2 to complete the proof. For another proof, check [28, p. 355].

2.1.18

P∞

n=1

H n xn

For |x| < 1, the following identity holds: ∞ X n=1

H n xn =

ln(1 + x) . 1−x

(2.28)

Proof. ln(1 + x) = (ln(1 + x)) 1−x



1 1−x



74

Chapter 2. Generating Functions and Powerful Identities

{expand both function in Taylor series} ! ∞ ! ! ∞ ! ∞ X X 1 X (−1)n−1 n (−1)n−1 n n−1 n x 1x x = x = n x n=1 n n=1 n=1 n=1   (−1)n−1 0 apply (2.80) with an = and bn = 1 = n n ! ! ∞ n ∞ n k−1 X X 1 X n+1 X (−1)k−1 (−1) (n − k + 1)0 = = x xn x n=1 k k n=1 k=1 k=1 ) ( n X (−1)k−1 = H n defined in (1.160) use k ∞ X

k=1

=

∞ X

H n xn .

n=1

2.1.19

P∞

n=1

Hn n x n

For |x| ≤ 1, x 6= 1, the following identity holds:     ∞ X Hn n 1−x 1 x = Li2 − Li2 (−x) − ln(2) ln(1 − x) − Li2 . (2.29) n 2 2 n=1 The following proof may be found in [8, p. 4]: Proof. Divide both sides of (2.28): ∞ X

H n xn =

n=1

by x then integrate using

R

xn−1 dx =

ln(1 + x) 1−x

xn n ,

Z Z Z ∞ X Hn n ln(1 + x) ln(1 + x) ln(1 + x) x = dx = dx + dx n x(1 − x) x 1−x n=1     1+x write ln(1 + x) = ln + ln(2) in the second integral 2  Z Z Z ln 1+x ln(1 + x) ln(2) 2 = dx − dx + dx x 1−x 1−x   Z  1−x = − Li2 (−x) − d Li2 − ln(2) ln(1 − x) 2   1−x = − Li2 (−x) − Li2 − ln(2) ln(1 − x) + C. 2

2.1. Generating Functions

75

The proof finishes on extracting C = Li2

2.1.20

1 2



by setting x = 0.

P∞

Hn n n=1 n2 x

For |x| ≤ 1, x 6= −1, the following identity holds:       ∞ X Hn n 2x x 1+x x = Li − Li − Li − Li3 (−x) 3 3 3 n2 1+x 1+x 2 n=1       1 1 1 − Li3 (x) + Li3 + ln(1 + x) Li2 (x) + Li2 + ln(2) ln(1 + x) . 2 2 2 (2.30)

Proof. Divide both sides of (2.29) by x then integrate using Z ∞ X Li2 Hn n x = 2 n n=1 |

1−x 2



− Li2 x {z

1 2

R

xn−1 dx =

xn n ,

 dx }

IBP

Li2 (−x) ln(1 − x) dx − ln(2) dx x x      Z 1−x ln(x) ln(1 + x) 1 = ln(x) Li2 dx − Li2 − 2 2 1−x Z Z Z Li2 (−x) ln(1 − x) ln(x) dx − dx − ln(2) dx + ln(2) 1−x x x      Z 1−x 1 ln(x) ln(1 + x) = ln(x) Li2 − Li2 − dx 2 2 1−x Z

Z

+ ln(2) Li2 (1 − x) − Li3 (−x) + ln(2) Li2 (x).

(2.31)

For the remaining integral, set a = ln(x) and b = ln(1 + x) in the algebraic identity: 2ab = a2 + b2 − (a − b)2 , 2

2

2

2 ln(x) ln(1 + x) = ln (x) + ln (1 + x) − ln



x 1+x

 .

Divide both sides by 1 − x then integrate, Z 2

ln(x) ln(1 + x) dx = 1−x

Z

2

Z

Z ln2

2

ln (x) ln (1 + x) dx + dx − 1−x 1−x | {z } | {z } | I1

I2



x 1+x

1−x {z I3

 dx . }

76

Chapter 2. Generating Functions and Powerful Identities

For I1 , expand

1 1−x

in series, ∞ Z X ln2 (x) dx = xn−1 ln2 (x)dx 1−x n=1  ∞  n X xn xn x IBP = − 2 ln(x) 2 + 2 3 ln2 (x) n n n n=1

Z

= − ln2 (x) ln(1 − x) − 2 ln(x) Li2 (x) + 2 Li3 (x).

(2.32)

For I2 , substitute 1 + x = y, ln2 (1 + x) dx = 1−x

ln2 (y) dy 2−y ) ( ∞ X y n−1 1 in Taylor series as expand 2−y 2n n=1 Z ∞ X 1 = y n−1 ln2 (y)dy n 2 n=1   ∞ X 1 yn yn yn 2 − 2 ln(y) = ln (y) + 2 2n n n2 n3 n=1     y y y = − ln2 (y) ln 1 − − 2 ln(y) Li2 + 2 Li3 2 2 2 {substitute y = 1 + x back}       1−x 1+x 1+x = − ln2 (1 + x) ln − 2 ln(1 + x) Li2 + 2 Li3 . 2 2 2 Z

For I3 , substitute Z ln2



x 1+x

1−x

x 1+x

Z

= t,

 Z dx =

ln2 (t) dt = 2 (1 − t)(1 − 2t)

Z

ln2 (t) dt − 1 − 2t

Z

ln2 (t) dt 1−t

= − ln2 (t) ln(1 − 2t) − 2 ln(t) Li2 (2t) + 2 Li3 (2t) + ln2 (t) ln(1 − t) + 2 ln(t) Li2 (t) − 2 Li3 (t)   1 − 2t 2 = − ln (t) ln − 2 ln(t) (Li2 (2t) − Li2 (t)) + 2 Li3 (2t) − 2 Li3 (t) 1−t   x substitute t = back 1+x       2x x x 2 = 2 Li3 − 2 Li3 − ln ln(1 − x) 1+x 1+x 1+x       x 2x x −2 ln Li2 − Li2 . 1+x 1+x 1+x

2.1. Generating Functions

77

Gather the three integrals then divide by 2,       Z x 2x 1+x ln(x) ln(1 + x) dx = Li3 − Li3 + Li3 1−x 1+x 1+x 2       x 2x x + Li3 (x) + ln Li2 − Li2 1+x 1+x 1+x   1+x 1 − ln(1 + x) Li2 − ln(x) Li2 (x) − ln2 (x) ln(1 − x) 2 2     1−x x 1 2 1 2 − ln (1 + x) ln + ln ln(1 − x). 2 2 2 1+x Substitute this integral in (2.31) then factor ln(1 + x), ln(x), and ln(2) out,       ∞ X Hn n 2x x 1+x x = Li − Li − Li − Li3 (−x) 3 3 3 n2 1+x 1+x 2 n=1        2x x 1+x − Li3 (x) + ln(1 + x) Li2 − Li2 + Li2 1+x 1+x 2        2x x 1−x − ln(x) Li2 − Li2 − Li2 − Li2 (x) 1+x 1+x 2     1−x 1 1 2 + Li2 + ln(2)[Li2 (x) + Li2 (1 − x)] + ln (1 + x) ln 2 2 2   1 2 x 1 2 − ln ln(1 − x) + ln (x) ln(1 − x) + C 2 1+x 2 {substitute the relations from (1.128), (1.129), and (1.119) }       2x x 1+x = Li3 − Li3 − Li3 − Li3 (−x) 1+x 1+x 2     1 − Li3 (x) + ln(1 + x) Li2 (x) + Li2 + ln(2) ln(1 + x) 2 − ln(x) [ln(1 − x) ln(1 + x) − ln(2) ln(1 − x)]   1 2 1−x + ln(2)[ζ(2) − ln(x) ln(1 − x)] + ln (1 + x) ln 2 2   1 x 1 − ln2 ln(1 − x) + ln2 (x) ln(1 − x) + C. 2 1+x 2 To extract the constant C, put x = 0, C = Li3

  1 − ln(2)ζ(2). 2

78

Chapter 2. Generating Functions and Powerful Identities

Therefore,       ∞ X 2x Hn n x 1+x x = Li − Li − Li − Li3 (−x) 3 3 3 n2 1+x 1+x 2 n=1       1 1 − Li3 (x) + Li3 + ln(1 + x) Li2 (x) + Li2 + ln(2) ln(1 + x) 2 2 − ln(x) [ln(1 − x) ln(1 + x) − ln(2) ln(1 − x)]   1−x 1 2 + ln(2)[ζ(2) − ln(x) ln(1 − x)] + ln (1 + x) ln 2 2   1 2 x 1 2 − ln ln(1 − x) + ln (x) ln(1 − x) − ln(2)ζ(2). 2 1+x 2 The proof completes on simplifying the last three lines to − 21 ln(2) ln2 (1 + x). A different proof may be found in [8, p. 9].

2.1.21

P∞

n=1

H n2 xn

The following identity is derived by Wolfgang Hintze (see [12]): ∞ X n=1

H n2 xn = −

2 ln(2)x 2 ln(1 − x) − , 1 − x2 1 − x2

|x| < 1.

(2.33)

Proof. His proof starts with considering the integral form of the harmonic number, !  Z 1 n ∞ ∞ Z 1 ∞ X X X 1−y2 1 √ n n n n H n2 x = dy dy x = x − (x y) 1−y 0 0 1−y n=1 n=1 n=1     Z 1 Z 1 √ √ x y y=u 1 x 2u x xu = − dy = − du. √ 2 1−x 1−x y 1 − x 1 − xu 0 1−y 0 1−u Let’s calculate this integral indefinitely:   Z 2u x xu − du 1 − u2 1 − x 1 − xu Z Z x 2u 2xu2 = du − du 2 1−x 1−u (1 − xu)(1 − u2 )   2xu2 x x write = + (1 − xu)(1 − u2 ) (1 − x)(1 − u) (1 + x)(1 + u)   2x − in the second integral (1 − x2 )(1 − xu)

2.1. Generating Functions

=−

79

Z Z Z x ln(1 − u2 ) x du x du 2x du − − + 2 1−x 1−x 1−u 1+x 1+u 1−x 1 − xu x ln(1 − u2 ) x ln(1 − u) x ln(1 + u) 2 ln(1 − xu) =− + − − 1−x 1−x 1+x 1 − x2 x ln(1 + u) x ln(1 + u) 2 ln(1 − xu) =− − − 1−x 1+x 1 − x2 2x ln(1 + u) 2 ln(1 − xu) =− − . 1 − x2 1 − x2

Thus, we have ∞ X

1

  xu 2u x du − 2 1 − x 1 − xu 0 1−u n=1 u=1 2 ln(2)x 2 ln(1 − x) 2x ln(1 + u) 2 ln(1 − xu) =− − − . =− 1 − x2 1 − x2 u=0 1 − x2 1 − x2

2.1.22

P∞

n=1

H n2 xn =

Hn 2

n

Z

xn

The following identity is also derived by Wolfgang Hintze (see [12]):     ∞ X H n2 n 1−x 1 1 x = Li2 − Li2 + 2 Li2 (x) + ln2 (1 − x) n 2 2 2 n=1   1−x + ln(1 + x) ln , |x| ≤ 1, x 6= 1. (2.34) 2

Proof. His approach begins with dividing both sides of (2.33): ∞ X

H n2 xn = −

n=1

by x then integrating using

R

2 ln(2)x 2 ln(1 − x) − 1 − x2 1 − x2

xn−1 dx =

xn n ,

Z Z ∞ X H n2 n dx 2 ln(1 − x) x = −2 ln(2) − dx 2 n 1 − x x(1 − x2 ) n=1 Z Z Z Z 2 dx ln(1 − x) ln(1 − x) ln(1 − x) = ln(2) − − 2 dx + dx − dx 1 − x2 x 1+x 1−x   Z 1−x ln(1 − x) 1 = ln(2) ln + 2 Li2 (x) + dx + ln2 (1 − x). 1+x 1+x 2

80

Chapter 2. Generating Functions and Powerful Identities

For the remaining integral, apply integration by parts, Z Z ln(1 + x) ln(1 − x) dx = ln(1 + x) ln(1 − x) + dx 1+x 1−x     1+x write ln(1 + x) = ln + ln(2) 2  Z Z ln 1+x ln(2) 2 dx + dx = ln(1 + x) ln(1 − x) + 1−x 1−x   1−x = ln(1 + x) ln(1 − x) + Li2 − ln(2) ln(1 − x) 2     1−x 1−x = Li2 + ln(1 + x) ln . 2 2 Plug this integral back in,   ∞ X H n2 n 1 1−x x = Li2 + 2 Li2 (x) + ln2 (1 − x) n 2 2 n=1   1−x + ln(1 + x) ln + C, 2 and the proof follows on extracting C = − Li2 ( 12 ) by setting x = 0.

2.1.23

P∞

Hn 2

n=1 n2

xn

For |x| ≤ 1, x 6= −1, the following identity holds:         ∞ X H n2 n 2x x 1+x 1 x = Li3 − Li3 − Li3 + Li3 2 n 1+x 1+x 2 2 n=1        1 1 1 x + ln(1 + x) Li2 + Li2 (x) + Li2 − ln(1 + x) ln 1+x 2 2 2 1 1 − Li3 (1 − x2 ) + ln(1 − x2 ) Li2 (1 − x2 ) + ln(2)[Li2 (−x) − Li2 (x)] 2 2   1 3 1 1 1 2 2 + Li3 (x) + ln (1 + x) + ln(x) ln (1 − x ) − ζ(3) + Li3 . 3 2 2 1+x (2.35) Proof. Bring back (2.34)     ∞ X H n2 n 1−x 1 x = Li2 − Li2 + 2 Li2 (x) − ln(2) ln(1 + x) n 2 2 n=1

2.1. Generating Functions

81

1 + ln2 (1 − x) + ln(1 + x) ln(1 − x) 2   1 2 1 2 1 2 2 use ln (1 − x) + ln(1 + x) ln(1 − x) = ln (1 − x ) − ln (1 + x) 2 2 2     1−x 1 = Li2 − Li2 + 2 Li2 (x) − ln(2) ln(1 + x) 2 2 1 1 + ln2 (1 − x2 ) − ln2 (1 + x). 2 2 Combine this identity and (2.29) then replace x by y, ∞ ∞ X H n2 n X Hn n 1 y − y = 2 Li2 (y) + Li2 (−y) + ln2 (1 − y 2 ) n n 2 n=1 n=1   1−y 1 . (2.36) − ln2 (1 + y) + ln(2) ln 2 1+y Rx n Divide both sides by y then integrate from y = 0 to x using 0 y n−1 dy = xn , ∞ ∞ X H n2 n X Hn n x − x 2 n n2 n=1 n=1 Z Z 1 x ln2 (1 − y 2 ) 1 x ln2 (1 + y) = 2 Li3 (x) + Li3 (−x) + dy − dy 2 0 y 2 0 y | {z } | {z } I1

I2

+ ln(2)[Li2 (−x) − Li2 (x)].

(2.37)

For I1 , substitute 1 − y 2 = t, Z

x

I1 = 0

1 ln2 (1 − y 2 ) dy = y 2

Z

1

1−x2

ln2 (t) dt 1−t

{recall the result from (2.32)} 1 1 2 = − ln (t) ln(1 − t) − 2 ln(t) Li2 (t) + 2 Li3 (t) 2 1−x2 

= ζ(3) + ln(x) ln2 (1 − x2 ) + ln(1 − x2 ) Li2 (1 − x2 ) − Li3 (1 − x2 ). For I2 , substitute Z I2 = 0

x

1 1+y

= t,

ln2 (1 + y) dy = y

Z

1 1 1+x

ln2 (t) dt = t(1 − t)

Z

1 1 1+x

ln2 (t) dt + t

Z

{recall the result from (2.32) for the second integral}

1 1 1+x

ln2 (t) dt 1−t

82

Chapter 2. Generating Functions and Powerful Identities

1 1 3 2 = ln (t) − ln (t) ln(1 − t) − 2 ln(t) Li2 (t) + 2 Li3 (t) 3 1 1+x     1 1 = 2ζ(3) − 2 Li3 − 2 ln(1 + x) Li2 1+x 1+x 2 + ln(x) ln2 (1 + x) − ln3 (1 + x). 3 Plug in the results of I1 and I2 along with the result from (2.30) in (2.37), the proof is finalized.

2.1.24

(2n n)

P∞

n=1

4n

Hn xn

For |x| < 1, the following identity holds:  √   ∞ 2n X 2 1+ 1−x n n √ √ H x = ln . n 4n 1−x 2 1−x n=1

(2.38)

The following proof may be found in [1, p. 5]: R1 n Proof. Using Hn = 0 1−y 1−y dy, we have ∞ X n=1

2n n 4n

   Z 1 ∞ X 1 2n 1 − yn Hn x = dy xn n 4 1 − y n 0 n=1 !   ∞ ∞ 2n 2n X X 1 n n n n x − (xy) dy 1 − y n=1 4n 4n n=1



1

Z = 0

n

{recall Taylor series for the two sums}   Z 1 1 1 1 √ −√ dy. = 1 − xy 1−x 0 1−y Let’s find both integrals indefinitely starting with the first one: Z 1 ln(1 − y) √ dy = − √ . (1 − y) 1 − x 1−x √ For the second integral, make the change of variable 1 − xy = t,   Z Z 1 dt 2 t √ √ √ dy = 2 = arctanh 1 − x − t2 (1 − y) 1 − xy 1−x 1−x p {substitute t = 1 − xy back} √  2 1 − xy =√ arctanh √ 1−x 1−x

2.1. Generating Functions

83





 1+z use 2 arctanh z = ln 1−z √  √ 1 − x + 1 − xy 1 =√ ln √ √ 1−x 1 − x − 1 − xy √   √ 1 − x + 1 − xy multiply the argument of the log by √ √ 1 − x + 1 − xy  √  √ 1 ( 1 − x + 1 − xy)2 =√ ln . −x(1 − y) 1−x Combine the two integrals,    Z 1 ∞ 2n X 1 1 1 n n √ Hn x = −√ dy 4n 1 − xy 1−x 0 1−y n=1   √ y=1 √ ln(1 − y) 1 ( 1 − x + 1 − xy)2 = − √ −√ ln −x(1 − y) 1−x 1−x y=0   √ y=1 √ 2 ( 1 − x + 1 − xy) 1 ln = −√ −x 1−x y=0 √   √    2 (2 1 − x) (1 + 1 − x)2 1 ln − ln = −√ −x −x 1−x √  √ 2   1 2 1−x 2 1+ 1−x √ √ = −√ ln =√ ln . 1−x 1+ 1−x 1−x 2 1−x

2.1.25

P∞

n=1

(2n n ) Hn 4n

n

xn

For |x| ≤ 1, the following identity holds:  √   ∞ 2n X 1− 1−x n Hn n √ x = 2 Li . 2 4n n 1+ 1−x n=1 Proof. Divide both sides of (2.38) by x then integrate using

R

xn−1 dx =

√   Z 2 1+ 1−x Hn n √ √ x = ln dx n 1−x 2 1−x n=1   Z ln 1+y Z √ y= 1−t 2y ln(1 − t) 1−x=y 1+t = −2 dy = −2 dt 2 1−y t √     1−y 1− 1−x √ = 2 Li2 (t) = 2 Li2 = 2 Li2 + C. 1+y 1+ 1−x ∞ X

2n n 4n



(2.39)

xn n ,

84

Chapter 2. Generating Functions and Powerful Identities

Setting x = 0 gives C = 0 and the proof is completed.

2.1.26

(2n n ) Hn

P∞

4n

n=1

n2

xn

For |x| ≤ 1, x 6= −1, the following identity holds: ∞ X n=1

2n n 4n

+2 Li3

! ! r r Hn n 2x 2x x = −4 ln 1 + Li2 n2 1+x 1+x √ ! r 2x Z 1+x 2x ln(1 − t) ln(1 + t) −4 dt. 1+x t 0



(2.40)

Proof. R y n−1Divideynboth sides of (2.39) by x then integrate from x = 0 to y using x dx = n , 0   √  Z y Li2 1−√1−x ∞ 2n X H 1+ 1−x n n n y =2 dx n n2 4 x 0 n=1   Z √1−y u Li2 1−u √ 1+u 1−x=u du = −4 2 1 − u 1 q 2y Z 1+y 1−u (1 − t) Li2 (t) 1+u =t = 2 dt t(1 + t) 0   1−t 1 2 write = − t(1 + t) t 1+t q q 2y 2y Z 1+y Z 1+y Li2 (t) Li2 (t) =2 dt − 4 dt t 1+t 0 |0 {z } IBP

q

= 2 Li3 (t)

0 q

2y 1+y

q

− 4 ln(1 + t) Li2 (t)

2y 1+y

2y 1+y

0

ln(1 − t) ln(1 + t) dt t 0 r r    r  2y 2y 2y = 2 Li3 − 4 ln 1 + Li2 1+y 1+y 1+y q 2y Z 1+y ln(1 − t) ln(1 + t) dt. −4 t 0 Z

−4

The proof follows on replacing y by x.

2.1. Generating Functions

P∞

2.1.27

n=1

85

2H2n −Hn 2n x n

For |x| < 1, the following identity holds: ∞ X 2H2n − Hn 2n x = 2 arctanh2 (x). n n=1

(2.41)

Proof. We begin with 1 arctanh(x) = − ln 2



1−x 1+x

 .

Squaring both sides,   1 2 1−x arctanh (x) = ln 4 1+x   1 1 2 1 2 1 2 2 use the algebric identity (a − b) = a + b − (a + b) 4 2 2 4 2

{with a = ln(1 − x) and b = ln(1 + x)} 1 2 1 1 = ln (1 − x) + ln2 (1 + x) − ln2 (1 − x2 ) 2 2 4 {expand all squared logs in series given in (2.6)} ∞ ∞ ∞ X Hn−1 n 1 X Hn−1 2n Hn−1 n X x + (−1)n x − x = n n 2 n=1 n n=1 n=1 ( use

∞ X

an +

n=1

∞ X

n

(−1) an = 2

n=1

∞ X H2n − 2n n=1

) a2n given in (1.5) for the first two sums

n=1

=2 =2

∞ X

1 2n

∞ ∞ X H2n−1 2n 1 X Hn−1 2n x − x 2n 2 n=1 n n=1

x2n −

∞ 1 X Hn − 2 n=1 n

1 n

x2n =

∞ 1 X 2H2n − Hn 2n x , 2 n=1 n

and the proof is finalized. If we replace x by ix in (2.41) then use arctanh2 (ix) = − arctan2 (x), we get ∞ X

(−1)n

n=1

2H2n − Hn 2n x = −2 arctan2 (x), n

|x| ≤ 1.

(2.42)

86

Chapter 2. Generating Functions and Powerful Identities

Furthermore, by differentiating (2.41) and (2.42) with respect to x, we find ∞ X

(2H2n − Hn )x2n−1 =

n=1 ∞ X

2 arctanh(x) , 1 − x2

(−1)n (2H2n − Hn )x2n−1 = −

n=1

2.1.28

2 arctan(x) . 1 + x2

(2.43) (2.44)

P∞

H2n 2n+1 n=1 2n+1 x

For |x| < 1, the following identity holds: ∞ X H2n 2n+1 1 x = − arctanh(x) ln(1 − x2 ). 2n + 1 2 n=1

(2.45)

Proof. 1 {ln(1 + x) − ln(1 − x)}{ln(1 + x) + ln(1 − x)} 2 1 1 = ln2 (1 + x) − ln2 (1 − x) 2 2 {expand both squared logs in series given in (2.6)} ∞ ∞ X Hn−1 n X Hn−1 n (−1)n x − x = n n n=1 n=1 ( ) ∞ ∞ ∞ X X X n use (−1) an − an = −2 a2n+1 given in (1.7)

arctanh(x) ln(1 − x2 ) =

n=1

n=1

n=0

∞ X H2n 2n+1 = −2 x 2n +1 n=0

{let the index start from 1, since H0 = 0} ∞ X H2n 2n+1 x , = −2 2n + 1 n=1 and the proof is finished. Let’s differentiate both sides of (2.45) with respect to x, ∞ X n=1

H2n x2n =

x arctanh(x) ln(1 − x2 ) − . 1 − x2 2(1 − x2 )

2.1. Generating Functions

P∞

2.1.29

n=1

87

(−1)n H2n 2n+1 x 2n+1

For |x| ≤ 1, the following identity holds: ∞ X 1 (−1)n H2n 2n+1 x = − arctan(x) ln(1 + x2 ). 2n + 1 2 n=1

(2.46)

Proof. Since −1 = i2 , we have 2

∞ X

(−1)n t2n H2n = 2

n=1

( use 2

∞ X

∞ X

(i)2n t2n H2n

n=1

a2n =

n=1

∞ X

an +

n=1

=

∞ X

∞ X

) n

(−1) an given in (1.5)

n=1 ∞ X

(it)n Hn +

n=1

(−it)n Hn

n=1

{make use of the generating function in (2.4)} ln(1 − it) ln(1 + it) − 1 − it 1 + it ln(1 − it) + ln(1 + it) + it(ln(1 − it) − ln(1 + it)) =− 1 + t2 2 ln(1 + t ) + it(−2i arctan t) =− 1 + t2 2 ln(1 + t ) 2t arctan t =− − . 1 + t2 1 + t2 =−

Therefore, 2

∞ X n=1

(−1)n t2n H2n = −

ln(1 + t2 ) 2t arctan t − . 1 + t2 1 + t2

Integrate both sides from t = 0 to x using 2

Rx 0

t2n dt =

x2n+1 2n+1 ,

 Z x ∞ X (−1)n H2n 2n+1 ln(1 + t2 ) 2t arctan t x =− + dt 2n + 1 1 + t2 1 + t2 0 n=1 Z x =− d(ln(1 + t2 ) arctan t) 0 x = − ln(1 + t2 ) arctan t 0 = − ln(1 + x2 ) arctan x.

Divide both sides by 2 to finish the proof. For a different proof, replace x by ix in (2.45) then use arctanh(ix) = i arctan(x).

88

Chapter 2. Generating Functions and Powerful Identities

2.1.30

P∞

n=1

Hn −H2n n

1 2n2



x2n

For |x| < 1, the following identity holds: ∞  X Hn − H2n

n

n=1

1 2n2



x2n = ln(1 − x) ln(1 + x).

Proof (i). Put a = ln(1 − x) and b = ln(1 + x) in the algebraic identity: ab =

1 1 (a + b)2 − (a − b)2 , 4 4

we have   1 2 1 2 1−x 2 ln(1 − x) ln(1 + x) = ln (1 − x ) − ln 4 4 1+x 1 2 2 = ln (1 − x2 ) − arctanh (x) 4 {expand the first squared log in series given in (2.6)}  and substitute the result of arctanh2 (x) given in (2.41) ∞ ∞ 1 X Hn−1 2n 1 X 2H2n − Hn 2n x − x = 2 n=1 n 2 n=1 n =

∞ ∞ 1 X Hn − n1 2n 1 X 2H2n − Hn 2n x − x 2 n=1 n 2 n=1 n  ∞  X Hn − H2n 1 − 2 x2n . = n 2n n=1

Proof (ii). Replace y by −y in (2.28): ∞ ln(1 + y) X = H n yn , 1−y n=1

we get

∞ ln(1 − y) X = (−1)n H n y n . 1+y n=1

Subtract the two generating functions, ∞ ∞ X ln(1 − y) ln(1 + y) X − = (−1)n H n y n − H n yn . 1+y 1−y n=1 n=1

(2.47)

2.2. Series Expansion of Powers of arcsin(z)

89

Integrate both sides from y = 0 to x using:  Z x ln(1 − y) ln(1 + y) − dy = ln(1 − x) ln(1 + x), 1+y 1−y 0 we have ∞ ∞ X (−1)n H n xn+1 X H n xn+1 − n+1 n+1 n=1 n=1 ) ( ∞ ∞ ∞ X X X n a2n−1 given in (1.8) an = −2 (−1) an − use

ln(1 − x) ln(1 + x) =

n=1

n=1

n=1

∞ X H 2n−1 2n = −2 x 2n n=1



(−1)n given in (1.161) use H n−1 = H n + n ∞ 1 X H 2n + 2n =− x2n , n n=1



and the proof follows on using H 2n = H2n − Hn given in (1.162). For a different proof, see [28, p. 334].

2.2 2.2.1

Series Expansion of Powers of arcsin(z) Series Expansion of arcsin(z)

For |z| ≤ 1, the following identity holds: arcsin(z) =

∞ X n=0

2n n 4n



z 2n+1 . 2n + 1

Proof. Differentiate arcsin(z) then integrate, Z Z 1 √ arcsin(z) = d(arcsin(z)) = dz 1 − z2 ( )  ∞ 2n X 1 2n n in Taylor series as z expand √ 4n 1 − z2 n=0 !    Z X ∞ ∞ 2n 2n Z X 2n 2n n n = z dz = z dz 4n 4n n=0 n=0

(2.48)

90

Chapter 2. Generating Functions and Powerful Identities

=

∞ X n=0

2n n 4n



z 2n+1 + C. 2n + 1

Extracting C = 0 by setting z = 0 completes the proof.

2.2.2

√ Series Expansion of arcsin(z) 2 1−z

For |z| < 1, the following identity holds: ∞ arcsin(z) 1 X 4n z 2n−1 √  = . 2 n=1 2n n 1 − z2 n

(2.49)

The following proof may be found in [41]: Proof. Multiply both sides of (1.46): Z

π 2

sin2n−1 (x)dx =

0

4n 1  2n 2n n

by z 2n−1 then take the summation over n ≥ 1, Z π2 ∞ 1 X 4n z 2n−1 1  = 2 n=1 2n n z sin(x) 0 n

∞ X

! 2

2

(z sin (x))

n

dx

n=1

{employ the geometric series formula}   2 2 Z π2 1 z sin (x) dx = z sin(x) 1 − z 2 sin2 (x) 0 Z π2 z sin(x) = dx 2 1 − z + z 2 cos2 (x) 0   x= π2 z cos(x) 1 arctan √ = −√ 1 − z2 1 − z 2 x=0   1 z arcsin(z) =√ arctan √ = √ . 2 2 1−z 1−z 1 − z2   z To justify the last step, differentiate arctan √1−z then integrate back. 2 If we integrate both sides of (2.49) from z = 0 to z, we get arcsin2 (z) =

∞ 1 X 4n z 2n  2, 2 n=1 2n n n

|z| ≤ 1.

(2.50)

2.2. Series Expansion of Powers of arcsin(z)

2.2.3

91

Series Expansion of arcsin3 (z)

For |z| ≤ 1, the following equality holds:   ∞ 2n  X 1 (2) z 2n+1 (2) 3 n arcsin (z) = 6 H2n − Hn . 4n 4 2n + 1 n=0

(2.51)

The following proof may be found in [41]: Proof. Let arcsin(z) = x in (2.48): arcsin(z) =

∞ X n=0

and write

2n n 4n



z 2n+1 2n + 1

  (2n)! 2n (2n)!2 = , = 2 2 n n! n! (2n)!

we have x=

∞ ∞ X (2n)!2 sin2n+1 (x) X 2 = bn fn (x), 4n n!2 (2n + 1)! n=0 n=0

where fn (x) =

sin2n+1 (x) , (2n + 1)!

bn =

(2.52)

(2n)! . 2n n!

Note that fn00 (x) = fn (x) − (2n + 1)2 fn (x),

(2.53)

bn+1 = (2n + 1)bn .

(2.54)

In light of (2.52), write x3 =

∞ X

an b2n fn (x).

(2.55)

n=0

Assuming a0 = 0 allows the index n to start from 1, x3 =

∞ X

an b2n fn (x).

n=1

Differentiate both sides of (2.56) with respect to x twice then divide by 6, x=

∞ 1X an b2n fn00 (x) 6 n=1

{substitute the result of fn00 (x) given in (2.53)}

(2.56)

92

Chapter 2. Generating Functions and Powerful Identities

=

∞ ∞ 1X 1X an b2n fn−1 (x) − (2n + 1)2 an b2n fn (x) 6 n=1 6 n=1

{shift the index n by +1 in the first sum} {and let n start from 0 in the second sum, since we assumed a0 = 0} ∞ ∞ 1X 1X = an+1 b2n+1 fn (x) − (2n + 1)2 an b2n fn (x) 6 n=0 6 n=0 {substitute bn+1 given in (2.54) in the first sum} ∞ ∞ X 1X 1 (2n + 1)2 an+1 b2n fn (x) − (2n + 1)2 an b2n fn (x) = 6 n=0 6 n=0 =

∞ X (2n + 1)2 [an+1 − an ] 2 bn fn (x). 6 n=0

(2.57)

By comparing the series in (2.52) and (2.57), we see that 1=

(2n + 1)2 [an+1 − an ] 6

or an+1 − an =

6 . (2n + 1)2

Employing the generalization in (1.149), we find   1 (2) an = 6 H2n − Hn(2) . 4

(2.58) (2)

Notice that an in (2.58) meets our assumption (a0 = 0), since H0 Substitute (2.58) in (2.55), x3 = 6

= 0.

 ∞  X 1 (2) H2n − Hn(2) b2n fn (x). 4 n=0

Finally, substitute fn (x) and bn back and let x = arcsin(z) to finish the proof.

2.2.4

Series Expansion of arcsin4 (z)

For |z| ≤ 1, the following identity holds: (2) ∞ 3 X 4n Hn−1 z 2n  arcsin (z) = . 2 n=1 2n n2 n 4

The following proof may be found in [41]:

(2.59)

2.2. Series Expansion of Powers of arcsin(z)

93

Proof. Set arcsin(z) = x in (2.50): arcsin2 (z) = we get

∞ 1 X 4n z 2n  2, 2 n=1 2n n n

∞ 1 X 4n sin2n (x)  x = . 2 n=1 2n n2 n 2

Since



2n n

we have x2 =

 =

(2n)! (2n)! (2n)! = = 2 , 2 2 (n!) (n(n − 1)!) n (n − 1)!2

∞ ∞ 1X n sin2n (x) 1X 2 (2 (n − 1)!)2 = b fn (x), 2 n=1 (2n)! 2 n=1 n

where fn (x) =

sin2n (x) , (2n)!

(2.60)

bn = 2n (n − 1)!.

Note that

In light of (2.60), write

fn00 (x) = fn−1 (x) − (2n)2 fn (x),

(2.61)

bn+1 = 2nbn .

(2.62)

∞ 1X x = an b2n fn (x). 2 n=1

(2.63)

4

Assuming a1 = 0 allows the index n to start from 2, ∞ 1X x = an b2n fn (x). 2 n=2 4

(2.64)

Differentiating both sides of (2.64) twice with respect to x then dividing by 12, ∞ 1 X x = an b2n fn00 (x) 24 n=2 2

{substitute the result of fn00 (x) from (2.61)} ∞ ∞ 1 X 1 X = an b2n fn−1 (x) − an (2n)2 b2n fn (x) 24 n=2 24 n=2 {shift the index n of the first sum by +1 } {and let n start from 1 in the second sum, since we assumed a1 = 0}

94

Chapter 2. Generating Functions and Powerful Identities

=

∞ ∞ 1 X 1 X an+1 b2n+1 fn (x) − an (2n)2 b2n fn (x) 24 n=1 24 n=1

{substitute bn+1 given in (2.62) in the first sum} ∞ ∞ 1 X 1 X (2n)2 an+1 b2n fn (x) − (2n)2 an b2n fn (x) 24 n=1 24 n=1 =

∞ 1 X (2n)2 [an+1 − an ] 2 bn fn (x). 2 n=1 12

(2.65)

By comparing the coefficients of fn (x) in (2.60) and (2.65), we see that 1=

(2n)2 [an+1 − an ] 12

or an+1 − an =

3 . n2

Using (1.145), we find (2)

an = 3Hn−1 .

(2.66)

Notice that an in (2.66) meets our assumption (a1 = 0). Substitute (2.66) in (2.63), we obtain x4 =

∞ 3 X (2) 2 H b fn (x). 2 n=1 n−1 n

Plugging fn (x) and bn back in and letting x = arcsin(z) completes the proof. By differentiating both sides of (2.51) and (2.59), we obtain   ∞ 2n  X arcsin2 (z) 1 (2) 2n (2) n √ =2 H − H z |z| < 1 (2.67) 2n 4n 4 n 1 − z2 n=1 and (2) ∞ arcsin3 (z) 3 X 4n Hn−1 z 2n−1 √  . |z| < 1. = 4 n=1 2n n 1 − z2 n

(2.68)

2.3. Identities by Beta Function

2.3

95

Identities by Beta Function

2.3.1

Expressing Beta Function as a Product

For n ∈ Z+ , the following identity holds: B(m, n) = Γ(n)

n−1 Y k=0

1 . k+m

(2.69)

Proof. Multiply both sides of (1.35): n Γ(z + n + 1) Y 1 Γ(z) = z z+k k=1

by

z Γ(z+n+1)

and use Γ(z + 1) = zΓ(z), we obtain n Y 1 Γ(z + 1) = . Γ(z + n + 1) z+k k=1

Next, replace z by m − 1 then multiply both sides by Γ(n), n

n−1

k=1

k=0

Y Y 1 1 Γ(m)Γ(n) := B(m, n) = Γ(n) = Γ(n) , Γ(m + n) k+m−1 k+m where the last form follows from shifting the index k by +1.

2.3.2

Evaluation of Four Logarithmic Integrals

For n ∈ Z+ , the following identities hold: Z

1

xn−1 ln(1 − x)dx = −

0

Z

xn−1 ln2 (1 − x)dx =

0

Hn2 + Hn ; n (2)

1

xn−1 ln3 (1 − x)dx = −

0

Z

(2.70)

(2)

1

0

Z

Hn ; n

xn−1 ln4 (1 − x)dx =

(3)

Hn3 + 3Hn Hn + 2Hn ; n (2)

1

(2.71)

(3)

(2)

(2.72) (4)

Hn4 + 6Hn2 Hn + 8Hn Hn + 2(Hn )2 + 6Hn . n (2.73)

The following proof may be found in [10, p. 157]:

96

Chapter 2. Generating Functions and Powerful Identities

Proof. Take the logarithm of both sides of (2.69), we have ln B(m, n) = ln Γ(n) + ln

n−1 Y k=0

n

use ln

Y

an =

X

= ln Γ(n) −

1 k+m

o ln(an ) given in (1.12)

n−1 X

ln(k + m).

k=0

Differentiate both sides with respect to m, n−1

∂ ∂m

X 1 B(m, n) =− B(m, n) k+m k=0

or

n−1

X 1 ∂ B(m, n) = − B(n, m) . ∂m k+m

(2.74)

k=0

Differentiate both sides of the latter identity with respect to m, n−1

n−1

k=0

k=0

X 1 X ∂2 ∂ 1 B(m, n) = − B(m, n) + B(m, n) . ∂m2 ∂m k+m (k + m)2 ∂ ∂m

B(n, m) from (2.74),   !2 n−1 n−1 2 X X ∂ 1 1 . B(m, n) = B(n, m)  + ∂m2 k+m (k + m)2

Substitute the result of

k=0

(2.75)

k=0

Differentiate (2.75) twice with respect to m,  !3 ! n−1 n−1 X 1 X 1 ∂3 B(m, n) = − B(m, n)  +3 ∂m3 k+m k+m k=0 k=0 ! # n−1 n−1 X X 1 1 + 2 , (2.76) 2 (k + m) (k + m)3 k=0 k=0  !4 ! n−1 n−1 X 1 X ∂4 1  B(m, n) = B(m, n) +6 ∂m4 k+m (k + m)2 k=0

n−1 X k=0

1 k+m

!2 +8

n−1 X k=0

k=0

1 k+m

!

n−1 X k=0

1 (k + m)3

!

2.3. Identities by Beta Function

97

n−1 X

+3

k=0

1 (k + m)2

!

n−1 X

+6

k=0

# 1 . (k + m)4

(2.77)

Now set m = 1 in (2.74), (2.75), (2.76), and (2.77) using: B(1, n) = and

n−1 X k=0

Γ(1)Γ(n) Γ(n)Γ(1) 1 = = Γ(n + 1) nΓ(n) n n

X 1 1 = = Hn(a) , a (k + 1) ka k=1

we get 

 Hn ∂ =− B(m, n) ; ∂m n m=1   2 (2) Hn2 + Hn ∂ = B(m, n) ; ∂m2 n m=1  3  (3) (2) Hn3 + 3Hn Hn + 2Hn ∂ = − B(m, n) ; ∂m3 n m=1   4 (4) (2) (3) (2) H 4 + 6Hn2 Hn + 8Hn Hn + 2(Hn )2 + 6Hn ∂ = n B(m, n) . 4 ∂m n m=1 On the other hand, by using the definition of the beta function: Z B(m, n) =

1

x

m−1

n−1

(1 − x)

dx

1−x→x

Z

=

0

1

xn−1 (1 − x)m−1 dx,

0

we have 

∂ B(m, n) ∂m

 m=1



∂ = ∂m

1

Z

x

n−1

m−1

(1 − x)

 dx

0

. m=1

A special case of Leibniz’s integral rule (see [43]) is differentiation under the integral sign theorem: Z b Z b ∂ d f (x, m)dx = f (x, m)dx. (2.78) dm a ∂m a Since the beta function has three variables (m, n, x), using (2.78), we have ∂ ∂m

Z

b

Z B(x, m, n)dx =

a

a

b

∂ B(x, m, n)dx. ∂m

(2.79)

98

Chapter 2. Generating Functions and Powerful Identities

Using this rule, we reach  Z 1   ∂ n−1 ∂ B(m, n) = x (1 − x)m−1 dx ∂m 0 ∂m m=1 m=1  Z 1 n−1 m−1 x ln(1 − x)(1 − x) dx = 0

m=1 1

Z

xn−1 ln(1 − x)dx;  2 Z 1 2   ∂ ∂ n−1 m−1 = B(m, n) x (1 − x) dx 2 ∂m2 0 ∂m m=1 m=1 Z 1  = xn−1 ln2 (1 − x)(1 − x)m−1 dx =

0

0

m=1

Z

1

xn−1 ln2 (1 − x)dx;  3  Z 1 3  ∂ ∂ n−1 m−1 B(m, n) = x (1 − x) dx 3 ∂m3 0 ∂m m=1 m=1 Z 1  = xn−1 ln3 (1 − x)(1 − x)m−1 dx =

0

0

m=1

Z

1

xn−1 ln3 (1 − x)dx;  Z 1 4   4 ∂ ∂ n−1 m−1 B(m, n) = x (1 − x) dx 4 ∂m4 0 ∂m m=1 m=1 Z 1  = xn−1 ln4 (1 − x)(1 − x)m−1 dx =

0

0

m=1

Z =

1

xn−1 ln4 (1 − x)dx,

0

and the proof follows on comparing the last eight results. A different proof, with no use of the beta function or the gamma function, may be found in [28, pp. 59–62].

2.4. Identities by Cauchy Product

2.4 2.4.1

99

Identities by Cauchy Product Cauchy Product of Two Power Series

P∞ P∞ Let n=1 an xn and n=1 bn xn be two power series. The Cauchy product of these two series is given by ! ∞ ! ! ∞ ∞ n X X X X n+1 n n x an x bn x = ak bn−k+1 . (2.80) n=1

n=1

n=1

k=1

Proof. ∞ X

∞ X

! an xn

n=1

! bn xn

n=1

= a1 x + a2 x2 + a3 x3 + · · ·



b1 x + b2 x2 + b3 x3 + · · ·



= x2 (a1 b1 ) + x3 (a1 b2 + a2 b1 ) + x4 (a1 b3 + a2 b2 + a3 b1 ) + · · · ! ! ! 1 2 3 X X X 2 3 4 =x ak b2−k + x ak b3−k + x ak b4−k + · · · k=1

k=1

=

∞ X

xn+1

n=1

k=1 n X

! ak bn−k+1

Following the steps above also gives ! ∞ ! ∞ ∞ X X X n n xn an x bn x = n=0

2.4.2

n=0

.

k=1

n=0

n X

! ak bn−k

.

(2.81)

k=0

Cauchy Product of − ln(1 − x) Li2 (x)

For |x| ≤ 1, x 6= 1, the following identity holds ∞ ∞ (2) X Hn n X Hn n − ln(1 − x) Li2 (x) = 2 x − 3 Li3 (x). x + n2 n n=1 n=1

Proof. We follow the same approach as in [28, p. 516]: Expand Li2 (x) and ln(1 − x) in series, ! ∞ ! ∞ X X xn xn (Li2 (x))(− ln(1 − x)) = n2 n n=1 n=1

(2.82)

100

Chapter 2. Generating Functions and Powerful Identities



1 1 employ the Cauchy product in (2.80) where an = 2 and bn = n n ! ∞ n X X 1 = xn+1 2 (n − k + 1) k n=1



k=1

{use the partial fraction decomposition for the inner sum} = (

∞ X

x

n=1 n X

n+1

n X k=1

1 1 1 + + (n + 1)k 2 (n + 1)2 k (n + 1)2 (n − k + 1) n

n

!

X1 X 1 1 use = = Hn given in (1.3) = Hn(2) and 2 k n−k+1 k k=1 k=1 k=1 ! ∞ (2) X Hn 2Hn n+1 x = + (n + 1) (n + 1)2 n=1 (2)

{let the index n start from 0, since H0 =

∞ X

(2)

xn+1

n=0

)

= H0 = 0} !

Hn 2Hn + n + 1 (n + 1)2

{shift the index n by −1} =

∞ X n=1

=

∞ X n=1

(2)

x

n

Hn−1 2Hn−1 + n n2 (2)

xn

Hn − n

1 n2

!

2Hn − + n2

2 n

!

∞ ∞ ∞ (2) X X X Hn n Hn n xn = x +2 x − 3 2 n n n3 n=1 n=1 n=1 ∞ ∞ (2) X X Hn n Hn n x − 3 Li3 (x). x +2 = n n2 n=1 n=1

2.4.3

Cauchy Product of Li22 (x)

For |x| ≤ 1, the following identity holds Li22 (x) = 4

∞ ∞ (2) X X Hn n Hn n x + 2 x − 6 Li4 (x). n3 n2 n=1 n=1

(2.83)

2.4. Identities by Cauchy Product

101

Proof. Divide both sides of (2.82): − ln(1 − x) Li2 (x) = 2 by x then integrate using 2

R

∞ ∞ (2) X Hn n X Hn n x + x − 3 Li3 (x) n2 n n=1 n=1

xn−1 dx =

xn n ,

∞ ∞ (2) X Hn n X Hn n x + x − 3 Li4 (x) n3 n2 n=1 n=1

Z =

1 − ln(1 − x) Li2 (x) dx = Li22 (x) + C. x 2

The proof finishes on extracting C = 0 by setting x = 0.

2.4.4

Cauchy Product of − ln(1 − x) Li3 (x)

For |x| ≤ 1, x 6= 1, the following identity holds − ln(1 − x) Li3 (x) = 2

∞ ∞ ∞ (2) (3) X Hn n X Hn n X Hn x + x + − 4 Li5 (x). n3 n2 n n=1 n=1 n=1 (2.84)

Proof. ! ∞ X xn (Li3 (x))(− ln(1 − x)) = n n=1   1 1 employ (2.80) where an = 3 and bn = n n ! ∞ n X X 1 = xn+1 3 k (n − k + 1) n=1 ∞ X xn n3 n=1

!

k=1

{make use of the partial fraction decomposition for the inner sum} =

∞ X

n X

1 1 1 + + 3 2 k2 (n + 1)k (n + 1) (n + 1)3 k n=1 k=1  1 + (n + 1)3 (n − k + 1) ! ∞ (3) (2) X Hn Hn 2Hn n+1 = x + + n + 1 (n + 1)2 (n + 1)3 n=1 x

n+1

n o (3) (2) let the index n start from 0, since H0 = H0 = H0 = 0

102

Chapter 2. Generating Functions and Powerful Identities

=

∞ X

(2)

(3)

2Hn Hn Hn + + n + 1 (n + 1)2 (n + 1)3

xn+1

n=0

!

{shift the index n by −1} ∞ X

=

(3)

x

n

n=1

=

=

=

∞ X

(3)

x

n=1 ∞ (3) X Hn n=1 ∞ X

Hn − n

n

n

xn +

(2)

Hn−1 H 2Hn−1 + + n−1 n n2 n3 1 n3

(2)

Hn − + n

1 n2

!

Hn − +2 n2

1 n

!

∞ ∞ ∞ (2) X X X Hn n Hn n xn x + 2 x − 4 2 3 n n n4 n=1 n=1 n=1

∞ ∞ (3) (2) X Hn n X Hn n Hn n x + x + 2 x − 4 Li4 (x). 2 n n n3 n=1 n=1 n=1

Cauchy Product of Li2 (x) Li3 (x)

2.4.5

For |x| ≤ 1, the following identity holds Li2 (x) Li3 (x) = 6

∞ ∞ ∞ (2) (3) X X Hn n X Hn n Hn n x + 3 x + x − 10 Li5 (x). 4 3 n n n2 n=1 n=1 n=1 (2.85)

Proof (i). Li3 (x) Li2 (x) = =

∞ X n=1

=

∞ X n=1

n X

xn+1

∞ X xn n3 n=1

n X k=1

k 3 (n

!

! ∞ X xn n2 n=1 !

1 − k + 1)2

1 2 1 + + (n + 1)2 k 3 (n + 1)3 k 2 (n + 1)3 (n − k + 1)2 k=1  3 3 + + (n + 1)4 k (n + 1)4 (n − k + 1) ! ∞ (3) (2) X Hn Hn Hn n+1 = x +3 +6 (n + 1)2 (n + 1)3 (n + 1)4 n=1 ! (3) (2) ∞ X H H H n−1 n−1 n−1 = xn +3 3 +6 4 n2 n n n=1

xn+1

2.4. Identities by Cauchy Product

=

=

=

∞ X

103 (3)

x

n

Hn − n2

1 n3

(2)

Hn − +3 n3

1 n2

Hn − +6 n4

n=1 ∞ ∞ ∞ (2) (3) X X X Hn n Hn n Hn n x x x + 3 + 6 2 3 n n n4 n=1 n=1 n=1 ∞ ∞ ∞ (2) (3) X X Hn n Hn n X Hn n 6 x + 3 x + x n4 n3 n2 n=1 n=1 n=1

− 10

1 n

!

∞ X xn n5 n=1

− 10 Li5 (x).

Proof (ii). Z Li2 (x) Li3 (x) = d(Li2 (x) Li3 (x)) Z  1 = Li22 (x) − ln(1 − x) Li3 (x) dx x {substitute the results from (2.83) and (2.84)} ! ∞ ∞ ∞ (2) (3) X X Hn n Hn n X Hn n = 6 x +3 x + x − 10 Li4 (x) dx n3 n2 n n=1 n=1 n=1   Z xn interchange integration and summation then use xn−1 dx = n ∞ ∞ ∞ (2) (3) X X X Hn n Hn n Hn =6 x +3 x + − 10 Li5 (x) + C. 4 3 n n n2 n=1 n=1 n=1 Z

1 x

The proof finalizes on finding C = 0.

2.4.6

Cauchy Product of Li23 (x)

For |x| ≤ 1, the following identity holds Li23 (x) = 12

∞ ∞ ∞ (2) (3) X X X Hn n Hn n Hn n x +6 x +2 x −20 Li6 (x). (2.86) n5 n4 n3 n=1 n=1 n=1

Proof. Divide both sides of (2.85): ∞ ∞ ∞ (2) (3) X X Hn n Hn n X Hn n Li2 (x) Li3 (x) = 6 x +3 x + x − 10 Li5 (x) 4 3 n n n2 n=1 n=1 n=1

by x then integrate, 6

∞ ∞ ∞ (2) (3) X X Hn n Hn n X Hn n x + 3 x + x − 10 Li6 (x) n5 n4 n3 n=1 n=1 n=1

104

Chapter 2. Generating Functions and Powerful Identities

Z =

Li2 (x) Li3 (x) 1 dx = Li23 (x) + C, x 2

and the proof follows as C = 0.

2.4.7

Cauchy Product of − ln(1 − x) Li4 (x)

For |x| ≤ 1, x 6= 1, the following identity holds − ln(1 − x) Li4 (x) = 2 +

∞ ∞ ∞ (2) (3) X Hn n X Hn n X Hn n x + x + x n4 n3 n2 n=1 n=1 n=1

∞ (4) X Hn n x − 5 Li5 (x). n n=1

(2.87)

The following proof may be found in [28, p. 516]: Proof. (Li4 (x))(− ln(1 − x)) = =

∞ X

x

n+1

n=1 ∞ X

n X k=1

(4)

(3)

1 n4

Hn − + n2

∞ X xn n4 n=1

!

∞ X xn n n=1 !

!

1 k 4 (n − k + 1) (2)

Hn Hn Hn Hn = xn+1 + + +2 2 3 n + 1 (n + 1) (n + 1) (n + 1)4 n=1 ! (4) (3) (2) ∞ X H H H H n−1 n−1 = xn + n−1 + n−1 +2 4 2 n n n3 n n=1 =

∞ X

(4)

x

n

n=1

=

Hn − n ∞ X

(3)

(4)

xn

n=1

Hn n

(3)

+

1 n3

(2)

Hn − + n3 (2)

Hn Hn + 3 2 n n

1 n2

!

Hn − +2 n4 ! Hn 5 +2 4 − 5 n n

1 n

!

∞ ∞ ∞ ∞ (2) (3) (4) X Hn n X Hn n X Hn n X Hn n =2 x + x + x + x − 5 Li5 (x), n4 n3 n2 n n=1 n=1 n=1 n=1

and we are done with the proof. Applying the Cauchy product, we also find, for |x| ≤ 1, the following identities: Li2 (x) Li4 (x) = 8

∞ ∞ ∞ (2) (3) X X X Hn n Hn n Hn n x + 4 x + 2 x n5 n4 n3 n=1 n=1 n=1

2.5. Identities by Abel’s Summation

105

+

Li3 (x) Li4 (x) = 20

∞ (4) X Hn n x − 15 Li6 (x); n2 n=1

(2.88)

∞ ∞ ∞ (2) (3) X X X Hn n Hn n Hn n x x x + 10 + 4 n6 n5 n4 n=1 n=1 n=1

∞ (4) X Hn n x − 35 Li7 (x); + n3 n=1

Li24 (x) = 40

(2.89)

∞ ∞ ∞ (2) (3) X X X Hn n Hn n Hn n x + 20 x + 8 x n7 n6 n5 n=1 n=1 n=1

+2

∞ (4) X Hn n x − 70 Li8 (x). n4 n=1

(2.90)

Note that (2.90) follows from dividing both sides of (2.89) by x then integrating.

2.5 2.5.1

Identities by Abel’s Summation Abel’s Summation

Given two finite sums n X

Pn

k=1

ak and

Pn

k=1 bk ,

ak bk = An bn + Am bm−1 −

k=m

define An =

n−1 X

Pn

i=1

Ak (bk+1 − bk ) .

ai . Then (2.91)

k=m

Proof. By the given sum, An = Ak =

k X

Pn

i=1

ai , one can write

ai = a1 + a2 + ... + ak−1 + ak

i=1

and so Ak−1 =

k−1 X

ai = a1 + a2 + ... + ak−1 .

i=1

Subtracting the two sums yields Ak − Ak−1 = ak .

(2.92)

Multiply both sides of (2.92) by bk then take the summation from k = m to n, n X k=m

a k bk =

n X k=m

(Ak − Ak−1 ) bk

106

Chapter 2. Generating Functions and Powerful Identities

=

n X

Ak bk −

k=m

For the first sum, use the fact that

n X

Ak−1 bk .

k=m n X

k=m n X k=m

and for the second sum, use

n−1 X

Ak bk ,

k=m n X

f (k) = f (m) +

k=m n X

f (k),

k=m

Ak bk = An bn +

n X

n−1 X

f (k) = f (n) +

f (k),

k=m+1 n X

Ak−1 bk = Am−1 bm +

k=m

Ak−1 bk

k=m+1

{shift the index k by +1} n−1 X

= Am−1 bm +

Ak bk+1 .

k=m

Combining the two sums, we obtain n X

ak bk = An bn + Am−1 bm +

k=m

n−1 X

Ak bk −

k=m

= An bn + Am−1 bm −

n−1 X

n−1 X

Ak bk+1

k=m

Ak (bk+1 − bk ) ,

k=m

and the proof is finished. For the two cases m = 0 and m = 1, we have Am−1 = 0. So, (2.91) becomes: n X

ak bk = An bn −

k=0 n X

n−1 X

Ak (bk+1 − bk ) ,

(2.93)

Ak (bk+1 − bk ) .

(2.94)

k=0

ak bk = An bn −

k=1

n−1 X k=1

Also note that the index k in the RHS of (2.94) can start from 0, since A0 = 0. Thus, n X k=1

ak bk = An bn −

n−1 X k=0

Ak (bk+1 − bk ) ,

An =

n X i=1

ai .

(2.95)

2.5. Identities by Abel’s Summation

107

You may find in [6, Theorem 2.20, p. 55] a proof for a similar formula: n X k=m

2.5.2

n X

ak bk = An bn+1 + Am−1 bm −

Ak (bk+1 − bk ) .

k=m

First Application

For integers p, q ≥ 2, the following identity holds: ∞ (p) X H k=1

Proof. Let ak =

1 kq

n (p) X H k=1

k kq



k kq

+

∞ (q) X H k

= ζ(p)ζ(q) + ζ(p + q).

kp

k=1

(2.96)

(p)

and bk = Hk in (2.95), =

n X 1 iq i=1

! Hn(p)

n−1 X k=0

k X 1 iq i=1

!



(p)

 1 given in (1.145) use = + (k + 1)p  n−1 X  (q)   1 (q) (p) = Hn Hn − Hk (k + 1)p (p) Hk+1

(p) Hk

k=0

{shift the index k by −1} = Hn(q) Hn(p) − = Hn(q) Hn(p) −

(q) n X Hk−1

kp

k=1 n X H (q) k

= Hn(q) Hn(p) −

k=1

1 kq

kp

k=1 n X

(q)

Hk 1 − p+q kp k

= Hn(q) Hn(p) + ζ(q + p) −

!

n (q) X H k

k=1

kp

.

Reorganize the terms, we have n (p) X H k=1

k kq

+

n (q) X H k

k=1

kp

(p)

Hk+1 − Hk

= Hn(q) Hn(p) + ζ(q + p).



108

Chapter 2. Generating Functions and Powerful Identities

Next, take the limit on both sides letting n → ∞, ∞ (p) X H k kq

k=1

+

∞ (q) X H k kp

k=1

= lim

n→∞

n o Hn(q) Hn(p) + ζ(q + p) ,

and the proof follows on using lim

n→∞

Hn(q)

n ∞ X X 1 1 = lim = = ζ(p). n→∞ kp kp k=1

k=1

For a different approach, see [28, p. 358]. Setting q = p in (2.96) yields ∞ (p) X H k=1

ζ 2 (p) + ζ(2p) . 2

(2.97)

7 ζ 2 (2) + ζ(4) = ζ(4); 2 4

(2.98)

k kp

=

Examples ∞ (2) X H k k2

k=1

=

∞ (3) X H k=1 ∞ (4) X Hk k4 k=1

ζ 2 (3) + ζ(6) ; 2

(2.99)

ζ 2 (4) + ζ(8) 13 = ζ(8), 2 12

(2.100)

k k3

=

=

where we used ζ 2 (2) = 25 ζ(4) and ζ 2 (4) = 76 ζ(8) given in (1.62) and (1.65).

2.5.3

Second Application

For integer p ≥ 2, the following identity holds:  2 (p) ∞ Hk X k=1

Proof. Let ak =  2 (p) n Hk X k=1

kp

1 kp

=

kp

∞ (p) X H k=1

k k 2p

=

ζ 3 (p) − ζ(3p) 3

(2.101)

 2 (p) and bk = Hk in (2.95), n X 1 p i i=1

!



Hn(p)

2

n−1 X k=0

k X 1 p i i=1

! 

(p) Hk+1

2



(p) Hk

2 

2.5. Identities by Abel’s Summation

 use =

Hn(p)

(p) Hk+1



Hn(p)

109

(p) Hk

= 2

 1 given in (1.145) + (k + 1)p

n−1 X

(p)

2Hk 1 + p (k + 1) (k + 1)2p

(p) Hk

k=0

!

{shift the index k by −1} ! (p) n  3 X 2H 1 (p) k−1 + 2p = Hn(p) − Hk−1 kp k k=1 !  (p) n   3 X 2Hk − k2p 1 1 (p) (p) Hk − p = Hn − + 2p k kp k k=1 2  (p) n n (p)  3 Hk X X Hk = Hn(p) − 2 + 3 − Hn(3p) . kp k 2p k=1

k=1

Rearranging the terms,

3

 2 (p) n Hk X k=1

−3

kp

n (p) X H k=1

k k 2p

 3 = Hn(p) − Hn(3p) .

Take the limit on both sides letting n → ∞,

3

 2 (p) ∞ Hk X k=1

kp

−3

∞ (p) X H k=1

k k 2p

= lim

n→∞



Hn(p)

3

 − Hn(3p) ,

and the proof completes on using lim Hn(3p) = ζ(3p). n→∞

Another approach may be found in [28, p. 359]. Examples  2 (2) ∞ Hk X

∞ (2) X H k k4

=

ζ 3 (2) − ζ(6) 9 = ζ(6); 3 8

k2 k=1  2 (3) ∞ ∞ (3) Hk X X Hk ζ 3 (3) − ζ(9) − = ; 3 6 k k 3 k=1 k=1  2 (4) ∞ ∞ (4) Hk X X Hk ζ 3 (4) − ζ(12) 493 − = = ζ(12), 4 8 k k 3 5528 k=1

k=1

k=1

(2.102)

(2.103)

(2.104)

110

Chapter 2. Generating Functions and Powerful Identities

where we used ζ 3 (2) = respectively.

2.5.4

35 8 ζ(6)

and ζ 3 (4) =

7007 5528 ζ(12)

given in (1.63) and (1.70)

Third Application ∞ X k=1

(q)

(p)

XH Hk −p −p q−1 k = (1 − 2 )ζ(q)ζ(p) + (2 − 2 ) (2k + 1)p kq k=1

∞ (p) X (−1)k Hk . −2q−1 kq n=1

(2.105)

Proof. We follow the same technique as in [31]: Let ak =

1 (2k−1)p

(q)

and bk = Hk − ζ(q) in (2.95), n (q) X H − ζ(q) k

(2k − 1)p

k=1

=

(Hn(q)

− ζ(q))

n X i=1

k n−1 X X 1 1 − p (2i − 1) (2i − 1)p i=1

!

  (q) (q) Hk+1 − Hk .

k=0

Let n → ∞ and write lim Hn(q) = ζ(q), n→∞

∞ (q) X H − ζ(q) k

k=1

1)p

(2k −

( use

k X i=1

=0−

∞ k X X i=1

k=0

1 (2i − 1)p

!



(q)

) 1 (p) −p (p) = H2k − 2 Hk given in (1.148) (2i − 1)p   1 (q) (q) and Hk+1 − Hk = (k + 1)q ∞ ∞ (p) (p) X X Hk H2k = 2−p − (k + 1)q (k + 1)q k=0

k=0

{shift both indexes by −1} = 2−p

(p) ∞ X Hk−1 k=1

−p

=2

(q)

Hk+1 − Hk

(p) ∞ X H − k

k=1

kq

1 kp

kq −

(p) ∞ X H2k−2

kq k=1 1 ∞ H (p) − X 2k (2k)p kq k=1

1 (2k−1)p



2.5. Identities by Abel’s Summation

= 2−p

111

∞ (p) X H k kq

k=1

∞ (p) X H k=1

2k kq

+

∞ X k=1

1 k q (2k − 1)p

{use (1.5) for the second sum} = 2−p

∞ (p) X H k nq

k=1

− 2q−1

∞ (p) X H k kq

k=1

= 2−p − 2q−1

∞ (p) X H k=1

k kq

− 2q−1

∞ (p) X (−1)k H k

kq

k=1

− 2q−1

∞ (p) X (−1)k H k

kq

k=1

+

+

∞ X k=1

∞ X k=1

1 k q (2k − 1)p

1 . (2.106) − 1)p

k q (2k

On the other hand, ∞ (q) X H − ζ(q) k

k=1

(2k − 1)p

=

∞ X k=1

(q)

X Hk 1 − ζ(q) p (2k − 1) (2k − 1)p k=1

{seperate the first term of the first sum} =1+

∞ X k=2

(q)

X Hk 1 − ζ(q) p (2k − 1) (2k − 1)p k=1

{shift both indexes by +1} =1+

∞ X k=1

=1+

=1+

∞ X k=1

∞ X

k=1 (q) Hk (2k + 1)p

(q)

∞ X Hk+1 1 − ζ(q) (2k + 1)p (2k + 1)p k=0

(q) Hk

1 + (k+1) q p (2k + 1)

+

∞ X k=1

− ζ(q)

∞ X k=0

1 (2k + 1)p ∞

X 1 1 − ζ(q) q p (k + 1) (2k + 1) (2k + 1)p k=0

{shift the index of the second sum by −1} ∞ X

(q)

X X Hk 1 1 + − ζ(q) p q p (2k + 1) k (2k − 1) (2k + 1)p k=1 k=2 k=0 ( ) ∞ ∞ X X 1 1 use the fact that =1+ k q (2k − 1)p k q (2k − 1)p

=1+

k=1

k=2

{and recall the result of the latter sum from (1.85)} =

∞ X k=1

(q)

X Hk 1 + − ζ(q)(1 − 2−p )ζ(p). (2k + 1)p k q (2k − 1)p k=1

Combining (2.106) and (2.107) completes the proof.

(2.107)

112

Chapter 2. Generating Functions and Powerful Identities

2.6

Identities By Fourier Series

2.6.1

Fourier Series

Let f (x) be a function with a period of 2p and integrable on the interval [−p, p]. Then its Fourier series is given by ∞ X

f (x) = a0 +

 an cos

n=1

nπx p

 +

∞ X

 bn sin

n=1

nπx p

 ,

(2.108)

where Z p 1 f (x)dx, a0 = 2π −p     Z Z 1 p nπx 1 p nπx an = f (x) cos dx, bn = f (x) sin dx. π −p p π −p p

py π

Proof. Suppose f series, f



is a 2π–periodic function and expand it in cosine and sine

 py  π

=

∞ X

An cos(ny) +

n=0

∞ X

Bn sin(ny),

n=0

where An and Bn are the coefficients of the two series. Separate the first term of both series and use cos(0) = 1 and sin(0) = 0, f

 py 

∞ X

= A0 +

π

An cos(ny) +

n=1

∞ X

Bn sin(ny).

(2.109)

n=1

To find A0 , integrate both sides of (2.109) from y = −π to π, Z

π

f

 py  π

−π

Z

π

dy =

Z A0 dy +

−π

π

∞ X

Z An cos(ny)dy +

−π n=1

π

∞ X

Bn sin(ny)dy

−π n=1

{interchange integration and summation} π Z π Z π ∞ ∞ X X = A0 y + An cos(ny)dy + An sin(ny)dy. −π

n=1

−π

n=1

−π

Since ( Ra 2 0 f (x)dx if f (x) is even function (f (−x) = f (x)), f (x)dx = 0 if f (x) is odd function (f (−x) = −f (x)) −a (2.110)

Z

a

2.6. Identities By Fourier Series

113

and cos(ny) is even an function and sin(ny) is an odd function, we have Z

π

f

 py  π

−π

dy = 2πA0 + 2

∞ X

Z

π

An

n=1

cos(ny)dy 0

∞ X

π sin(ny) = 2πA0 + 2 = 2πA0 + 0. An n 0 n=1 Divide both sides by 2π, A0 =

1 2π

Z

π

f

 py  π

−π

dy.

(2.111)

To find An , multiply both sides of (2.109) by cos(ny) then integrate from y = −π to π, Z π   py cos(ny)dy f π −π Z π Z π Z π ∞ ∞ X X = A0 cos(ny)dy + An cos2 (ny)dy + Bn sin(ny) cos(ny)dy −π

−π

n=1

n=1

−π

{the last integral is 0, since the integrand is an odd function} π π ∞ X 2ny + sin(2ny) sin(ny) + An = A0 n 4n −π

=

−π

n=1 ∞ X

2A0 sin(nπ) An + n n=1



sin(2nπ) +π 2n



{write sin(nπ) = sin(2nπ) = 0, since n is an integer} ∞ X =0+π An . n=1

Divide both sides by π, An =

1 π

Z

π

f −π

 py  π

cos(ny)dy.

(2.112)

To find Bn , multiply both sides of (2.109) by sin(ny) then integrate from x = −π to π, Z π   py f sin(ny)dy π −π Z π Z π Z π ∞ ∞ X X = A0 sin(ny)dy + An cos(ny) sin(ny)dy + Bn sin2 (ny)dy −π

n=1

−π

n=1

−π

114

Chapter 2. Generating Functions and Powerful Identities

{the first two integrals are 0, since their integrand is an odd function} π   ∞ ∞ ∞ X X X sin(2nπ) 2ny − sin(2ny) = B π − = π Bn , = Bn n 4n 2n −π

n=1

n=1

n=1

where we used sin(2nπ) = 0 for integer n. Divide both sides by π, Z 1 π  py  f sin(ny)dy. Bn = π −π π

(2.113)

Plugging the results from (2.111), (2.112), and (2.113) in (2.109) yields f

 py  π

1 = 2π

Z

py π

 py 

f

π

−π

+ Substitute

π

dy +

∞  Z X 1

π

n=1

π

f

 py  π 

−π

 cos(ny)dy cos(ny)

Z ∞  X 1 π  py  − sin(ny)dy sin(ny). f π −π π n=1

(2.114)

= x in (2.114),

1 f (x) = 2p |

Z

∞  Z X 1

p

f (x)dx + p {z } n=1 |

−p

+

ao ∞  X n=1 |

= a0 +

1 p

Z

p

p



−p

f (x) cos {z



nπx p





∞ X

an

f (x) sin −p {z bn

∞ X n=1

 an cos

nπx p

nπx p

+







dx cos }

nπx p



   nπx dx sin p }  bn sin

n=1

nπx p

 .

   Since we assumed f py has a period of 2π and so f py = f πp (y + 2π) = π π  py f py π + 2p , and since we substituted π = x, we have f (x) = f (x + 2p), which indicates that f (x) has a period of 2p and the proof is finished.

2.6.2

Fourier Series of Even Function

Let f (x) be an even function with a period of 2p and integrable on the interval [−p, p]. Then its Fourier series is given by f (x) = a0 +

∞ X n=1

 an cos

nπx p

 ,

(2.115) (continued)

2.6. Identities By Fourier Series

115

where a0 =

1 π

Z

p

f (x)dx,

an =

0

2 π

p

Z

 f (x) cos

0

nπx p

 dx.

Proof. Let’s recall the definitions of a0 , an , and bn in (2.108): Z p 1 f (x)dx a0 = 2π −p {the integrand is an even function} Z 1 p = f (x)dx. π 0 1 an = π

p

Z



nπx p

f (x) cos −p

 dx

{the integrand is an even function}   Z 2 p nπx dx. = f (x) cos π 0 p

bn =

1 π

Z

p

 f (x) sin

−p

nπx p

 dx

{the integrand is an odd function} = 0. The proof follows on plugging a0 , an , and bn in (2.108).

2.6.3

Fourier Series of Odd Function

Let f (x) be an odd function with a period of 2p and integrable on the interval [−p, p]. Then its Fourier series is given by f (x) =

∞ X

 bn sin

n=1

where

2 bn = π

Z

p

 f (x) sin

0

Proof. We follow the previous approach:

nπx p



nπx p



,

dx.

(2.116)

116

Chapter 2. Generating Functions and Powerful Identities

1 2π

a0 =

Z

p

f (x)dx −p

{the integrand is an odd function} = 0.

an =

1 π

p

Z



nπx p

f (x) cos −p

 dx

{the integrand is an odd function} = 0.

bn =

1 π

Z

p

 f (x) sin

−p

nπx p

 dx

{the integrand is an even function}   Z 2 p nπx = dx, f (x) sin π 0 p and the proof completes on plugging a0 , an , and bn in (2.108).

2.6.4

Fourier Series of cos(zx)

The following identity holds: " # ∞ X 2z sin(πz) 1 (−1)n cos(nx) cos(zx) = − , π 2z 2 n=1 n2 − z 2

z∈ / Z.

(2.117)

Proof. Since cos(zx) is an even function, we recall (2.115) f (x) = a0 +

∞ X

 an cos

n=1

where a0 =

1 π

Z

p

f (x)dx,

an =

0

2 π

nπx p



p

Z

,

 f (x) cos

0

nπx p

 dx.

Since cos(zx) = cos(zx + 2π), which indicates that the period of the function is 2π and so p = π, its Fourier expansion is given by cos(zx) = a0 +

∞ X n=1

an cos(nx).

(2.118)

2.6. Identities By Fourier Series

117

Let’s find a0 and an : 1 a0 = π

Z

π

sin(πz) . πz

cos(zx)dx = 0

Z 2 π cos(zx) cos(nx)dx an = π 0 {make use of 2 cos(x) cos(y) = cos(x − y) + cos(x + y)} Z 1 π = [cos((z − n)x) + cos((z + n)x)]dx π 0  π 1 sin((z − n)x) sin((z + n)x) + = π z−n z+n  0 1 sin((z − n)π) sin((z + n)π) = + π z−n z+n {use sin(x ± y) = sin(x)cos(y) ± cos(x) sin(y)}   2 n cos(πz) sin(πn) − z sin(πz) cos(πn) = π n2 − z 2 {write cos(nπ) = (−1)n and sin(nπ) = 0, since n is an integer} =−

2(−1)n z sin(πz) . π(n2 − z 2 )

Substituting the results of a0 and an in (2.118) completes the proof.

2.6.5

Fourier Series of sin(zx)

The following equality holds: sin(zx) = −

∞ 2 sin(πz) X (−1)n n sin(nx) , π n2 − z 2 n=1

z∈ / Z.

(2.119)

Proof. Since sin(zx) is an odd function, we recollect (2.116) f (x) =

∞ X n=1

 bn sin

nπx p

 ,

2 bn = π

Z

p

 f (x) sin

0

nπx p

 dx.

Since sin(zx) = sin(zx + 2π), which indicates that the period of the function is 2π and so p = π, its Fourier expansion is given by sin(zx) =

∞ X n=1

bn sin(nx).

(2.120)

118

Chapter 2. Generating Functions and Powerful Identities

Let’s find bn : Z 2 π sin(zx) sin(nx)dx bn = π 0 {make use of 2 sin(x) sin(y) = cos(x − y) − cos(x + y)} Z 1 π = [cos((z − n)x) − cos((z + n)x)]dx π 0  π 1 sin((z − n)x) sin((z + n)x) = − π z−n z+n  0 1 sin((z − n)π) sin((z + n)π) − = π z−n z+n {use sin(x ± y) = sin(x)cos(y) ± cos(x) sin(y)}   2 z cos(πz) sin(πn) − n sin(πz) cos(πn) = π n2 − z 2 {write cos(nπ) = (−1)n and sin(nπ) = 0 for integer n} =−

2(−1)n n sin(πz) . π(n2 − z 2 )

Plugging bn in (2.120) yields the proof.

2.6.6

Fourier Series of ln(sin x)

For 0 < x < π, we have ln(sin x) = − ln(2) −

∞ X cos(2nx) . n n=1

(2.121)

Proof (i). Since ln | sin x| is an even function and has a period of π as ln | sin x| = ln | sin(x + π)| and so p = π/2. Thus, based on (2.115), its Fourier expansion is given by ∞ X ln | sin x| = a0 + an cos(2nx), (2.122) n=1

where a0 =

2 π

Z

π 2

ln | sin x|dx,

an =

0

4 π

Z

π 2

ln | sin x| cos(2nx)dx. 0

We have 2 a0 = π

Z

π 2

ln | sin x|dx 0

2.6. Identities By Fourier Series

119

{note that ln | sin x| = ln(sin x) for 0 < x < π} Z π2 2 = ln(sin x)dx π 0 {this integral is given in (3.107)}  2 π − ln(2) = − ln(2) = π 2 and an =

4 π

Z

π 2

Z π2 4 ln(sin x) cos(2nx)dx π 0 Z π2 1 − sin(2nx) cot(x)dx 2n 0

ln | sin x| cos(2nx)dx = 0

π2 4 = sin(2nx) ln(sin x) 2πn 0 {z } |

IBP

0

{the latter integral is given in (3.106)} 1 π π =− =− . 2n 2 4n Plug a0 and an in (2.122), ln | sin x| = − ln(2) −

∞ X cos(2nx) . n n=1

The proof follows on noticing that ln | sin x| = ln(sin x) for 0 < x < π. Proof (ii). By considering the real parts of Euler’s formula in (1.16), we have cos(x) = Reix . Therefore, ∞ ∞ ∞ X X X e2inx (e2ix )n cos(2nx) =R =R n n n n=1 n=1 n=1  2ix = −R ln 1 − e

{write e2ix = cos(2x) + i sin(2x)} 

= −R ln (1 − cos(2x) − i sin(2x))  y 1 use ln(x + iy) = ln(x2 + y 2 ) + i arctan , x > 0 given in (1.15) 2 x 

= −R

{since 1 − cos(2x) > 0 for 0 < x < π}   1 − sin(2x) ln[(1 − cos(2x))2 + sin2 (2x)] + i arctan 2 1 − cos(2x)

120

Chapter 2. Generating Functions and Powerful Identities

{use 1 − cos(2x) = 2 sin2 x and sin(2x) = 2 sin x cos x}    −2 sin x cos x 1 ln[4 sin4 x + 4 sin2 x cos2 x] + i arctan = −R 2 2 sin2 x   1 = −R ln[4 sin2 x(sin2 x + cos2 x)] + i arctan (− cot x) 2 n  π  π o note that arctan(− cot x) = arctan − tan −x =− −x 2 2  π  1 ln(2 sin x)2 − i −x = −R 2 2 n π o = R − ln(2 sin x) + i −x (2.123) 2 = − ln(2 sin x) = − ln(2) − ln(sin x), and the proof is complete. Further, by considering the imaginary parts in (2.123), we see that ∞ ∞ X X π e2inx sin(2nx) −x=J = . 2 n n n=1 n=1

Replace x by x/2 ∞ X sin(nx) π x = − . n 2 2 n=1

(2.124)

The latter identity is very useful. To show that, integrate both sides, −

∞ X cos(nx) π x2 = x − + C. n2 2 4 n=1

To find the constant, C, set x = 0, C=− Therefore,

∞ X 1 π2 = −ζ(2) = − . n2 6 n=1

∞ X cos(nx) x2 π π2 = − x + , n2 4 2 6 n=1

0 ≤ x ≤ 2π.

(2.125)

Integrating the latter identity from x = 0 to x gives ∞ X sin(nx) x3 π 2 π2 = − x + x, n3 12 4 6 n=1

and so on.

0 ≤ x ≤ 2π,

(2.126)

2.6. Identities By Fourier Series

2.6.7

121

Fourier Series of ln(cos x)

For |x|
0 given in (1.15) use ln(x + iy) = ln(x2 + y 2 ) + i arctan 2 x

 = −R

{since 1 + cos(2x) > 0 for |x| < π/2}   1 sin(2x) ln[(1 + cos(2x))2 + sin2 (2x)] + i arctan 2 1 + cos(2x)

{use 1 + cos(2x) = 2 cos2 x and sin(2x) = 2 sin x cos x}    1 2 sin x cos x = −R ln[4 cos4 x + 4 sin2 x cos2 x] + i arctan 2 2 cos2 x   1 = −R ln[4 cos2 x(cos2 x + sin2 x)] + i arctan (tan x) 2   1 = −R ln(2 cos x)2 + ix 2 = R {− ln(2 cos x) − ix} = − ln(2 cos x) = − ln(2) − ln(cos x).

(2.128)

122

2.6.8

Chapter 2. Generating Functions and Powerful Identities

Fourier Series of ln(tan x)

For 0 < x
1, q 6= p, and p + q is even > 6, there ∞ (p) X Hn does not exist a closed form for the series . nq n=1

238

4.3

Chapter 4. Harmonic Series

Alternating Harmonic Series

4.3.1

P∞

n=1

(−1)n Hn n

Show that

Solution

∞ X (−1)n Hn 1 1 = ln2 (2) − ζ(2). n 2 2 n=1

(4.135)

Set a = 0 in (3.90),

Z 1 ∞ X (−1)n Hn ln(1 + x) =− dx n x(1 + x) 0 n=1   Z 12 y 12 x= 1−y ln(1 − y) 1 = − dy = Li2 (y) 0 = − Li2 , y 2 0  and the solution finalizes on using Li2 12 = 12 ζ(2) − 12 ln2 (2) given in (1.120). For a different approach, set x = −1 in (2.7) and use Li2 (−1) = − 12 ζ(2).

4.3.2

P∞

n=1

Show that

Solution (i)

(−1)n H2n n

∞ X 5 1 (−1)n H2n = − ζ(2) + ln2 (2). n 8 4 n=1

(4.136)

We have ∞ ∞ X X (−1)n H2n (−1)n H2n =2 n 2n n=1 n=1



Hn let an = in the series tramsformation in (1.9) n ∞ n X i Hn = 2R n n=1 {set x = i in the generating function in (2.7)}   1 2 = 2 R Li2 (i) + ln (1 − i) . 2

The values of these two terms are given in (1.108) and (1.25).



4.3. Alternating Harmonic Series

239

Set x = 1 in (2.42):

Solution (ii)

∞ X

(−1)n

n=1

2H2n − Hn 2n x = −2 arctan2 (x), n

we get ∞ X

2H2n − Hn = −2 arctan2 (1) n n=1  π 2 3 π2 = −2 = − ζ(2). =− 4 8 4 (−1)n

Substitute the result from (4.135) to finalize the solution.

4.3.3

P∞

n=1

Show that

Solution (i)

(−1)n Hn n2

∞ X (−1)n Hn 5 = − ζ(3). 2 n 8 n=1

(4.137)

Setting x = −1 in (2.8),

∞ X (−1)n Hn 1 = Li3 (−1) − Li3 (2) + ln(2) Li2 (2) + ln(−1) ln2 (2) + ζ(3). 2 n 2 n=1

The values of Li3 (−1), Li3 (2), and Li2 (2) are given in (1.103), (1.141), and (1.140). Solution (ii)

Replace x by −x in (2.7), ∞ X (−)n Hn n 1 x = Li2 (−x) + ln2 (1 + x). n 2 n=1

Divide both sides by x then integrate using

R1 0

xn−1 dx =

1 n,

Z 1 Z ∞ X 1 1 ln2 (1 + x) (−1)n Hn Li2 (−x) = dx + dx n2 x 2 0 x 0 n=1 Z 1 1 ln2 (1 + x) = Li3 (−1) + dx. 2 0 x This integral is calculated in (3.38) and Li3 (−1) = − 34 ζ(3) given in (1.103). An alternative solution may be found in [28, pp. 508–509] and (4.22).

240

Chapter 4. Harmonic Series

P∞

4.3.4

n=1

Show that

(−1)n H2n n2

∞ X 23 (−1)n H2n = ζ(3) − πG. 2 n 16 n=1

(4.138)

Solution ∞ ∞ X X (−1)n H2n (−1)n H2n = 4 2 n (2n)2 n=1 n=1   Hn let an = 2 in (1.9) n ∞ n X i Hn = 4R n2 n=1

{set x = i in the generating function in (2.8)}  = 4 R Li3 (i) − Li3 (1 − i) + ln(1 − i) Li2 (1 − i) +

 1 ln(i) ln2 (1 − i) + ζ(3) . 2

By using the values in (1.122), (1.25), and (1.26), we find:  3  π2 π π 2 2 ln(i) ln (1 − i) = ln(2) − − ln (2) i; (4.139) 8 32 8   π π2 1 π3 π 2 ln(1 − i) Li2 (1 − i) = − G − ln(2) − ln(2)G + + ln (2) i. 4 32 2 64 8 (4.140) Collect the values from (1.109), (4.139), and (4.140) to finish the solution.

4.3.5

P∞

n=1

Show that

Solution (i)

(2)

(−1)n Hn n

∞ (2) X (−1)n Hn 1 = ln(2)ζ(2) − ζ(3). n 2 n=1

(4.141)

Setting x = −1 produces

− ln(2) Li2 (−1) = 2

∞ ∞ (2) X (−1)n Hn X (−1)n Hn + − 3 Li3 (−1). n2 n n=1 n=1

The solution completes on collecting the values from (4.137), (1.102), and (1.103).

4.3. Alternating Harmonic Series

Solution (ii)

241

Put a = 1 in (3.92),  Z 1 ∞ (2) X ln(x) ln 1+x (−1)n Hn 2 =− dx n 1−x 0 n=1 Z 1 Z 1 ln(x) ln(1 + x) ln(x) =− dx + ln(2) dx. 1 − x 1 −x 0 0

These two integrals are given in (3.125) and (3.2).

P∞

4.3.6

n=1

(3)

(−1)n Hn n

Show that

∞ (3) X (−1)n Hn 3 19 = ln(2)ζ(3) − ζ(4). n 4 16 n=1

(4.142)

Setting x = −1 in (2.17),

Solution

∞ (3) X (−1)n Hn 1 = Li4 (−1) − ln(2) Li3 (−1) − Li22 (−1). n 2 n=1

The values of these polylogarithm functions are given (1.104), (1.103), and (1.102). For another solution, see (4.81).

P∞

4.3.7

n=1

(−1)n Hn n3

Show that   ∞ X (−1)n Hn 1 11 7 1 1 = 2 Li4 − ζ(4)+ ln(2)ζ(3)− ln2 (2)ζ(2)+ ln4 (2). 3 n 2 4 4 2 12 n=1 (4.143) Put a = 3 in (3.90),

Solution

1 = 2

Z |0

1 2

Z ∞ X (−1)n Hn 1 1 ln2 (x) ln(1 + x) =− dx n3 2 0 x(1 + x) n=1   Z 1 2 y y x= 1−y 1 2 ln 1−y ln(1 − y) = dy 2 0 y Z 1 Z 12 ln2 (y) ln(1 − y) 1 2 ln3 (1 − y) ln(y) ln2 (1 − y) dy + dy − dy . y 2 0 y y {z } | {z } |0 {z } I1

I2

I3

242

Chapter 4. Harmonic Series

For I1 + I2 , integrate I1 by parts and let 1 − y = x in I2 , I1 + I2 =

1 1 4 ln (2) + 3 3

1 2

Z 0

ln3 (x) dx + 1−x

1

Z

1

|2

ln3 (x) dx 1−x {z }

R1 0

Z

1

3

Z

1 2

R 1/2

0

3

1 4 ln (x) 2 ln (x) ln (2) + dx − dx 3 1 − x 3 1−x 0 0 {collect the results from (3.4) and (3.34)}   2 7 1 = 4 Li4 − 6ζ(4) + ln(2)ζ(3) − ln2 (2)ζ(2) + ln4 (2). 2 2 3 =

For I3 , IBP

I3 =

1−y→y

=

1 4 ln (2) + 2

Z

1 4 ln (2) + 2

1 2

0

Z

1 1 2

ln2 (y) ln(1 − y) dy 1−y ln(y) ln2 (1 − y) dy y

) ln(y) ln2 (1 − y) dy to both sides then divide by 2 add I3 := y 0 Z 1 Z 1 1 2 ln(y) ln2 (1 − y) 1 1 ln(y) ln2 (1 − y) = ln4 (2) + dy + dy 4 2 0 y 2 21 y Z 1 1 1 ln(y) ln2 (1 − y) = ln4 (2) + dy 4 2 0 y | {z } (

Z

1 2

1−y=x

Z 1 1 1 ln(1 − x) ln2 (x) = ln4 (2) + dx 4 2 0 1−x {set a = 2 in (3.87) to get the latter integral} ∞ X 1 4 Hn = ln (2) + 2ζ(4) − 2 2 n3 n=1 {substitute the result from (4.5)} 1 1 = ln4 (2) − ζ(4). 4 4 Gather the three integrals to complete the solution. A different method is by putting x = −1 in (2.15). For another approach, check [29].

Remark: For odd q > 3, there does not exist a closed form for

∞ X (−1)n Hn . nq n=1

4.3. Alternating Harmonic Series (2)

P∞

4.3.8

243

n=1

(−1)n Hn n2

Show that   ∞ (2) X 1 51 7 1 (−1)n Hn = −4 Li4 + ζ(4)− ln(2)ζ(3)+ln2 (2)ζ(2)− ln4 (2). 2 n 2 16 2 6 n=1 (4.144) Put x = −1 in (2.83),

Solution

∞ ∞ (2) X X (−1)n Hn (−1)n Hn 1 2 (−1) − 2 = 3 Li (−1) + Li . 4 2 2 n 2 n3 n=1 n=1

Substituting the results from (4.143), (1.102), and (1.104) completes the solution. Also Check [28, pp. 505–506] for a different method.

P∞

4.3.9

n=1

2 (−1)n Hn n2

Show that   ∞ X (−1)n Hn2 41 7 1 1 1 = 2 Li − ζ(4)+ ln(2)ζ(3)− ln2 (2)ζ(2)+ ln4 (2). 4 2 n 2 16 4 2 12 n=1 (4.145) Replace x by −x in (2.6),

Solution

∞ X

(−1)n

n=1

1 Hn−1 n x = ln2 (1 + x). n 2

then integrating from x = 0 to 1, Multiplying both sides by − ln(1−x) x −

1 2

1

Z 0

=

Z ∞ X ln2 (1 + x) ln(1 − x) (−1)n Hn−1 1 dx = −xn−1 ln(1 − x)dx x n 0 n=1 ∞ X

(−1)n

n=1

Hn − n

1 n

{make use of (2.70)}   X ∞ ∞ Hn (−1)n Hn2 X (−1)n Hn = − . n n2 n3 n=1 n=1

Rearranging the terms, Z ∞ ∞ X X (−1)n Hn2 (−1)n Hn 1 1 ln2 (1 + x) ln(1 − x) = − dx. n2 n3 2 0 x n=1 n=1

244

Chapter 4. Harmonic Series

The values of these two terms are computed in (3.117) and (4.143). Also check [28, pp. 506–508] for a different solution. (2)

P∞

4.3.10

n=1

(−1)n Hn Hn n

Show that   ∞ (2) X (−1)n Hn Hn 7 1 1 = −2 Li4 + ζ(4) − ln(2)ζ(3) + ln2 (2)ζ(2) n 2 8 4 n=1 −

Solution

1 4 ln (2). 12

(4.146)

Multiply both sides of (2.72): Z

(2)

1

xn−1 ln3 (1 − x)dx = −

0

(3)

Hn3 + 3Hn Hn + 2Hn n

by (−1)n−1 then take the summation, ∞ ∞ ∞ (2) (3) X X X (−1)n Hn Hn (−1)n Hn (−1)n Hn3 +3 +2 n n n n=1 n=1 n=1 Z 1 Z ∞ 1 X ln3 (1 − x) = (−x)n−1 dx = dx ln3 (1 − x) 1+x 0 0 n=1

{substitute the result from (3.29)}   1 = −6 Li4 . 2 On the other hand, replace x by −x in (2.22), ∞ X

  ln3 (1 + x) . (−1)n Hn3 − 3Hn Hn(2) + 2Hn(3) xn = − 1+x n=1 Divide both sides by x then integrate using

R1 0

xn−1 dx =

1 n,

∞ ∞ ∞ (2) (3) X X X (−1)n Hn3 (−1)n Hn Hn (−1)n Hn −3 +2 n n n n=1 n=1 n=1 Z 1 3 Z 1 3 Z 1 3 ln (1 + x) ln (1 + x) ln (1 + x) =− dx = dx − dx x(1 + x) 1+x x 0 0 0

{the first integral is ln4 (2)/4 and the second one is given in (3.39)}

(4.147)

4.3. Alternating Harmonic Series

245

  1 21 3 1 = 6 Li4 − 6ζ(4) + ln(2)ζ(3) − ln2 (2)ζ(2) + ln4 (2). 2 4 2 2

(4.148)

Take the difference of (4.147) and (4.148) then divide by 6 to finalize the solution.

P∞

4.3.11

n=1

3 (−1)n Hn n

Show that ∞ X (−1)n Hn3 5 9 3 1 = − ζ(4) + ln(2)ζ(3) − ln2 (2)ζ(2) + ln4 (2). (4.149) n 8 8 4 4 n=1

Solution

∞ ∞ (3) X X (−1)n Hn3 (−1)n Hn +4 n n n=1 n=1

= −6ζ(4) +

21 3 1 ln(2)ζ(3) − ln2 (2)ζ(2) + ln4 (2). 4 2 2

The remaining sum is calculated in (4.142).

P∞

4.3.12

n=1

Show that

(−1)n Hn n4

∞ X (−1)n Hn 1 59 = ζ(2)ζ(3) − ζ(5). 4 n 2 32 n=1

(4.150)

Putting a = 3 in (3.90) gives

Solution

Z ∞ X (−1)n Hn 1 1 ln3 (x) ln(1 + x) dx. = n4 6 0 x(1 + x) n=1

(4.151)

Let’s calculate the integral: Z 0

1

ln3 (x) ln(1 + x) dx = x(1 + x) Z = 0

3

Z 0

ln3 (x) ln(1 + x) dx − x(1 + x)

ln (x) ln(1 + x) dx + x(1 + x)

Z 0

1

3

Z |1

ln3 (x) ln(1 + x) dx x(1 + x) {z }

ln (x) ln(1 + x) dx − 1+x

x→1/x

Z 0

1

ln4 (x) dx. 1+x

246

Chapter 4. Harmonic Series

R1 3 R 1 3 ln(1+x) ln3 (x) ln(1+x) dx := 0 ln (x) ln(1+x) dx − 0 ln (x)1+x dx to both x(1+x) x R 1 ln3 (x) ln(1+x) notice 0 dx nicely cancels out from both sides, we obtain 1+x R1 0

sides

ln3 (x) ln(1 + x) dx x(1 + x) 0 Z 1 4 Z 1 3 ln3 (x) ln(1 + x) ln (x) ln (x) ln(1 + x) dx − dx + dx x(1 + x) 1 + x x 0 0 {z } | 1

Z

2

Z = 0

IBP

Z

3

5 ln (x) ln(1 + x) dx − x(1 + x) 4 {z }

= |0

Z 0

1

4

ln (x) dx. 1+x

I

To evaluate I, make the substitution x = 1−t t ,   Z 1 4 Z 1 3 Z 1 ln3 t ln(t) 1−t ln (t) ln (t) ln(1 − t) dt = dt − 3 dt I= 1−t 1−t 0 1−t 0 0 Z 1 2 Z 1 ln(t) ln3 (1 − t) ln (t) ln2 (1 − t) +3 dt − dt 1−t 1−t 0 0 | {z } | {z } IBP

ln4 (t) dt − 3 1−t

IBP

ln3 (t) ln(1 − t) = dt 1−t 0 0 Z Z 1 3 1 1 ln4 (1 − t) ln (1 − t) ln(t) dt − dt +2 t 4 0 t 0 {z } {z } | | =

Z

1

3 4

Z 0

1

1−t→t 4

ln (t) dt − 1−t

Substitute the result of I back, Z

Z

1

Z

1

1−t→t

0

3

ln (t) ln(1 − t) dt. 1−t

ln3 (x) ln(1 + x) dx x(1 + x) 0 Z Z Z 3 1 ln4 (x) 5 1 ln4 (x) 1 1 ln3 (x) ln(1 − x) = dx − dx − dx 8 0 1−x 8 0 1+x 2 0 1−x {gather the results from (3.5), (3.11), and (3.124)} 177 = 3ζ(2)ζ(3) + ζ(5). 16 1

Plugging this integral in (4.151) finalizes the solution. For another method, see [29].

4.3. Alternating Harmonic Series (2)

P∞

4.3.13

247

n=1

(−1)n Hn n3

Show that

∞ (2) X 11 (−1)n Hn 5 = ζ(5) − ζ(2)ζ(3). 3 n 32 8 n=1

(4.152)

Replace x by −x in (2.83),

Solution

Li22 (−x)

∞ ∞ (2) X X (−1)n Hn n (−1)n Hn n =4 x +2 x − 6 Li4 (−x). 3 2 n n n=1 n=1

Divide both sides by x then integrate using 1

Z 0

R1 0

xn−1 dx =

1 n,

∞ ∞ (2) X X (−1)n Hn (−1)n Hn Li22 (−x) 45 dx = 4 + 2 + ζ(5). 4 3 x n n 8 n=1 n=1

Substitute the relation involving the integral from (3.153) then rearrange the terms, ∞ ∞ ∞ (2) X (−1)n Hn 45 5 7 X Hn X (−1)n Hn = ζ(5) + ζ(2)ζ(3) + − . n3 16 16 16 n=1 n4 n4 n=1 n=1

These two series are computed in (4.6) and (4.150).

4.3.14

P∞

n=1

2 (−1)n Hn n3

Show that     ∞ X (−1)n Hn2 1 1 19 11 − 4 ln(2) Li4 + ζ(5) + ζ(2)ζ(3) = −4 Li5 3 n 2 2 32 8 n=1 7 2 2 5 − ln2 (2)ζ(3) + ln3 (2)ζ(2) − ln (2). 4 3 15 Solution

(4.153)

Replace x by −x in (2.12) then multiply through by −1, ∞ X

  2 ln2 (1 + x) 2Hn 2 (2) + 2 xn−1 = − . (−1) Hn − Hn − n n 1+x n=1 n

Multiply both sides by ∞ X

1 2

ln2 (x) then integrate using

(−1)

n=1

n



Hn2

Hn(2)

1 2

R1 0

2Hn 2 − + 2 n n

xn−1 ln2 (x)dx =



1 n3



1 n3 ,

248

Chapter 4. Harmonic Series

=−

1 2

1

Z 0

ln2 (1 + x) ln2 (x) IBP 1 dx = 1+x 3

Z

1

0

ln(x) ln3 (1 + x) dx. x

Distribute then reorder the terms, ∞ ∞ ∞ (2) X X X (−1)n Hn2 (−1)n Hn (−1)n Hn = −2 Li (−1) + + 2 5 3 3 n n n4 n=1 n=1 n=1 Z 1 1 ln(x) ln3 (1 + x) + dx. 3 0 x

These terms are calculated in (1.105), (4.152), (4.150), and (3.123) respectively. Check [28, pp. 517–519] for an alternative approach.

4.3.15

P∞

n=1

(4)

(−1)n Hn n

Show that ∞ (4) X (−1)n Hn 7 3 = −2ζ(5) + ln(2)ζ(4) + ζ(2)ζ(3). n 8 8 n=1

Solution

(4.154)

Set a = 4 then replace x by −x in (2.2), ∞ X

(−1)n Hn(4) xn =

n=1

Divide both sides by x then integrate using

R1 0

Li4 (−x) . 1+x

xn−1 dx =

1 n,

Z 1 Z 1 Z 1 ∞ (4) X (−1)n Hn Li4 (−x) Li4 (−x) Li4 (−x) = dx = dx − dx n x 1+x 0 x(1 + x) 0 n=1 |0 {z } IBP

Z

1

ln(1 + x) Li3 (−x) = Li5 (−1) − ln 2 Li4 (−1) + dx x 0 Z 1 2 Li2 (−x) IBP = Li5 (−1) − ln(2) Li4 (−1) − Li2 (−1) Li3 (−1) + dx x 0 Z 1 2 15 7 3 Li2 (−x) = − ζ(5) + ln(2)ζ(4) − ζ(2)ζ(3) + dx 16 8 8 x 0 {recall the relation involving the latter integral from (3.153)} ∞ ∞ X 15 7 1 7 X Hn (−1)n Hn = − ζ(5) + ln(2)ζ(4) + ζ(2ζ(3) + + 2 . (4.155) 4 16 8 4 8 n=1 n n4 n=1

4.3. Alternating Harmonic Series

249

The solution finalizes on recalling the results from (4.6) and (4.150). For different approaches, see [28, p. 516] and (4.82).

(3)

(−1)n Hn n=1 n2

P∞

4.3.16

Show that

Solution

∞ (3) X (−1)n Hn 21 3 = ζ(5) − ζ(2)ζ(3). 2 n 32 4 n=1

(4.156)

Replace x by −x in (2.17),

∞ (3) X (−1)n Hn n 1 x = Li4 (−x) − ln(1 + x) Li3 (−x) − Li22 (−x). n 2 n=1

Divide both sides by x then integrate using

R1 0

xn−1 dx =

1 n,

∞ (3) X (−1)n Hn n2 n=1

Z = 0

1

Li4 (−x) dx − x

1

Z |0

ln(1 + x) Li3 (−x) 1 dx − x 2 {z }

Z

1

0

Li22 (−x) dx x

IBP

Z 3 1 Li22 (−x) = Li5 (−1) + Li2 (−1) Li3 (−1) − dx 2 0 x Z 3 15 3 1 Li22 (−x) = ζ(2)ζ(3) − ζ(5) − dx. 8 16 2 0 x The latter integral is given in (3.152). Also, check [28, pp. 513–515] for different methods for both series in (4.156) and (4.152).

4.3.17

(2)

(−1)n Hn Hn n=1 n2

P∞

Show that     ∞ (2) X (−1)n Hn Hn 1 1 23 7 = 4 Li + 4 ln(2) Li − ζ(5) + ln2 (2)ζ(3) 5 4 2 n 2 2 8 4 n=1 2 15 2 5 ln (2). − ln3 (2)ζ(2) − ζ(2)ζ(3) + 3 16 15

(4.157)

250

Chapter 4. Harmonic Series

Solution

Multiply both sides of (2.72): (2)

1

Z

xn−1 ln3 (1 − x)dx = −

0

(3)

Hn3 + 3Hn Hn + 2Hn n

n

by − (−1) then consider the summation, n ∞ ∞ ∞ (2) (3) X X X (−1)n Hn3 (−1)n Hn Hn (−1)n Hn + 3 + 2 n2 n2 n2 n=1 n=1 n=1 ! Z 1 3 ∞ ln (1 − x) X (−x)n =− dx x n 0 n=1 Z 1 3 ln (1 − x) ln(1 + x) = dx. x 0

(4.158)

On the other hand, replace x by −x in (2.22), ∞ X

  ln3 (1 + x) (−1)n Hn3 − 3Hn Hn(2) + 2Hn(3) xn = − . 1+x n=1 Multiply through by − ln(x) x then integrate using −

R1 0

xn−1 ln(x)dx =

1 n2 ,

∞ ∞ ∞ (2) (3) X X X (−1)n Hn3 (−1)n Hn Hn (−1)n Hn − 3 + 2 n2 n2 n2 n=1 n=1 n=1 Z 1 3 ln (1 + x) ln(x) dx = x(1 + x) 0 Z 1 3 Z 1 3 ln (1 + x) ln(x) ln (1 + x) ln(x) = dx − dx x 1+x 0 |0 {z } IBP

Z = 0

1

ln3 (1 + x) ln(x) 1 dx + x 4

Z 0

1

ln4 (1 + x) dx. x

(4.159)

Take the difference of the two relations in (4.158) and (4.159) then divide by 6, Z Z 1 4 ∞ (2) X (−1)n Hn Hn 1 1 ln3 (1 − x) ln(1 + x) 1 ln (1 + x) = dx + dx 2 n 6 x 24 x 0 0 n=1 Z 1 1 ln3 (1 + x) ln(x) + dx. 6 0 x These three integrals are given in (3.119), (3.40), and (3.123).

4.3. Alternating Harmonic Series

4.3.18

P∞

n=1

251

3 (−1)n Hn 2 n

Show that     ∞ X 1 (−1)n Hn3 1 9 27 = −6 Li − 6 ln(2) Li + ζ(5) + ζ(2)ζ(3) 5 4 2 n 2 2 4 16 n=1 − Solution

1 21 2 ln (2)ζ(3) + ln3 (2)ζ(2) − ln5 (2). 8 5

(4.160)

Combine (4.158) and (4.159) then divide by 2,

Z Z ∞ X (−1)n Hn3 1 1 ln3 (1 − x) ln(1 + x) 1 1 ln4 (1 + x) = dx − dx n2 2 0 x 8 0 x n=1 Z ∞ (3) X (−1)n Hn 1 1 ln3 (1 + x) ln(x) − dx − 2 . 2 0 x n2 n=1 These terms are given in (3.119), (3.40), (3.123), and (4.156) respectively. To see the two series in (4.152) and (4.160) computed differently, check [28, pp. 520–523].

252

4.4

4.4.1

Chapter 4. Harmonic Series

Harmonic Series with Powers of 2 in the Denominator P∞

Hn n=1 n2n

Show that

∞ X Hn 1 = ζ(2). n n2 2 n=1

(4.161)

Set x = 1/2 in (2.7),

Solution (i)

  ∞ X Hn 1 1 = Li + ln2 (2), 2 n n2 2 2 n=1 and the value of Li2 ( 12 ) is given in (1.120). Set a = 0 in (3.91),

Solution (ii)

Z 1 ∞ X 1 Hn ln(1 + x) 1 = dx = − Li2 (−x) 0 = − Li2 (−1) = ζ(2). n n2 x 2 0 n=1

4.4.2

P∞

Hn n=1 n2 2n

Show that ∞ X Hn 1 = ζ(3) − ln(2)ζ(2). 2 n n 2 2 n=1

Solution (i)

Set x = 1/2 in (2.8),   ∞ X 1 1 Hn = − ln(2) Li2 − ln3 (2) + ζ(3). 2 n n 2 2 2 n=1

Plug in the value of Li2 ( 12 ) given in (1.120). Solution (ii)

By integration by parts, Z − 0

1 2

xn−1 ln(x)dx =

ln(2) 1 + 2 n. n2n n 2

(4.162)

4.4. Harmonic Series with Powers of 2 in the Denominator

253

Multiply both sides by Hn then take the summation over n ≥ 1, Z 12 ∞ ∞ X X Hn ln(x) Hn + =− ln(2) n 2 2n n2 n x 0 n=1 n=1

∞ X

! Hn x

n

dx

n=1

{recall the generating function in (2.4)} Z 21 ln(x) ln(1 − x) dx = x(1 − x) 0 {make use of (3.86)} Z 1 ln(x) ln(1 − x) dx = x 0 Z 1 2 ln (x) IBP 1 = dx. 2 0 1−x Therefore,

Z ∞ ∞ X X Hn 1 1 ln2 (x) Hn = dx − ln(2) . 2 2n n n 2 1 − x n2 0 n=1 n=1

Gather the results from (3.3) and (4.161) to finish the solution. A different approach may be found in [28, p. 500].

4.4.3

P∞

(2)

Hn n=1 n2n

Show that

Solution (i)

∞ (2) X Hn 5 = ζ(3). n n2 8 n=1

Put x = 1/2 in (2.82) then reorganize the terms,     ∞ ∞ (2) X X Hn 1 1 Hn = 3 Li + ln(2) Li − 2 . 3 2 n n2 2 2 n2 2n n=1 n=1

Collect the values from (4.162), (1.120), and (1.132). Solution (ii)

Set a = 1 in (3.91), Z 1 ∞ (2) X Hn ln(1 − x) ln(1 + x) = − dx. n n2 x 0 n=1

This integral is given in (3.115).

(4.163)

254

Chapter 4. Harmonic Series 2 Hn n=1 n2n

P∞

4.4.4

Show that

Solution

∞ X 7 Hn2 = ζ(3). n n2 8 n=1

(4.164)

Divide both sides of (2.71): (2)

1

Z

xn−1 ln2 (1 − x)dx =

0

Hn Hn2 + n n

by 2n then consider the summation over n ≥ 1, Z 1 2 ∞ ∞ (2) X X Hn ln (1 − x) Hn2 + = n n n2 n2 x 0 n=1 n=1 1

Z = 0

∞  n X x n=1

!

2

dx

{use the geometric series formula}  x  Z 1 2 ln (1 − x) ln (y) 1−x=y 2 dy. dx = x x 1− 2 0 1+y 2

Rearrange the terms, Z 1 2 ∞ ∞ (2) X X ln (y) Hn2 Hn = dy − . n2n n2n 0 1+y n=1 n=1 Substituting the results from (3.9) and (4.163) completes the solution. An alternative solution is by setting x = 1/2 in the generating function in (2.14).

P∞

Hn n=1 n3 2n

4.4.5

Show that   ∞ X Hn 1 1 1 4 1 + ζ(4) − ln(2)ζ(3) + ln (2). = Li 4 3 n n 2 2 8 8 24 n=1

Solution

By integration by parts, Z 0

1 2

xn−1 ln2 (x)dx =

ln2 (2) 2 ln(2) 2 + 2 n + 3 n. n2n n 2 n 2

(4.165)

4.4. Harmonic Series with Powers of 2 in the Denominator

255

Multiply through by Hn then take the summation over n ≥ 1, ln2 (2)

∞ ∞ ∞ X X X Hn Hn Hn + 2 ln(2) + 2 n 2 2n 3 2n n2 n n n=1 n=1 n=1 ! Z 21 2 ∞ ln (x) X n = Hn x dx x 0 n=1

{recall the generating function in (2.4)} Z 21 2 ln (x) ln(1 − x) dx =− x(1 − x) 0 Z 1/2 2 Z 12 2 ln (x) ln(1 − x) ln (x) ln(1 − x) =− dx − dx x 1−x 0 |0 {z } IBP

1 1 = − ln4 (2) − 3 3

Z

1 2

0

ln3 (x) dx − 1−x

Z 0

1 2

ln2 (x) ln(1 − x) dx. 1−x

Reordering the terms, ∞ ∞ ∞ X X X Hn 1 4 1 2 Hn Hn = − ln (2) − ln (2) − ln(2) 3 n n n 2 6 2 n2 n2 2n n=1 n=1 n=1 Z 1 Z 1 1 2 ln2 (x) ln(1 − x) 1 2 ln3 (x) dx − dx. − 6 0 1−x 2 0 1−x

Grouping the results from (4.161), (4.162), (3.34), and (3.142) finalizes the solution. Check [28, pp. 500–501] for another solution.

4.4.6

P∞

(2)

Hn n=1 n2 2n

Show that   ∞ (2) X Hn 1 1 1 1 4 1 = Li4 + ζ(4) + ln(2)ζ(3) − ln2 (2)ζ(2) + ln (2). 2 n n 2 2 16 4 4 24 n=1 (4.166) Solution

let x = 1/2 in (2.83),     ∞ ∞ (2) X X Hn 1 1 2 1 Hn = 3 Li − Li − 2 , 4 2 2 n n 2 2 2 2 n3 2 n n=1 n=1

and the solution finishes on collecting the values from (4.165) and (1.120).

256

Chapter 4. Harmonic Series

4.4.7

2 Hn n=1 n2 2n

P∞

Show that   ∞ X 1 37 7 1 1 4 Hn2 = − Li4 + ζ(4) − ln(2)ζ(3) + ln2 (2)ζ(2) − ln (2). 2 2n n 2 16 4 4 24 n=1 (4.167) Solution

Divide both sides of (2.71): Z

(2)

1

xn−1 ln2 (1 − x)dx =

0

Hn2 Hn + n n

n

by n2 then take the summation over n ≥ 1, ! Z 1 2 ∞ ∞ ∞ (2) X X Hn ln (1 − x) X (x/2)n Hn2 + = dx n2 2n n=1 n2 2n x n 0 n=1 n=1 Z 1 2 ln (1 − x) ln(1 − x/2) dx =− x 0  Z 1 2 ln (y) ln 1+y 1−x=y 2 = − dy 1−y 0 {set a = 2 in (3.92)} ∞ (3) X (−1)n Hn = −2 . n n=1

Then, we have ∞ ∞ ∞ (2) (3) X X X Hn2 Hn (−1)n Hn = − − 2 . n2 2n n2 2n n n=1 n=1 n=1

These two sums are calculated in (4.166) and (4.142).

4.4.8

P∞

(3)

Hn n=1 n2n

Show that   ∞ (3) X Hn 1 5 7 1 1 4 = Li4 − ζ(4) + ln(2)ζ(3) − ln2 (2)ζ(2) + ln (2). n n2 2 16 8 4 24 n=1 (4.168)

4.4. Harmonic Series with Powers of 2 in the Denominator

257

Take x = 1/2 in the generating function in (2.17),

Solution (i)

      ∞ (3) X 1 1 1 2 1 Hn = Li4 + ln(2) Li3 − Li2 , n2n 2 2 2 2 n=1 where the values of Li3 ( 21 ) and Li2 ( 12 ) are given in (1.120) and (1.132). Put a = 2 in (3.91),

Solution (ii)

Z ∞ (3) X Hn 1 1 ln2 (1 − x) ln(1 + x) = dx, n2n 2 0 x n=1 and this integral is given in (3.118).

P∞

Hn n=1 n4 2n

4.4.9

Show that     ∞ X Hn 1 1 1 1 = 2 Li5 + ln(2) Li4 − ln3 (2)ζ(2) + ln2 (2)ζ(3) 4 n n 2 2 2 6 2 n=1 1 1 1 5 1 ln (2). − ln(2)ζ(4) − ζ(2)ζ(3) + ζ(5) + 8 2 32 40 Solution

(4.169)

By integration by parts, we have Z −

1 2

xn−1 ln3 (x)dx =

0

6 6 ln(2) 3 ln2 (2) ln3 (2) + + + . n4 2 n n3 2n n2 2n n2n

Multiply both sides by Hn then consider the summation over n ≥ 1, 6

∞ ∞ ∞ ∞ X X X X Hn Hn Hn Hn 2 3 + 6 ln(2) + 3 ln (2) + ln (2) 4 2n 3 2n 2 2n n n n n n2 n=1 n=1 n=1 n=1 ! Z 12 3 ∞ ln (x) X n = −Hn x dx x 0 n=1

{substitute the generating function in (2.4)} Z 21 3 ln (x) ln(1 − x) = dx x(1 − x) 0 Z 12 3 Z 12 3 ln (x) ln(1 − x) ln (x) ln(1 − x) = dx + dx x 1−x |0 {z } |0 {z } IBP

1−x→x

258

Chapter 4. Harmonic Series

1 1 = − ln5 (2) + 4 4

Z 0

1 2

ln4 (x) dx + 1−x

Z

1 1 2

ln3 (1 − x) ln(x) dx x

{recall the relation involving of the latter integral from (3.146)} Z 1 Z 93 1 5 1 2 ln4 (x) 1 1 ln3 (x) ln(1 − x) = − ζ(5) − ln (2) − dx + dx 16 10 2 0 1−x 2 0 1−x {plug in the results from (3.35) and (3.124)}     3 21 2 1 1 + 12 ln(2) Li4 + ζ(5) + ln (2)ζ(3) − 2 ln3 (2)ζ(2) = 12 Li5 2 2 16 4 2 −3ζ(2)ζ(3) + ln5 (2). 5 Put together the results from (4.161), (4.162), and (4.165) to finish the solution. For a different approach, see [28, pp. 501–502].

P∞

(4)

Hn n=1 n2n

4.4.10

The following sum is proposed by Cornel Vălean (see[33]):     ∞ (4) X Hn 21 81 1 1 = 6 Li5 + 6 ln(2) Li4 − ζ(5) − ζ(2)ζ(3) n n2 2 2 16 8 n=1 +

Solution

1 21 2 ln (2)ζ(3) − ln3 (2)ζ(2) + ln5 (2). 8 5

(4.170)

Put a = 3 in (3.91), we obtain Z ∞ (4) X Hn 1 1 ln3 (1 − x) ln(1 + x) dx. = − n2n 6 0 x n=1

This integral is calculated in (3.119) and the solution is complete.

4.4.11

P∞

(2)

Hn n=1 n3 2n

The following sum is proposed by Cornel Vălean (see [34]):     ∞ (2) X Hn 1 1 23 1 = −2 Li − 3 ln(2) Li + ζ(5) − ln(2)ζ(4) 5 4 3 n n 2 2 2 64 16 n=1 +

23 23 2 7 3 13 5 ζ(2)ζ(3) − ln (2)ζ(3) + ln (2)ζ(2) − ln (2). 16 16 12 120

(4.171)

4.4. Harmonic Series with Powers of 2 in the Denominator

Solution (i)

259

Let x = 1/2 in (2.87),

    ∞ ∞ ∞ ∞ (3) (4) (2) X X X 1 Hn n 1 Hn Hn X Hn + x = 5 Li − . +ln(2) Li −2 5 4 n3 2n n=1 n2 2n 2 2 n4 2n n=1 n2n n=1 n=1 (4.172) Now put x = 1/2 in (2.85),       ∞ ∞ ∞ (2) (3) X X X Hn Hn 1 1 1 Hn + = Li . Li + 10 Li − 6 2 3 5 3 2n 2 2n 4 2n n n 2 2 2 n n=1 n=1 n=1 (4.173) Take the difference of (4.172) and (4.173), 3

        ∞ (2) X Hn 5 1 1 1 1 1 1 = Li5 − ln(2) Li4 + Li2 Li3 3 n n 2 2 2 2 2 2 2 2 n=1 −2

∞ ∞ (4) X Hn 1 X Hn + . n4 2 n 2 n=1 n2n n=1

Gather the values from (1.120), (1.132), (4.169), and (4.170) to finalize the solution. Solution (ii)

Divide both sides of (2.83): Li22 (x) = 4

∞ ∞ (2) X X Hn n Hn n x + 2 x − 6 Li4 (x) n3 n2 n=1 n=1

by x then integrate from x = 0 to 1/2, using

R

1 2

0

xn−1 dx =

1 n2n ,

  Z 1 ∞ ∞ (2) X X Hn Hn 1 2 Li22 (x) 1 dx − 2 = + 3 Li . 5 3 2n 4 2n n 2 x n 2 0 n=1 n=1 The first two terms are calculated in (3.154) and (4.169).

4.4.12

P∞

(3)

Hn n=1 n2 2n

The following sum is also proposed by Cornel Vălean (see [34]):     ∞ (3) X Hn 1 1 81 5 = 4 Li + 3 ln(2) Li − ζ(5) + ln(2)ζ(4) 5 4 2 n n 2 2 2 64 16 n=1 7 7 5 3 11 5 ln (2)ζ(2) + ln (2). − ζ(2)ζ(3) + ln2 (2)ζ(3) − 8 8 12 120

(4.174)

260

Chapter 4. Harmonic Series

Solution

Combine (4.172) and (4.173),

        ∞ ∞ (4) (3) X 5 Hn 1 3 1 1 1 1 3 X Hn = . Li + ln(2) Li − Li Li − 5 4 2 3 n2 2n 2 2 2 2 2 2 2 2 n=1 n2n n=1 Put together the values from (1.120), (1.132), and (4.170) to finalize the solution. 2 Hn n=1 n3 2n

P∞

4.4.13

Show that     ∞ X Hn2 1 1 279 37 = −2 Li5 − ln(2) Li4 + ζ(5) − ln(2)ζ(4) 3 n n 2 2 2 64 16 n=1 −

7 2 1 3 1 5 9 ζ(2)ζ(3) + ln (2)ζ(3) + ln (2)ζ(2) − ln (2). 16 16 12 40

Solution

Multiply both sides of (2.11): ∞ X

(Hn2 − Hn(2) )xn =

n=1 2

by

ln (x) x

(4.175)

ln2 (1 − x) . 1−x

then integrate from x = 0 to 1/2,

∞   Z 12 X ln2 (1 − x) ln2 (x) dx = Hn2 − Hn(2) xn−1 ln2 (x)dx x(1 − x) 0 0 n=1  ∞    ln2 (2) 2 ln(2) X 2 IBP = Hn2 − Hn(2) + + n2n n2 2n n3 2 n n=1   ∞ ∞ (2) X X ln(2) 2 Hn2 − Hn 2 (2) = ln(2) (Hn − Hn ) + + 2 n2n n2 2 n n3 2n n=1 n=1 ( ) Z 12 ln(2) 2 n−1 write + 2 n =− x (ln(2) + 2 ln(x))dx n2n n 2 0 ! Z 21 ∞ ∞ (2) X X Hn2 − Hn n−1 (2) 2 = ln(2) (Hn − Hn ) x (ln(2) + 2 ln(x))dx + 2 n3 2n 0 n=1 n=1

Z

1 2

Z = ln(2) 0

1 2

{reverse the order of integration and summation} ! ∞ ∞ (2) X ln(2) + 2 ln(x) X (2) Hn2 − Hn 2 n (Hn − Hn )x dx + 2 x n3 2 n n=1 n=1 {recall the generation function in (2.11)}

4.4. Harmonic Series with Powers of 2 in the Denominator

261

  ∞ (2) X ln(2) + 2 ln(x) ln2 (1 − x) Hn2 − Hn − dx + 2 = ln(2) x 1−x n3 2n 0 n=1 Z 12 2 Z 21 ln (1 − x) ln(x) ln2 (1 − x) = − ln2 (2) dx − 2 ln(2) dx x(1 − x) x(1 − x) 0 0 ∞ ∞ (2) X X Hn2 Hn +2 − 2 n3 2 n n3 2n n=1 n=1 1 2

Z

Reorganizing the terms, we have Z 1 ∞ ∞ (2) X X Hn2 Hn 1 2 ln2 (1 − x) ln2 (x) dx = + n3 2 n n3 2n 2 0 x(1 − x) n=1 n=1 | {z } I1

1 2

Z

2

1 2

ln(x) ln (1 − x) ln2 (1 − x) 1 dx + ln2 (2) dx . x(1 − x) 2 x(1 − x) 0 | {z } {z }

+ ln(2) 0

|

Z

I2

(4.176)

I3

For I1 , set a = 2 in (3.86), 1 2

Z I1 = 0

ln2 (1 − x) ln2 (x) dx = x(1 − x)

1

Z 0

ln2 (1 − x) ln2 (x) dx x

{expand ln2 (1 − x) in series given in (2.6)} Z ∞ X Hn−1 1 n−1 2 =2 x ln (x)dx n 0 n=1   ∞ ∞ X X Hn − n1 2! Hn − 4ζ(5) =2 = 4 3 n n n4 n=1 n=1 {substitute the result from (4.6)} = 8ζ(5) − 4ζ(2)ζ(3).

(4.177)

For I2 , Z

1 2

I2 = 0

ln(x) ln2 (1 − x) 1−x=y dx = x(1 − x)

Z

1 1

|2

ln(1 − y) ln2 (y) dy y(1 − y) {z } R1 0

R 1/2 0

Z 21 ln(1 − y) ln2 (y) ln(1 − y) ln2 (y) = dy − dy y(1 − y) y(1 − y) 0 0 {set a = 2 in (3.88) to get the first integral}   1 1 1 and write = + in the second one y(1 − y) y 1−y Z

1

262

Chapter 4. Harmonic Series

Z 12 Z 12 ∞ X ln(1 − y) ln2 (y) ln(1 − y) ln2 (y) Hn − dy dy = −2 − n3 y 1−y 0 n=1 |0 {z } IBP

Z 1 Z 21 ∞ X 1 4 Hn 1 2 ln3 (y) ln(1 − y) ln2 (y) − = −2 ln (2) − dy − dy n3 3 3 0 1−y 1−y 0 n=1 {recall the results from (4.5), (3.34), and (3.142)}   9 7 1 1 4 1 = 2 Li4 − ζ(4) + ln2 (2)ζ(3) + − ln2 (2)ζ(2) + ln (2). 2 4 4 2 12 For I3 , 1 2

Z 1 ln2 (1 − x) ln2 (y) 1−x=y I3 = dx = dy 1 (1 − y)y x(1 − x) 0 2 Z 1 2 Z 1 2 ln (y) ln (y) dy + dy = 1 1 y 1−y 2 2 | {z } Z

R1 0

R 1/2 0 1

2 2 ln (y) ln (y) 1 dy − dy = ln3 (2) + 3 1−y 0 0 1−y {collect the results from (3.3) and (3.33)} 1 = ζ(3). 4

Z

1

2

Z

Substitute the results of I1 , I2 , and I3 along with the result from (4.171) in (4.176) to complete the solution. For a different method, check [21].

4.4.14

(2)

P∞

n=1

Hn Hn n2 2n

The following sum is proposed by Cornel Vălean (see [?]):     ∞ (2) X Hn Hn 1 31 1 1 + ln(2) Li − ζ(5) + ln(2)ζ(4) = 2 Li 4 5 2 n n 2 2 2 32 8 n=1 1 1 3 1 5 + ζ(2)ζ(3) − ln (2)ζ(2) + ln (2). 8 12 40 Solution

(4.178)

Divide both sides of (2.72): Z − 0

(2)

1

xn−1 ln3 (1 − x)dx =

(3)

Hn3 + 3Hn Hn + 2Hn n

4.4. Harmonic Series with Powers of 2 in the Denominator

263

by n2n then consider the summation over n ≥ 1, Z 1 3 ∞ ∞ ∞ (2) (3) X X X Hn Hn Hn ln (1 − x) Hn3 +3 +2 =− 2 2n 2 2n 2 2n n n n x 0 n=1 n=1 n=1 1

Z = 0

∞ X (x/2)n n n=1

! dx

{use the geometric series formula}  Z 1 3 ln (y) ln 1+y ln (1 − x) ln(1 − x/2) 1−x=y 2 dx = dy x 1−y 0 {set a = 3 in (3.92)} 3

= −6

∞ (4) X (−1)n Hn . n n=1

Substituting the result of the latter sum from (4.154), ∞ ∞ ∞ (2) (3) X X X Hn3 Hn Hn Hn 21 9 + 3 + 2 = 12ζ(5) − ln(2)ζ(4) − ζ(2)ζ(3). 2 2n 2 2n 2 2n n n n 4 4 n=1 n=1 n=1

(4.179) Next, divide both sides of (2.22): ∞  ln3 (1 − x) X n  3 = x Hn − 3Hn Hn(2) + 2Hn(3) − 1−x n=1

by x, then integrate using

R

1 2

0

xn−1 dx =

1 n2n ,

Z 12 3 ∞ (3) (2) X Hn3 − 3Hn Hn + 2Hn ln (1 − x) = − dx. n n2 x(1 − x) 0 n=1

(4.180)

On the other hand, multiply both sides of (2.22) by − ln(x) x then integrate, 1 2

ln3 (1 − x) ln(x) dx x(1 − x) 0 ! Z 12 ∞   X n−1 3 (2) (3) = Hn − 3Hn Hn + 2Hn − x ln(x)dx Z

0

n=1 IBP

=

∞  X

Hn3 − 3Hn Hn(2) + 2Hn(3)

n=1



ln(2) 1 + 2 n n2n n 2

Distribute then rearrange the terms, ∞ ∞ ∞ (2) (3) X X X Hn3 Hn Hn Hn − 3 + 2 n2 2 n n2 2n n2 2n n=1 n=1 n=1

 .

264

Chapter 4. Harmonic Series ∞ (2) (3) X ln3 (1 − x) ln(x) Hn3 − 3Hn Hn + 2Hn dx − ln(2) x(1 − x) n2n 0 n=1 ( ) Z 12 3 ln (1 − x) the latter sum is equal to − dx given in (4.180) x(1 − x) 0 Z 21 3 Z 21 3 ln (1 − x) ln(x) ln (1 − x) = dx + ln(2) dx . x(1 − x) x(1 − x) {z } {z } |0 |0 1 2

Z

=

I1

I2

For I1 , 1 2

Z 0

Z 1!

1

Z

=

ln3 (1 − x) ln(x) dx + x

1 2

0

1

Z = |0

1 2

Z |0

ln3 (1 − x) ln(x) dx 1−x {z }

3

IBP

ln (1 − x) ln(x) 1 1 dx + ln5 (2) + x 4 4

3

ln (1 − x) ln(x) dx − x {z } Z

1

|0

1 2

ln4 (1 − x) dx x {z } 1−x→x

3

1

Z

Z

1 ln (1 − x) ln(x) dx + ln5 (2) x 4

1 2

1−x→x

1 + 4

1

|2

ln4 (x) dx 1−x {z }

R1 0

R 1/2 0

1 ln3 (1 − x) ln(x) ln (x) ln(1 − x) dx − dx + ln5 (2) 1 1 − x x 4 0 2 Z 1 4 Z 1 1 ln (x) 1 2 ln4 (x) + dx − dx 4 0 1−x 4 0 1−x {recall the relation involving the second integral from (3.146)} Z Z 1 Z 1 1 ln3 (x) ln(1 − x) 1 2 ln4 (x) 1 1 ln4 (x) = dx − dx + dx 2 0 1−x 2 0 1−x 4 0 1−x 3 1 5 − ζ(5) + ln (2) 8 10 {put together the results from (3.124), (3.35), and (3.5)}     1 285 21 2 1 − 12 ln(2) Li4 + ζ(5) − 3ζ(2)ζ(3) − ln (2)ζ(3) = −12 Li5 2 2 16 4 2 +2 ln3 (2)ζ(2) − ln5 (2). 5 Z

=

1

3

Z

1

4.4. Harmonic Series with Powers of 2 in the Denominator

265

For I2 , Z

1 2

0

ln3 (1 − x) 1−x=y dx = x(1 − x)

Z

1 1 2

ln3 (y) dy = y(1 − y)

Z

1 1 2

ln3 (y) dy + y

1

Z

1

|2

ln3 (y) dy 1−y {z }

R1 0

Z

1

3

Z

1 2

R 1/2 0

3

1 ln (y) ln (y) = − ln4 (2) + dy − dy 4 1 − y 1−y 0 0 {collect the results from (3.4) and (3.34)}   3 1 1 21 = 6 Li4 ln(2)ζ(3) − ln2 (2)ζ(2) + ln4 (2). − 6ζ(4) + 2 4 2 4 Combine the results of I1 and I2 ,     ∞ ∞ ∞ (2) (3) X X X Hn3 Hn Hn Hn 1 1 − 3 + 2 = −12 Li − 6 ln(2) Li 5 4 2 2n 2 2n 2 2n n n n 2 2 n=1 n=1 n=1 +

1 3 5 285 ζ(5) − 6 ln(2)ζ(4) − 3ζ(2)ζ(3) + ln3 (2)ζ(2) − ln (2). 16 2 20

(4.181)

Taking the difference of (4.179) and (4.181) then dividing by 6 finishes the solution. 3 Hn n=1 n2 2n

P∞

4.4.15

The following sum is proposed by Cornel Vălean (see [?]):     ∞ X Hn3 279 25 1 1 = −14 Li5 − 9 ln(2) Li4 + ζ(5) − ln(2)ζ(4) 2 2n n 2 2 16 4 n=1 7 7 13 3 31 5 − ζ(2)ζ(3) − ln2 (2)ζ(3) + ln (2)ζ(2) − ln (2). 8 4 12 120 Solution

(4.182)

Combining (4.179) and (4.181) yields

    ∞ ∞ (3) X X 1 477 Hn3 Hn 1 2 − 6 ln(2) Li4 + ζ(5) +4 = −12 Li5 2 2n 2 2n n n 2 2 16 n=1 n=1 −

45 21 1 3 5 ln(2)ζ(4) − ζ(2)ζ(3) + ln3 (2)ζ(2) − ln (2). 4 4 2 20

Plug in the result from (4.174) then divide by 2 to finish the solution.

266

Chapter 4. Harmonic Series

Harmonic Series with Powers of 2n + 1 in the Denominator

4.5

P∞

4.5.1

n=0

(−1)n H2n+1 2n+1

Show that

Solution

∞ X (−1)n H2n+1 π = G − ln(2). 2n + 1 8 n=0

Set an =

Hn n in

(4.183)

(1.11):

∞ X

n

(−1) a2n+1 = J

n=0

∞ X

in an ,

n=1

we have ∞ ∞ n X X (−1)n H2n+1 i Hn =J 2n + 1 n n=0 n=1

{let x = i in the generating function in (2.7)}   1 2 = J Li2 (i) + ln (1 − i) . 2 Gather the values from (1.108) and (1.25) to finish the solution.

P∞

4.5.2

n=0

(−1)n H2n+1 (2n+1)2

Show that ∞ X (−1)n H2n+1 π 2 1 = −J Li3 (1 − i) − ln (2) − ln(2)G. 2 (2n + 1) 16 2 n=0

Solution

Let an =

Hn n2

(4.184)

in (1.11), ∞ ∞ n X X (−1)n H2n+1 i Hn = J 2 (2n + 1) n2 n=0 n=1

{set x = i in the generating function in (2.8)} 

 1 2 = J Li3 (i) − Li3 (1 − i) + ln(1 − i) Li2 (1 − i) + ln(i) ln (1 − i) + ζ(3) . 2 Gather the values from (1.109), (4.140), (1.26), and (1.25) to finish the solution.

4.5. Harmonic Series with Powers of 2n + 1 in the denominator

4.5.3

P∞

267

(2)

(−1)n H2n+1

n=0

2n+1

Show that (2) ∞ X (−1)n H2n+1 17π 3 π 2 1 = 2 J Li3 (1 − i) + + ln (2) + ln(2)G. (4.185) 2n + 1 192 8 2 n=0

Set an =

Solution

(2) Hn n

in (1.11),

(2) ∞ ∞ n (2) X X (−1)n H2n+1 i Hn =J 2n + 1 n n=0 n=1

{employ the generating function in (2.10)} = J {Li3 (i) + 2 Li3 (1 − i) − ln(1 − i) Li2 (1 − i) − ζ(2) ln(1 − i) − 2ζ(3)} . These terms are given in (1.109), (4.140), and (1.25).

4.5.4

P∞

Hn n=1 (2n+1)2

Show that

∞ X

Hn 7 3 = ζ(3) − ln(2)ζ(2). 2 (2n + 1) 4 2 n=1 Solution

Replace x by x2 in (2.4), ∞ X n=1

Hn x2n = −

ln(1 − x2 ) . 1 − x2

Multiply both sides by − ln(x) then integrate using: Z − 0

1

x2n ln(x)dx =

1 , (2n + 1)2

we obtain ∞ X

Z 1 Hn ln(x) ln(1 − x2 ) = dx (2n + 1)2 1 − x2 0 n=1 Z 1 √ x= y 1 ln(y) ln(1 − y) √ = dy. 4 0 y 1−y This integral is calculated in (3.110). A different way to calculate this sum may be found in (4.74).

(4.186)

268

Chapter 4. Harmonic Series (−1)n Hn n=0 (2n+1)2

P∞

4.5.5

Show that ∞ X 3π 3 π (−1)n Hn = 2 J Li3 (1 − i) + + ln2 (2) − ln(2)G. 2 (2n + 1) 32 8 n=0

Solution

Substitute H2n = H2n+1 −

1 2n+1

ln(2) + Hn − H2n+1 + Multiply both sides by ln(2)

(−1)n (2n+1)2

(4.187)

in (3.96),

1 = 2n + 1

Z

1

0

x2n dx. 1+x

then take the summation over n ≥ 0,

∞ X

∞ ∞ ∞ X (−1)n (−1)n Hn X (−1)n H2n+1 X (−1)n + − + (2n + 1)2 n=0 (2n + 1)2 n=0 (2n + 1)2 (2n + 1)3 n=0 n=0 ! Z 1 ∞ X (−1)n x2n 1 = dx (2n + 1)2 0 1+x n=0

{multiply the sum by x/x} Z = 0

( use

∞ X

1

1 1+x

∞ 1 X (−1)n x2n+1 x n=0 (2n + 1)2

n

(−1) a2n+1 = J

n=0 1

1 1+x

Z

1

Z = 0

= 0

∞ X

! dx )

n

i an given n=1 ! ∞ n n X

in (1.11)

1 i x J dx x n=1 n2   1 Li2 (ix) J dx 1+x x

{make use of (1.111) for Li2 (ix)}  Z 1  Z 1 1 i ln(y) = J − dy dx 1 − ixy 0 1+x 0   i 1 use the fact that J = 1 − ixy 1 + x2 y 2 Z 1  Z 1 ln(y) 1 = − dy dx 1 + x2 y 2 0 0 1+x Z 1Z x ln(x/t) xy=t = dt dx x(1 + x)(1 + t2 ) 0 0

4.5. Harmonic Series with Powers of 2n + 1 in the denominator

269

{change the order of integration} Z 1  Z 1 ln(x/t) 1 dx dt = 2 t x(1 + x) 0 1+t {evaluate the inner integral by partial fraction decoposition}   Z 1 1 1 2 π2 = Li2 (−t) + ln (t) + ln(2) ln(t) + dt 2 2 12 0 1+t Z 1 Z Z 1 Z 1 1 ln2 (t) ln(t) π 2 1 dt Li2 (−t) dt + dt + ln(2) dt + = 2 1 + t2 2 0 1 + t2 12 0 1 + t2 0 1+t 0   1 in series in the second and third integrals expand 1 + t2 Z 1 Z 1 ∞ Li2 (−t) 1X n = dt + (−1) x2n ln2 (t)dt 2 1 + t 2 0 0 n=0 Z 1 ∞ X π2  π  (−1)n + ln(2) x2n ln(t)dt + 12 4 0 n=0 Z 1 ∞ ∞ n X (−1) X (−1)n Li2 (−t) π3 = dt + − ln(2) + . 1 + t2 (2n + 1)3 (2n + 1)2 48 0 n=0 n=0 Reorganize the terms, ∞ ∞ ∞ X X X (−1)n Hn (−1)n H2n+1 (−1)n = − 2 ln(2) (2n + 1)2 (2n + 1)2 (2n + 1)2 n=0 n=0 n=0 Z 1 π3 Li2 (−t) dt + . + 2 1+t 48 0

(4.188)

Put together the results from (4.184), (1.205), and (3.147) to end the solution.

4.5.6

(2)

(−1)n Hn n=0 2n+1

P∞

∞ (2) X (−1)n Hn 5π 3 π = 4 J Li3 (1 − i) + + ln2 (2) + 2 ln(2)G. 2n + 1 48 4 n=0

Solution

(4.189)

We begin with expanding arctan x in series in the following integral: Z 0

1

Z ∞ X ln(x) arctan x (−1)n 1 x2n ln(x) dx = dx x(1 + x) 2n + 1 0 1+x n=0

270

Chapter 4. Harmonic Series

=

Z ∞ X (−1)n 1 ∂ 1 x2n dx 2n + 1 0 ∂n 2 1 + x n=0

{use differentiation under the integral sign theorem given in (2.78)} Z 1 2n ∞ x 1 X (−1)n d dx = 2 n=0 2n + 1 dn 0 1 + x {recall the result from (3.96)} ∞ 1 X (−1)n d = (Hn − H2n + ln 2) 2 n=0 2n + 1 dn {use the derivative of the harmonic number given in (1.157)} ∞  1 X (−1)n  (2) = 2H2n − Hn(2) − ζ(2) 2 n=0 2n + 1 ∞ ∞ ∞ (2) (2) X X 1 X (−1)n Hn 1 (−1)n (−1)n H2n − − ζ(2) 2n + 1 2 n=0 2n + 1 2 2n + 1 n=0 n=0   1 (2) (2) write H2n = H2n+1 − in the first sum (2n + 1)2 (2) ∞ ∞ ∞ ∞ (2) X X X (−1)n H2n+1 (−1)n (−1)n 1 X (−1)n Hn 1 = − − − ζ(2) . 3 2n + 1 2 n=0 2n + 1 (2n + 1) 2 2n + 1 n=0 n=0 n=0

=

(4.190) On the other hand, we have Z 1 Z 1 Z 1 ln(x) arctan x ln(x) arctan x ln(x) arctan x dx = dx − dx x(1 + x) x 1+x 0 0 0 {expand arctan x in series in the first integral} Z 1 ∞ X (−1)n Z 1 ln(x) arctan x = x2n ln(x)dx − dx 2n + 1 1+x 0 0 n=0 Z 1 ∞ X (−1)n ln(x) arctan x =− dx. (4.191) − 3 (2n + 1) 1+x 0 n=0 Combine (4.190) and (4.191), Z 1 (2) ∞ ∞ ∞ (2) X X X (−1)n H2n+1 (−1)n Hn (−1)n ln(x) arctan x =2 −ζ(2) +2 dx. 2n + 1 2n + 1 2n + 1 1+x 0 n=0 n=0 n=0 Gather the results from (4.185), (1.93), and (3.149) to finish the solution.

4.5. Harmonic Series with Powers of 2n + 1 in the denominator

P∞

4.5.7

n=0

271

(−1)n H2n+1 (2n+1)3

Show that ∞ X (−1)n H2n+1 35π = 2β(4) − ζ(3). 3 (2n + 1) 128 n=0

Solution

(4.192)

Multiply both sides of (2.44): ∞ arctan x 1X (−1)n (Hn − 2H2n )x2n−1 = 1 + x2 2 n=0

by x ln2 (x) then integrate from x = 0 to 1, Z 0

1

Z 1 ∞ x ln2 (x) arctan x 1X n (−1) (H − 2H ) dx = x2n ln2 (x)dx n 2n 1 + x2 2 n=0 0   ∞ 2 1X (−1)n (Hn − 2H2n ) = 2 n=0 (2n + 1)3 =

∞ ∞ X X (−1)n H2n (−1)n Hn − 2 3 (2n + 1) (2n + 1)3 n=0 n=0



 1 in the last series 2n + 1 ∞ ∞ ∞ n X X X (−1)n H2n+1 (−1)n (−1) Hn = − 2 + 2 . (2n + 1)3 (2n + 1)3 (2n + 1)4 n=0 n=0 n=0 use H2n = H2n+1 −

Rearrange the terms, 2

∞ ∞ ∞ X X X (−1)n Hn (−1)n (−1)n H2n+1 = + 2 (2n + 1)3 (2n + 1)3 (2n + 1)4 n=0 n=0 n=0 Z 1 x ln2 (x) arctan x − dx 1 + x2 0 ) ( Z 1 2 ∞ X (−1)n ln (x) arctan x write 2 = dx, (2n + 1)4 x 0 n=0

{which follows from expanding arctan x in series then integrating} Z 1 2 Z 1 ∞ X (−1)n Hn ln (x) arctan x x ln2 (x) arctan x = + dx − dx (2n + 1)3 x 1 + x2 0 0 n=0

272

Chapter 4. Harmonic Series

=

 Z 1 ∞ X 1 x (−1)n Hn + ln2 (x) arctan x dx − 3 2 (2n + 1) x 1 + x 0 n=0 Z 1 2 ∞ n X ln (x) arctan x (−1) Hn + = dx. 3 (2n + 1) x(1 + x2 ) 0 n=0

Collect the results from (4.93) and (3.150) then divide by 2 to finish the solution.

P∞

4.5.8

(2)

(−1)n H2n+1

n=0

(2n+1)2

Show that (2) ∞ X (−1)n H2n+1 35π π2 = −β(4) + ζ(3) − G. 2 (2n + 1) 64 48 n=0

Set x = i in (2.83) then take the imaginary parts of both sides,

Solution

J Li22 (i) = 4 J

Use J

(4.193)

∞ X

in an =

∞ X

∞ ∞ (2) X X Hn i Hn i x + 2 J x − 6 J Li4 (i). n3 n2 n=1 n=1

(−1)n a2n+1 given in (1.11),

n=0

n=1

(2) ∞ ∞ X X (−1)n H2n+1 (−1)n H2n+1 1 2 (i) − 2 = 3 J Li (i) + J Li . 4 2 2 (2n + 1) 2 (2n + 1)3 n=0 n=0

Using the value from (1.108), we have Li22 (i)

5 = ζ(4) − G2 − 128



 π2 G i. 24

Collect this result along with these from (1.110) and (4.192), the solution is finished.

4.5.9

P∞

(2)

Hn n=1 (2n+1)2

Show that ∞ X

(2)

Hn = 8 Li4 (2n + 1)2 n=1

  1 121 1 − ζ(4)+7 ln(2)ζ(3)−2 ln2 (2)ζ(2)+ ln4 (2). 2 16 3 (4.194)

4.5. Harmonic Series with Powers of 2n + 1 in the denominator

Solution (i)

273

We have 1 =− (2n + 1)2

Z

1

x2n ln(x)dx.

0

(2)

Multiply both sides by Hn then take the summation over n ≥ 1, Z 1 ∞ (2) X Hn (2) = − H x2n ln(x)dx n 2 (2n + 1) 0 n=1 n=1 ! Z 1 ∞ X =− ln(x) (x2 )n Hn(2) dx ∞ X

0

n=1

{replace x by x2 in the generating function in (2.2)} Z 1 ln(x) Li2 (x2 ) dx =− 1 − x2 0  write Li2 (x2 ) = 2 Li2 (x) + 2 Li2 (−x) given in (1.115) Z 1 Z 1 ln(x) Li2 (−x) ln(x) Li2 (x) dx − 2 dx = −2 2 1−x 1 − x2 0 0 Z 1 Z 1 Z 1 ln(x) Li2 (x) ln(x) Li2 (x) ln(x) Li2 (−x) =− dx − dx − dx 1−x 1+x 1−x |0 {z } |0 {z } |0 {z } I1

I2

Z − 0

|

1

I3

ln(x) Li2 (−x) dx . 1+x {z } I4

For I1 , Z

1

ln(x) Li2 (x) dx 1−x 0 ( ) ∞ Li2 (x) X (2) n−1 given in (2.3) write = Hn−1 x 1−x n=1 Z 1 (2) ∞ ∞ X X Hn−1 (2) = Hn−1 xn−1 ln(x)dx = − n2 0 n=1 n=1 I1 =

=−

(2) ∞ X Hn − n2 n=1

1 n2

=−

∞ (2) X Hn + ζ(4). n2 n=1

For I2 , Z I2 = 0

1

ln(x) Li2 (x) dx 1+x

274

Chapter 4. Harmonic Series

{expand 1/(1 + x) in series} Z 1 n =− xn−1 ln(x) Li2 (x)dx (−1) ∞ X

=−

0

n=1 ∞ X

(2)

2Hn 2ζ(2) Hn + 3 − n2 n n2

n

(−1)

n=1

! .

For I3 , Z

1

ln(x) Li2 (−x) dx 1−x 0 {expand Li2 (−x) in series } Z ∞ X (−1)n 1 xn ln(x) = dx n2 1−x 0 n=1 I3 =

{set a = 2 in (1.153) to get the integral} ∞  X (−1)n  (2) = H − ζ(2) . n n2 n=1 For I4 , Z

1

ln(x) Li2 (−x) dx 1+x ( ) ∞ Li2 (x) X (2) n−1 replace x by −x in Hn−1 x given in (2.3) = 1−x n=1 Z 1 ∞ X (2) =− (−1)n Hn−1 xn−1 ln(x)dx I4 =

0

0

k=1

=

∞ X

(−1)n

n=1

=

(2) (2) ∞ X Hn−1 n Hn − = (−1) n2 n2 n=1

1 n2

∞ (2) X 7 (−1)n Hn + ζ(4). 2 n 8 n=1

Gathering the four integrals, ∞ X

∞ ∞ ∞ (2) (2) (2) X X Hn (−1)n Hn X (−1)n Hn Hn 5 = 2 − + − ζ(4). 2 3 2 2 (2n + 1) n n n 8 n=1 n=1 n=1 k=1

Combining the results from (4.143), (4.144), and (2.98) completes the solution.

4.5. Harmonic Series with Powers of 2n + 1 in the denominator

275

Solution (ii) Set p = q = 2 in the third application of Abel’s summation in (2.105), ∞ ∞ (2) (2) (2) 5 1 X (−1)n Hn 1 X Hn Hn = − , ζ(4) − (2n + 1)2 8 4 n=1 n2 2 n=1 n2 n=1 ∞ X

and the solution finalizes on substituting the results from (4.144) and (2.98).

4.5.10

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

P∞

Show that ∞ X

Hn2 = 8 Li4 (2n + 1)2 n=1 Solution

  61 1 1 − ζ(4) + ln2 (2)ζ(2) + ln4 (2). (4.195) 2 16 3

Replace x by x2 in (2.11), ∞ ln2 (1 − x2 ) X 2 = (Hn − Hn(2) )x2n . 1 − x2 n=1

Multiply both sides by − ln(x) then integrate from x = 0 to 1, Z 1 ∞ (2) X Hn2 − Hn ln(x) ln2 (1 − x2 ) = − dx (2n + 1)2 1 − x2 0 n=1 Z √ 1 1 ln(y) ln2 (1 − y) x=y dy. = − √ 4 0 y(1 − y) Reorder the terms, ∞ X

Z ∞ (2) X Hn2 Hn 1 1 ln(y) ln2 (1 − y) = − dy. √ (2n + 1)2 (2n + 1)2 4 0 y(1 − y) n=1 n=1 Substitute the results from (4.194) and (3.141) to complete the solution.

4.5.11

P∞

(2)

Hn n=1 (2n+1)3

Show that ∞ X

(2)

Hn 49 93 = ζ(2)ζ(3) − ζ(5). 3 (2n + 1) 8 8 n=1

(4.196)

276

Chapter 4. Harmonic Series

Solution

Set q = 2 and p = 3 in (2.105),

∞ ∞ (3) (3) (2) X 7 15 X Hn (−1)n Hn Hn = − 2 . ζ(2)ζ(3) − (2n + 1)3 8 8 n=1 n2 n2 n=1 n=1 ∞ X

The last two sums are given in (4.105) and (4.156).

P∞

(3)

Hn n=1 (2n+1)2

4.5.12

Show that

∞ X

(3)

Hn 31 = ζ(5) − 8ζ(2)ζ(3). 2 (2n + 1) 2 n=1 Solution

(4.197)

Set q = 3 and p = 2 in (2.105),

∞ X

∞ ∞ (3) (2) (2) X Hn 3 15 X Hn (−1)n Hn = ζ(2)ζ(3) − − 4 . (2n + 1)2 4 4 n=1 n3 n3 n=1 n=1

The last two sums are given in (4.102) and (4.152).

P∞

(3)

Hn n=1 (2n+1)3

4.5.13

+4

P∞

n=1

(3)

(−1)n Hn n3

Show that ∞ X

∞ (3) (3) X Hn (−1)n Hn 17 31 + 4 = − ζ 2 (3) − ζ(6). 3 3 (2n + 1) n 16 16 n=1 n=1

Solution

Set q = p = 3 in (2.105), ∞ X

∞ ∞ (3) (3) (3) X Hn 7 2 31 X Hn (−1)n Hn ζ (3) − = − 4 . (2n + 1)3 8 8 n=1 n3 n3 n=1 n=1

The first sum is given in (2.99).

(4.198)

4.6. Skew Harmonic Series

4.6

277

Skew Harmonic Series P∞

4.6.1

n=1

(−1)n H n n

Show that

Solution

∞ X (−1)n H n 1 1 = − ζ(2) − ln2 (2). n 2 2 n=1

(4.199)

Replace x by −x in (2.28), ∞ X

(−1)n H n xn =

ln(1 − x) . 1+x

R1

xn−1 dx =

n=1

Divide both sides by x then integrate using

0

(4.200) 1 n,

Z 1 Z 1 Z 1 ∞ X (−1)n H n ln(1 − x) ln(1 − x) ln(1 − x) = dx = dx − dx n x 1+x 0 x(1 + x) 0 n=1 |0 {z } 1−x=y

1

Z = 0

ln(y) dy − 1−y

1

Z 0

ln(1 − x) dx. 1+x

Put together the results from (3.2) and (3.27) to complete the solution. For an alternative solution, set x = −1 in the generating function in (2.29).

P∞

Hn n=1 n3

4.6.2

Show that   ∞ X Hn 1 3 1 1 4 − ζ(4) − ln2 (2)ζ(2) + ln (2). = 2 Li4 3 n 2 2 2 12 n=1 Solution

Multiply both sides of (2.28) by 1 2

Z 0

ln2 (x) 2x

then integrate using

1

xn−1 ln2 (x)dx =

1 , n3

Z ∞ X Hn 1 1 ln(1 + x) ln2 (x) = dx n3 2 0 x(1 − x) n=1

(4.201)

278

Chapter 4. Harmonic Series

=

1 2

1

Z ln(1 + x) ln2 (x) 1 1 ln(1 + x) ln2 (x) dx + dx x 2 0 1−x {z } | {z }

Z

1

Z |0

1 =− 6

IBP

0

IBP

Z 1 ln3 (x) ln(x) ln(1 + x) ln(1 − x) dx + dx 1+x x 0 Z 1 1 ln(1 − x) ln2 (x) + dx. 2 0 1+x

Put together the results from (3.10), (3.116), and (3.130) to finish the solution. For a different method, see (4.13).

4.6.3

P∞

n=1

(−1)n H n n3

Show that   ∞ X 3 1 1 4 (−1)n H n 1 = 2 Li4 − ζ(4) − ln2 (2)ζ(2) + ln (2). (4.202) 3 n 2 2 2 12 n=1 Solution (i)

Multiply both sides of (4.200) by 1 2

Z

ln2 (x) 2x

1

xn−1 ln2 (x)dx =

0

then integrate using

1 , n3

Z ∞ X (−1)n H n 1 1 ln(1 − x) ln2 (x) = dx n3 2 0 x(1 + x) n=1 Z Z 1 1 ln(1 − x) ln2 (x) 1 1 ln(1 − x) ln2 (x) dx − dx = 2 0 x 2 0 1+x | {z } IBP

1 = 6

Z 0

1

ln3 (x) 1 dx − 1−x 2

Z 0

1

ln(1 − x) ln2 (x) dx, 1+x

and the solution completes on putting together the results from (3.4) and (3.131). Solution (ii)

Put a = 3 in (4.21), ∞ ∞ ∞ ∞ X X (−1)n H n (−1)n Hn X H n 3 X Hn = − + . n3 n3 n3 4 n=1 n3 n=1 n=1 n=1

These sums are given in (4.143), (4.201) and (4.5).

4.6. Skew Harmonic Series

279

(−1)n H n Hn n=1 n

P∞

4.6.4

Show that ∞ X 1 1 (−1)n H n Hn = ln3 (2) − ln(2)ζ(2) − ζ(3). n 3 4 n=1

Solution (i)

(4.203)

Multiply both sides of (1.164): Z H n = ln(2) − 0

1

(−x)n dx 1+x

n

by

(−1) Hn n

then take the summation,

Z 1 ∞ ∞ X X (−1)n H n Hn (−1)n Hn 1 = ln(2) − n n 0 1+x n=1 n=1

∞ X Hn n x n n=1

! dx

{recall the generating function in (2.7)}   Z 1 (−1)n Hn 1 1 2 = ln(2) − ln (1 − x) + Li2 (x) dx n 2 0 1+x n=1 Z Z 1 ∞ X (−1)n Hn 1 1 ln2 (1 − x) Li2 (x) = ln(2) − dx − dx n 2 1 + x 0 0 1+x n=1 | {z } ∞ X

IBP

Z ∞ X (−1)n Hn 1 1 ln2 (1 − x) = ln(2) − dx n 2 0 1+x n=1 Z 1 ln(1 + x) ln(1 − x) − ln(2)ζ(2) − dx. x 0 The solution finalizes on combining the results from (4.135), (3.28), and (3.115). Solution (ii) Multiply both sides of (4.200) by − ln(1−x) then integrate using x R1 − 0 xn−1 ln(1 − x)dx = Hnn ,

Z = 0

1

Z 1 2 ∞ X ln (1 − x) (−1)n H n Hn =− dx n x(1 + x) 0 n=1 Z 1 2 Z 1 2 Z 1 2 ln2 (1 − x) ln (1 − x) ln (1 − x) ln (y) dx − dx = dx − dy. 1+x x 1+x 0 0 1−y |0 {z } 1−x=y

These two integrals are calculated in (3.28) and (3.3).

280

Chapter 4. Harmonic Series

P∞

4.6.5

n=1

H n Hn n2

Show that   ∞ X 1 43 3 1 H n Hn = −3 Li4 + ζ(4) + ln2 (2)ζ(2) − ln4 (2). (4.204) 2 n 2 16 4 8 n=1

The following solution can be found in [18]: Solution

Multiply both sides of (1.164): Z

1

H n = ln(2) − 0

by

Hn n2

(−x)n dx 1+x

then take the summation, Z 1 ∞ ∞ X X Hn H n Hn 1 = ln(2) − 2 2 n n 1 + x 0 n=1 n=1

∞ X Hn (−x)n 2 n n=1

! dx.

For the integral, utilize the generating function in (2.8) for the sum, ! Z 1 Z 1 Z 1 ∞ X Hn (−x)n 1 Li3 (−x) Li3 (1 + x) dx = dx − dx 2 n 1+x 1+x 0 1+x n=1 |0 {z } |0 {z } I1

Z + |0

1

I2

Z 1 Z 1 1 ln(−x) ln2 (1 + x) ζ(3) ln(1 + x) Li2 (1 + x) dx + dx + dx . 1+x 2 0 1+x 1+x {z } {z } | 0 {z } | I3

I4

I5

For I1 , Z

1

ln(1 + x) Li2 (−x) dx x 0 1 3 1 3 5 = − ln(2)ζ(3) + Li22 (−x) 0 = − ln(2)ζ(3) + ζ(4). 4 2 4 16 1 I1 = ln(1 + x) Li3 (−x) 0 − IBP

For I2 ,

1 I2 = Li4 (1 + x) 0 = Li4 (2) − ζ(4).

For I3 , Z 1 Li3 (1 + x) I3 = Li3 (1 + x) ln(1 + x) 0 − dx 1+x 1 = ln(2) Li3 (2) − Li4 (1 + x) 0 = ln(2) Li3 (2) − Li4 (2) + ζ(4). IBP

4.6. Skew Harmonic Series

281

For I4 , 1 I4 = − Li2 (1 + x) ln (1 + x) 0 + 2 IBP

2

IBP

Z 0

1

Li2 (1 + x) ln(1 + x) dx 1+x

2

= − ln (2) Li2 (2) + 2 ln(2) Li3 (2) − 2 Li4 (2) + 2ζ(4).

For I5 ,

1 I5 = ζ(3) ln(1 + x) 0 = ln(2)ζ(3).

Group all integrals (I1 to I5 ) along with the result from (4.4), ∞ X 7 53 1 H n Hn = ln(2)ζ(3) − ζ(4) + 3 Li4 (2) − 2 ln(2) Li3 (2) + ln2 (2) Li2 (2). 2 n 4 16 2 n=1

Collect the values of Li4 (2), Li3 (2), and Li2 (2) given in (1.142), (1.141), and (1.140) respectively to finalizes the solution.

P∞

4.6.6

n=1

(−1)n H n Hn n2

Show that   ∞ X (−1)n H n Hn 29 3 1 1 = 3 Li − ζ(4) − ln2 (2)ζ(2) + ln4 (2). 4 2 n 2 16 4 8 n=1 (4.205)

Solution

Multiply both sides of (1.164) by

(−1)n Hn n2

Z 1 ∞ ∞ X X (−1)n H n Hn (−1)n Hn = ln(2) − n2 n2 0 n=1 n=1

then take the summation, ! ∞ X 1 Hn n x dx. 1 + x n=1 n2

For the integral, make use of the generating function in (2.8), ! Z 1 Z 1 Z 1 ∞ X Hn xn Li3 (1 − x) 1 Li3 (x) dx − dx dx = 2 1 + x n 1 + x 1+x 0 0 0 n=1 | {z } | {z } I1

Z + |0

1

1

I2

2

Z 1 ln(1 − x) Li2 (1 − x) 1 ln(x) ln (1 − x) ζ(3) dx + dx + dx . 1+x 2 0 1+x 1 +x 0 {z } | {z } | {z } Z

I3

I4

I5

282

Chapter 4. Harmonic Series

For I1 , expand

1 1+x

in series, Z 1 ∞ X Li3 (x) n dx = − xn−1 Li3 (x)dx (−1) 1+x 0 n=1

1

Z I1 = 0

{recall the result from (3.104)}   ∞ X ζ(3) ζ(2) Hn − 2 + 3 =− (−1)n n n n n=1 ∞ X (−1)n Hn 5 = ln(2)ζ(3) − ζ(4) − . 4 n3 n=1

For I2 , Z

1

Li3 (1 − x) 1−x=y dx = 1+x

Z

1

Li3 (y) dy 0 0 2−y ) ∞ X y n−1 1 expand in series as 2−y 2n n=1 Z ∞ 1 X 1 = y n−1 Li3 (y)dy n 2 0 n=1

I2 = (

{recall the result of the integral from (3.104) }   ∞ X 1 ζ(3) ζ(2) Hn = − + 2n n n2 n3 n=1   X ∞ Hn 1 = ln(2)ζ(3) − ζ(2) Li2 + . n n3 2 2 n=1 For I3 , 1

Z I3 = 0

Z 1 ln(1 − x) Li2 (1 − x) ln(y) Li2 (y) 1−x=y dx = dy 1+x 2−y 0 Z 1 ∞ X 1 = y n−1 ln(y) Li2 (y)dy n 2 0 n=1 Z 1 ∞ X 1 ∂ n−1 y Li2 (y)dy = n 2 ∂n 0 n=1

{use differentiation under the integral sign theorem given in (2.78)} Z 1 ∞ X 1 d = y n−1 Li2 (y)dy n dn 2 0 n=1 {recall the result of the integral from (3.103)}

(4.206)

4.6. Skew Harmonic Series

283

=

  ∞ X 1 d ζ(2) Hn − 2n dn n n2 n=1

{use the derivative of the harmonic number given in (1.157)} ! ∞ (2) X 1 2Hn Hn 2ζ(2) = + 2 − 2n n3 n n2 n=1   ∞ ∞ (2) X X 1 Hn Hn + − 2ζ(2) Li . =2 2 n n3 n n2 2 2 2 n=1 n=1

(4.207)

For I4 , Z 1 ln(x) ln2 (1 − x) ln(1 − y) ln2 (y) 1−x=y dx = dy 1+x 2−y 0 0 Z 1 ∞ X 1 = y n−1 ln(1 − y) ln2 (y)dy n 2 0 n=1 Z 1 2 ∞ X 1 ∂ = y n−1 ln(1 − y)dy n 2 2 ∂n 0 n=1 Z 1 ∞ X 1 d2 = y n−1 ln(1 − y)dy n dn2 2 0 n=1   ∞ X 1 d2 Hn = − 2n dn2 n n=1 ! ∞ (2) (3) X 1 2ζ(3) 2ζ(2) 2Hn 2Hn 2Hn + = − 3 − − 2n n n2 n n2 n n=1   ∞ ∞ ∞ (2) (3) X X X 1 Hn Hn Hn = 2 ln(2)ζ(3) + 2ζ(2) Li2 −2 − 2 − 2 . 2 2n n3 2n n2 2n n n=1 n=1 n=1 Z

1

I4 =

(4.208) For I5 , I5 = ζ(3) ln(1 + x)|10 = ln(2)ζ(3). Combine the results of all integrals (I1 to I5 ) and rearrange the terms, ∞ X (−1)n H n Hn 5 = ζ(4) − 2 ln(2)ζ(3) 2 n 4 n=1

+ ln(2)

∞ ∞ ∞ (3) X (−1)n Hn X (−1)n Hn X Hn + + . n2 n3 2n n n=1 n=1 n=1

Grouping the results from (4.137), (4.143), and (4.168) finishes the solution.

284

4.6.7

Chapter 4. Harmonic Series

P∞

n=1

H 2n H2n n2

Show that

Solution

∞ X 7 H 2n H2n = ζ(4). 2 n 4 n=1

Put an =

H n Hn n2 ∞ X n=1

we have

in (1.5):

a2n =

∞ ∞ 1X 1X an + (−1)n an , 2 n=1 2 n=1

∞ ∞ ∞ X H 2n H2n 1 X H n Hn 1 X (−1)n H n Hn = + . (2n)2 2 n=1 n2 2 n=1 n2 n=1

Gather the results from (4.204) and (4.205) then multiply through by 4.

(4.209)

4.7. Harmonic Series with Rational Argument

4.7 4.7.1

285

Harmonic Series with Rational Argument P∞

n=1

Show that

Solution (i)

(−1)n H n 2

n

∞ X (−1)n H n2 1 = ln2 (2) − ζ(2). n 2 n=1

(4.210)

Replace n by n/2 in (3.97), 1

Z

xn−1 ln(1 + x)dx =

0

Hn − H n2 . n

(4.211)

Multiply both sides by (−1)n then consider the summation, Z 1 ∞ ∞ ∞ X (−1)n Hn X (−1)n H n2 ln(1 + x) X − = (−x)n n n x 0 n=1 n=1 n=1   Z 1 −x ln(1 + x) dx = x 1+x 0 Z 1 1 ln(1 + x) =− dx = − ln2 (2). 1 + x 2 0

! dx

Recall the result from (4.135) to finalize the solution. Solution (ii)

Consider (2.70) Z 1 Z 1 Hn x=y 2 =− xn−1 ln(1 − x)dx = −2 y 2n−1 ln(1 − y 2 )dy. n 0 0

Replace n by n/2 to have H n2 =− n

Z

1

y n−1 ln(1 − y 2 )dy.

0

Next, multiply both sides by (−1)n then take the summation, ! Z 1 ∞ ∞ X (−1)n H n2 ln(1 − y 2 ) X n =− (−y) dy n y 0 n=1 n=1   Z 1 Z 1 ln(1 − y 2 ) −y ln(1 − y 2 ) =− dy = dy y 1+y 1+y 0 0 {write ln(1 − y 2 ) = ln(1 − y) + ln(1 + y)}

(4.212)

286

Chapter 4. Harmonic Series

Z

1

= 0

ln(1 + y) dy + 1+y

Z 0

1

ln(1 − y) dy. 1+y

2

The first integral is 21 ln (2) and the second one is given in (3.27). For a different approach, set x = −1 in the generating function in (2.34).

4.7.2

Hn

P∞

2

n=1 n2

Show that

Solution

∞ X H n2 11 = ζ(3). 2 n 8 n=1

(4.213)

Divide both sides of (4.212) by n then take the summation, Z 1 ∞ ∞ X H n2 ln(1 − y 2 ) X y n = − dy n2 y n 0 n=1 n=1 Z 1 ln(1 − y 2 ) ln(1 − y) dy = y 0 {write ln(1 − y 2 ) = ln(1 − y) + ln(1 + y)} Z 1 2 Z 1 ln (1 − y) ln(1 + y) ln(1 − y) dy + dy = y y 0 |0 {z } 1−y=x

Z

1

= 0

ln2 (x) dx + 1−x

Z 0

1

ln(1 + y) ln(1 − y) dy, y

and these two integrals are given in (3.3) and (3.115). For a different approach, see (4.36).

4.7.3

P∞

Show that

Solution

(−1)n H n 2

n=1

n2

∞ X (−1)n H n2 3 = − ζ(3). 2 n 8 n=1

Multiply both sides of (4.212) by

(−1)n n

(4.214)

then take the summation,

Z 1 ∞ ∞ X (−1)n H n2 ln(1 − y 2 ) X (−y)n = − dy n2 y n 0 n=1 n=1

4.7. Harmonic Series with Rational Argument

287

1

ln(1 − y 2 ) ln(1 + y) dy y 0 {write ln(1 − y 2 ) = ln(1 − y) + ln(1 + y)} Z 1 2 Z 1 ln (1 + y) ln(1 + y) ln(1 − y) = dy + dy, y y 0 0 Z

=

Gathering the results from (3.38) and (3.115) finishes the solution. Check (4.42) for a different approach.

4.7.4

P∞

Hn 2

n=1 n3

Show that   ∞ X H n2 7 1 1 4 1 11 = −2 Li4 ln (2). + ζ(4) − ln(2)ζ(3) + ln2 (2)ζ(2) − 3 n 2 4 4 2 12 n=1 (4.215)

Solution

Divide both sides of (4.211): Z 1 Hn − H n2 xn−1 ln(1 + x)dx = n 0

by n2 then consider the summation, ! Z 1 ∞ ∞ ∞ X Hn X H n2 ln(1 + x) X xn − = dx n3 n3 x n2 0 n=1 n=1 n=1 Z 1 ln(1 + x) Li2 (x) = dx x 0 {expand ln(1 + x) in series} ∞ X (−1)n−1 Z 1 = xn−1 Li2 (x)dx n 0 n=1 {recall the result of the integral from (3.103)}   ∞ X (−1)n−1 ζ(2) Hn = − 2 n n n n=1 = ζ(2)η(2) +

∞ X (−1)n Hn . n3 n=1

288

Chapter 4. Harmonic Series

Since

P∞

Hn n=1 n3

= 54 ζ(4) (given in (4.5)) and ζ(2)η(2) = 54 ζ(4), we have ∞ ∞ X X H n2 (−1)n Hn = − . n3 n3 n=1 n=1

The latter sum is calculated in (4.143).

4.7.5

P∞

n=1

(−1)n H n 2

n3

Show that   ∞ X (−1)n H n2 7 1 1 1 27 = 2 Li4 − ζ(4)+ ln(2)ζ(3)− ln2 (2)ζ(2)+ ln4 (2). 3 n 2 8 4 2 12 n=1 (4.216) Solution

Multiply both sides of (4.211) by

(−1)n n2

then consider the summation,

Z 1 ∞ ∞ ∞ X (−1)n Hn X (−1)n H n2 ln(1 + x) X (−x)n − = dx n3 n3 x n2 0 n=1 n=1 n=1 Z 1 5 1 ln(1 + x) Li2 (−x) dx = Li22 (−1) = ζ(4), =− x 2 8 0 and the solution is finalized on plugging in the result from (4.143).

4.7.6

P∞

n=1

Hn H n 2

n2

Show that   ∞ X Hn H n2 97 21 3 1 1 = −3 Li + ζ(4)− ln(2)ζ(3)+ ln2 (2)ζ(2)− ln4 (2). 4 2 n 2 16 8 4 8 n=1 (4.217) Solution

Multiply both sides of (4.211) by Z 1 ∞ ∞ X Hn2 X Hn H n2 − = n2 n2 0 n=1 n=1

Hn n

then consider the summation, ! ∞ ln(1 + x) X Hn n x dx x n n=1

{recall the generating function in (2.7)}   Z 1 ln(1 + x) 1 2 = ln (1 − x) + Li2 (x) dx x 2 0 Z 1 Z 1 2 1 ln(1 + x) ln (1 − x) ln(1 + x) Li2 (x) = dx + dx. 2 0 x x 0

4.7. Harmonic Series with Rational Argument

289

Apply integration by parts in the second integral, Z

1

0

Z 1 Li2 (−x) ln(1 − x) ln(1 + x) Li2 (x) dx = − Li2 (−x) Li2 (x)|10 − dx x x 0 Z ∞ X 5 (−1)n 1 n−1 x ln(1 − x)dx = ζ(4) − 4 n2 0 n=1 =

∞ X 5 (−1)n Hn ζ(4) + . 4 n3 n=1

Substituting this integral back yields ∞ ∞ ∞ X X Hn H n2 Hn2 X (−1)n Hn 5 ζ(4) + = − − n2 4 n2 n3 n=1 n=1 n=1 Z 1 1 ln(1 + x) ln2 (1 − x) dx. − 2 0 x

The solution finalizes on collecting the results from (4.99), (4.143), and (3.118).

4.7.7

P∞

(−1)n Hn H n 2

n=1

n2

Show that   ∞ X (−1)n Hn H n2 1 35 5 21 = 5 Li4 ln(2)ζ(3) − ln2 (2)ζ(2) − ζ(4) + 2 n 2 4 8 4 n=1 +

Solution

5 4 ln (2). 24

Multiply both sides of (4.211) by

(−1)n Hn n

Z 1 ∞ ∞ X (−1)n Hn2 X (−1)n Hn H n2 − = n2 n2 0 n=1 n=1

(4.218)

then take the summation, ! ∞ ln(1 + x) X Hn n (−x) dx x n n=1

{replace x by −x in (2.7) to get the sum}   Z 1 ln(1 + x) 1 2 = ln (1 + x) + Li2 (−x) dx x 2 0 Z 1 3 Z 1 ln (1 + x) ln(1 + x) Li2 (−x) 1 = dx + dx 2 0 x x 0 Z 1 1 ln3 (1 + x) 1 = dx − Li22 (−x)|10 2 0 x 2

290

Chapter 4. Harmonic Series

=

1 2

Z 0

1

ln3 (1 + x) 5 dx − ζ(4). x 16

Reorder the terms, Z ∞ ∞ X X (−1)n Hn H n2 1 1 ln3 (1 + x) (−1)n Hn2 5 = − dx + ζ(4). 2 2 n n 2 x 16 0 n=1 n=1 Substituting the results from (4.145) and (3.39) finishes the solution.

4.7.8

P∞

n=1

Show that

Solution

(−1)n H n 2

n4

∞ X (−1)n H n2 1 25 = ζ(2)ζ(3) − ζ(5). 4 n 8 32 n=1

Multiply both sides of (4.211) by

(−1)n n4

(4.219)

then take the summation,

Z 1 ∞ ∞ ∞ X (−1)n Hn X (−1)n H n2 ln(1 + x) X (−x)n − = dx n4 n4 x n3 0 n=1 n=1 n=1 Z 1 2 Z 1 3 Li2 (−x) ln(1 + x) Li3 (−x) IBP dx = − ζ(2)ζ(3) + dx. = x 8 x 0 0 Recall the relation involving the latter integral from (3.153), ∞ ∞ ∞ X (−1)n H n2 1 7 X Hn X (−1)n Hn = − ζ(2)ζ(3) − − . n4 4 8 n=1 n4 n4 n=1 n=1

Collecting the results from (4.6) and (4.150) completes the solution. For a different solution, see (4.43).

4.8. Harmonic Series with Binomial Coefficient in the Numerator

4.8

4.8.1

291

Harmonic Series with Binomial Coefficient in the Numerator (2n n ) Hn

P∞

n=1

4n

n

Show that

∞ X

2n n 4n



n=1

Solution (i)

Hn = 2ζ(2). n

(4.220)

By Taylor series,   ∞ ∞ 2n 2n X X 1 n n n √ = x =1+ xn n 4 1 − x n=0 4n n=1

or

∞ X

2n n 4n



1 − 1. (4.221) 1 −x n=1 R1 Multiply through by − ln(1−x) then integrate using − 0 xn−1 ln(1 − x)dx = Hnn , x ∞ X n=1

2n n 4n



Z

Hn =− n Z

|0 1

1

xn = √

Z 1 ln(1 − x) ln(1 − x) √ dx dx + x x 1−x {z } {z } |0 √

1−x=y

1−x=y

Z

1

ln(y) ln(y) = −4 dy + dy 2 0 1−y 0 1−y   2 1 1 write = + 1 − y2 1−y 1+y Z 1 Z 1 ln(y) ln(y) =− dy − 2 dy, 0 1−y 0 1+y and these two integrals are given in (3.2) and (3.8). Solution (ii)

we have

Setting x = 1 in (2.39):  √   ∞ 2n X 1− 1−x n Hn n √ x = 2 Li2 , 4n n 1+ 1−x n=1 ∞ X n=1

2n n 4n



Hn = 2 Li2 (1) = 2ζ(2). n

292

4.8.2

Chapter 4. Harmonic Series n (2n n ) (−1) Hn

P∞

4n

n=1

Show that

n

∞ X n=1

2n n 4n



 √  (−1)n Hn = 2 Li2 2 2 − 3 , n

(4.222)

Proof. Set x = −1 in (2.39), ∞ X



n=1

4.8.3

2n n 4n

P∞

n=1

Show that

(−1)n Hn = 2 Li2 n

(2n n ) Hn 4n

n2

∞ X n=1

Solution (i)

√ !  √  1− 2 √ = 2 Li2 2 2 − 3 . 1+ 2

2n n 4n



Hn 9 = ζ(3) − 4 ln(2)ζ(2). 2 n 2

Divide both sides of (4.221) by x then integrate,  Z Z ∞ 2n n X dx dx n x √ = . − n 4 n x x 1−x n=1

√ To evaluate the first integral, set 1 − x = y, Z Z dx 2 √ =− dy 1 − y2 1−x √     1− 1−x 1−y √ = ln = ln 1+y 1+ 1−x √   1+ 1−x √ multiply the argument of the log by 1+ 1−x   x √ = ln (1 + 1 − x)2 √ = ln(x) − 2 ln(1 + 1 − x). Substitute this integral back and note that the second integral is ln(x),  ∞ 2n n X √ n x = −2 ln(1 + 1 − x) + C, n n 4 n=1

(4.223)

4.8. Harmonic Series with Binomial Coefficient in the Numerator

293

where C = 2 ln(2) if we set x = 0. Then, we have  ∞ 2n n X √ n x = −2 ln(1 + 1 − x) + 2 ln(2). n 4 n n=1 Multiply through by − ln(1−x) then integrate using − x ∞ X n=1

2n n 4n



Hn =2 n2

1

Z

ln(1 +

|0

Z

R1 0

(4.224)

xn−1 ln(1 − x)dx =

Hn n ,

Z 1 ln(1 − x) 1 − x) ln(1 − x) dx −2 ln(2) dx x x 0 {z } {z } |

1−x=y

1−x=y

1

Z

1

y ln(1 + y) ln(y) ln(y) dy − 2 ln(2) dy 2 1−y 0 0 1−y 1 1 2y = − in the first integral} {write 1 − y2 1−y 1+y Z 1 Z 1 ln(1 + y) ln(y) ln(1 + y) ln(y) ln(y) dy − 4 dy −2 ln(2) dy 1−y 1 + y 1 −y 0 |0 {z } =8

1

Z =4 0

IBP

1

Z =4 0

ln(1 + y) ln(y) dy + 2 1−y

1

Z 0

2

ln (1 + y) dy − 2 ln(2) y

Z

1

0

ln(y) dy. 1−y

Putting together the results from (3.125), (3.38), and (3.2) finalizes the solution. Solution (ii) Set x = 1 in (2.40),  Z 1 ∞ 2n X ln(1 + t) ln(1 − t) n Hn = −4 ln(2) Li2 (1) + 2 Li3 (1) − 4 dt. n n2 4 t 0 n=1 The remaining integral is given in (3.115).

4.8.4

P∞

(2) (2n n ) Hn

n=1

4n

n

Show that

∞ X n=1

Solution

2n n 4n



(2)

Hn n

=

3 ζ(3). 2

Multiply both sides of (4.54): (2)

Hn Hn + n2 n

ζ(2) − = n

Z 0

1

xn−1 ln(x) ln(1 − x)dx

(4.225)

294

by

Chapter 4. Harmonic Series

(2n n) 4n

then take the summation over n ≥ 1, ∞ X

 (2)  ∞ ∞ 2n 2n X Hn X n Hn n 1 + − ζ(2) n2 4n n 4n n n=1 n=1 n=1 !  Z 1 ∞ 2n ln(x) ln(1 − x) X n n x = dx x 4n 0 n=1 2n n 4n



{recall the generating function in (4.221)}   Z 1 ln(x) ln(1 − x) 1 √ = − 1 dx x 1−x 0 Z 1 Z 1 ln(x) ln(1 − x) ln(x) ln(1 − x) √ = dx dx − x x 1 − x 0 0 {z } {z } | | 1−x→x

1

1

IBP 2

ln(x) ln(1 − x) 1 ln (x) √ dx − dx. 2 x(1 − x) 0 0 1−x {substitute the results from (3.110) and (3.3)} Z

Z

=

= 6ζ(3) − 6 ln(2)ζ(2), and the solution completes on collecting the result from (4.223) and writing  ∞ 2n X n 1 = 2 ln(2), 4n n n=1 which follows from (4.224) on setting x = 1.

4.8.5

P∞

n=1

Show that

(2) (2n n ) H2n

4n

n

∞ X n=1

Solution

2n n 4n



(2)

H2n 31 = 2πG + ζ(3). n 8

Divide both sides of (2.67):   ∞ 2n  X arcsin2 (x) 1 (2) (2) n √ =2 H2n − Hn x2n . n 4 4 1 − x2 n=1

(4.226)

4.8. Harmonic Series with Binomial Coefficient in the Numerator

by x then integrate using 2 ∞ X n=1

2n n 4n

(2)

R1 0

x2n−1 dx =

(2)

295

1 n,

Z π2 arcsin2 (x) x=sin u √ dx = u2 csc u du 2 0 x 1−x 0 Z π2   u    u  π2 IBP 2 −2 u ln tan du = u ln tan 2 2 0 0 | {z }



H2n − 14 Hn n

Z

1

=

0

{use the Fourier series of ln(tan x) given in (2.129)} ! Z π2 ∞ X cos((2n + 1)u) = −2 du u −2 2n + 1 0 n=0 Z π2 ∞ X 1 =4 u cos((2n + 1)u)du 2n + 1 0 n=0   ∞ X 1 π cos(nπ) sin(nπ) 1 =4 − − 2n + 1 2(2n + 1) (2n + 1)2 (2n + 1)2 n=0 {write cos(nπ) = (−1)n and sin(nπ) = 0, since n is an integer} ∞ ∞ X X 1 (−1)n +4 = 2π 2 (2n + 1) (2n + 1)3 n=0 n=0 {the first sum is the definition of the Catalan’s constant (see 1.205)} {and the second sum can be obtained from (1.85)} 7 = 2πG + ζ(3). 2 Thus,

∞ X n=1

 (2) ∞ (2) 2n 7 1 X n Hn H2n = 2πG + ζ(3) + . n 2 4 n=1 4n n

2n n 4n



The latter sum is given in (4.225).

4.8.6

P∞

n=1

2 (2n n ) Hn

4n

n

Show that

∞ X n=1

Solution (i)

2n n 4n



Hn2 21 = ζ(3). n 2

(4.227)

Multiply both sides of (2.71): Z 0

(2)

1

xn−1 ln2 (1 − x)dx =

Hn2 + Hn n

296

by

Chapter 4. Harmonic Series 1 2n 4n n



∞ X n=1

then take the summation, 2n n 4n



 (2) Z 1 2 ∞ 2n Hn2 X n Hn ln (1 − x) + = n n 4 n x 0 n=1

∞ X n=1

2n n 4n

!

 x

n

dx

{recall the generating function in (4.221)}   Z 1 2 ln (1 − x) 1 √ = − 1 dx x 1−x 0 Z 1 2 Z 1 2 ln (1 − x) ln (1 − x) √ = dx − dx x x 1−x 0 0 {z } | {z } | √

=8 Z =4 0

1−x=y

1−x=y

Z 0 1

1

2

ln (y) dy − 1 − y2

Z

ln2 (y) dy + 3 1+y

Z

1

ln2 (y) dy 1−y

1

ln2 (y) dy. 1−y

0

0

Rearranging the terms,   (2) Z 1 2 Z 1 2 ∞ ∞ 2n 2n 2 X X ln (y) ln (y) n Hn n Hn =4 dy + 3 dy − . 4n n 4n n 0 1+y 0 1−y n=1 n=1 Gathering the results from (3.9), (3.3), and (4.225) completes the solution. Solution (ii)

Multiply both sides of (2.38):  √   ∞ 2n X 2 1+ 1−x n n √ √ H x = ln , n 4n 1−x 2 1−x n=1

then integrate using − by − ln(1−x) x ∞ X n=1

=

0

xn−1 ln(1 − x)dx =

Hn n ,

√   ln(1 − x) 1+ 1−x √ √ ln dx 2 1−x 0 x 1−x   Z 1 √ ln(y) 1+y 1−x=y = −8 ln dy 2 2y 0 1−y Z 1 2 Z 1 ln(1 − x) ln(1 + x) ln (1 − x) 4 dx −4 dx x x 0 |0 {z }

2n n 4n

y= 1−x 1+x

R1

|x| < 1.



Hn2 = −2 n

Z

1

Z

1

1−x=y

=4 0

2

ln (y) dx − 4 1−y

Z 0

1

ln(1 − x) ln(1 + x) dx. x

These two integrals are given in (3.3) and (3.115).

4.8. Harmonic Series with Binomial Coefficient in the Numerator 2 (2n n ) Hn

P∞

4.8.7

Show that  ∞ 2n X H2 n=1

4n

n=1

n 4n

Solution

297

n n2

n2

= 32 Li4

  1 4 −14ζ(4)+7 ln(2)ζ(3)−8 ln2 (2)ζ(2)+ ln4 (2). 2 3 (4.228)

Multiply both sides of (2.39):  √   ∞ 2n X 1− 1−x n Hn n √ x = 2 Li2 4n n 1+ 1−x n=1

by − ln(1−x) then integrate using − x ∞ X

2n n 4n



n=1

Hn2 = −2 n2 1−x=y

=

R1

xn−1 ln(1 − x)dx = Hnn ,   √ Z 1 ln(1 − x) Li2 1−√1−x 1+ 1−x 0

0

Z −8

1

y ln(y) Li2

x 

1−y 1+y

dx



dy 1 − y2   Z 1 (1 − x) ln 1−x Li2 (x) 1−y 1+x 1+y =x = −4 dx x(1 + x) 0    Z 1 1−x 1 2 = −4 − ln Li2 (x)dx x 1+x 1+x 0 Z 1 Z 1 ln(1 − x) Li2 (x) ln(1 + x) Li2 (x) dx +4 dx = −4 x x |0 {z } |0 {z } 0

I1

Z −8 |0

1

I2

Z 1 ln(1 − x) Li2 (x) ln(1 + x) Li2 (x) dx +8 dx . 1+x 1+x {z } |0 {z } I3

I4

For I1 ,

1 1 5 I1 = − Li22 (x)|10 = − ζ 2 (2) = − ζ(4). 2 2 4 For I2 , expand ln(1 + x) in series, I3 =

Z ∞ X (−1)n−1 1 n−1 x Li2 (x)dx n 0 n=1

{this integral is given in (3.103)}

298

Chapter 4. Harmonic Series

=

  ∞ ∞ X X (−1)n Hn (−1)n−1 ζ(2) Hn − 2 = ζ(2)η(2) + n n n n3 n=1 n=1

{collect the result from (4.143)}   1 3 7 1 1 4 = 2 Li4 − ζ(4) + ln(2)ζ(3) − ln2 (2)ζ(2) + ln (2). 2 2 4 2 12 For I3 , perform integration by parts, I3 =

Z 1 1 ln2 (1 + x) ln(1 − x) 1 2 ln (2)ζ(2) + dx 2 2 0 x {this integral is given in (3.117)} 1 3 = ln2 (2)ζ(2) − ζ(4). 2 16

For I4 , make use of the dilogarithm reflection formula in (1.119), 1

ln(1 − x)[ζ(2) − ln(x) ln(1 − x) − Li2 (1 − x)] dx 1+x 0 Z 1 Z 1 Z 1 ln(x) ln2 (1 − x) ln(1 − x) Li2 (1 − x) ln(1 − x) dx − dx − dx = ζ(2) 1 + x 1 + x 1+x 0 0 0 {these three integrals are given in (3.27), (4.208), and (4.207)} Z

I4 =

∞ ∞ (3) (2) X X 5 1 Hn Hn = − ζ(4) − 2 ln(2)ζ(3) + ln2 (2)ζ(2) + 2 + 4 2 n2n n2 2 n n=1 n=1

{collect the results from (4.168) and (4.166)}   29 1 1 1 − ζ(4) − ln2 (2)ζ(2) + ln4 (2). = 3 Li4 2 16 4 8 The solution completes on grouping the four integrals.

4.9. Harmonic Series with Binomial Coefficient in the Denominator

4.9

299

Harmonic Series with Binomial Coefficient in the Denominator P∞

4.9.1

n=1

4n Hn (2n) n2 n

Show that

Solution (i)

∞ X 4n Hn 7  2n n2 = 6 ln(2)ζ(2) + 2 ζ(3). n=1 n

(4.229)

Let a = b = n in the beta function in (1.52): Z 1 a−1 x + xb−1 dx = B(a, b), (1 + x)a+b 0

we have

1

Z 0

Γ2 (n) 2 2xn−1 dx = = 2n (1 + x)2n Γ(2n) n n

or Z

1 n

2n n

=

1

0

xn−1 dx = (1 + x)2n

Z 0

1

1 x



x (1 + x)2

n dx.

(4.230)

4n Hn n

then consider the summation over n ≥ 1,  n ! Z 1 ∞ ∞ X 4n Hn 4x 1 X Hn  dx 2n n2 = n (1 + x)2 0 x n=1 n n=1

Multiply both sides by

{make use of (2.7) for the sum}     4x 1 2 4x 1 Li2 + ln 1 − dx = (1 + x)2 2 (1 + x)2 0 x   Z 1 2 + 2x 1−x IBP = ln(x) ln dx 1+x 0 x(1 − x)    Z 1 4 1−x 2 + ln(x) ln dx = x 1−x 1+x 0 Z 1 Z 1 ln(x) ln(1 − x) ln(x) ln(1 − x) ln(x) ln(1 + x) dx + 4 dx −2 dx x 1 − x x 0 |0 {z } Z

Z =2 0

1

1



1−x→x

Z −4 0

1

ln(x) ln(1 + x) dx 1−x

300

Chapter 4. Harmonic Series

Z =6 |0

1

Z 1 Z 1 ln(x) ln(1 + x) ln(x) ln(1 + x) ln(x) ln(1 − x) dx −2 dx −4 dx x x 1−x 0 0 {z } {z } | IBP 1

Z =3

0

IBP

ln2 (x) dx + 1−x

Z

1

0

ln2 (x) dx − 4 1+x

Z 0

1

ln(x) ln(1 + x) dx, 1−x

and the solution completes on grouping the results from (3.3), (3.9), and (3.125). Solution (ii)

Set x = 1 in (2.40),

Z 1 ∞ X ln(1 − t) ln(1 + t) 4n Hn  = −4 ln(2) Li (1) + 2 Li (1) − 4 dt. 2 3 2n n2 t 0 n=1 n This integral is given in (3.115). Solution (iii)

Replace z by

x in (2.49),

√ 1 ∞ arcsin( x) 1 X 4n xn− 2 √  = . 2 n=1 2n n 1−x n √ Multiply both sides by −2 ln(1−x) then integrate using − x

R1 0

xn−1 ln(1−x)dx =

√ Z 1 ∞ X 4n Hn arcsin( x) ln(1 − x) √  = −4 dx √ 2n n2 x 1−x 0 n=1 n Z π2 √ x=sin u = −8 u ln(cos u)du 0

{use the Fourier series of ln(cos u) given in (2.127)} # Z π2 " ∞ X (−1)n cos(2nu) = −8 u − ln(2) − du n 0 n=1 Z π2 Z π2 ∞ X (−1)n = 8 ln(2) u du + 8 u cos(2nu)du n 0 0 n=1    2 ∞ X (−1)n 1 π cos(nπ) π sin(nπ) = 8 ln(2) +8 − 2+ + 8 n 4n 4n2 4n n=1 {we have cos(nπ) = (−1)n and sin(nπ) = 0 for integer n} ∞ ∞ X X (−1)n 1 = 6 ln(2)ζ(2) − 2 +2 3 3 n n n=1 n=1 = 6 ln(2)ζ(2) − 2 Li3 (−1) + 2ζ(3). The solution completes on writing Li3 (−1) = − 34 ζ(3) given in (1.103).

Hn n ,

4.9. Harmonic Series with Binomial Coefficient in the Denominator

4.9.2

P∞

n=1

4n H2n (2n) n2 n

Show that

Solution

301

∞ X 35 4n H2n  2 = 3 ln(2)ζ(2) + ζ(3). 2n n 4 n=1 n

(4.231)

Multiply both sides of (2.49): ∞ arcsin(x) 1 X 4n x2n−1 √  = 2 n=1 2n n 1 − x2 n

by −4 ln(1 − x) then integrate using −

R1 0

x2n−1 ln(1 − x)dx =

H2n 2n ,

Z 1 ∞ X 4n H2n arcsin x ln(1 − x) √  = −4 dx 2n 2 n 1 − x2 0 n=1 n Z π2 x=sin u u ln(1 − sin u)du = −4 0 π 2

Z = −4 0 u π 4 − 2 =t

Z

=

  π u  u ln 2 sin2 − du 4 2 π 4

(16t − 4π) ln(2 sin2 t)dt

0

Z

π 4

Z (16t − 4π)dt + 2

= ln(2) 0

π 4

(16t − 4π) ln(sin t)dt 0

{use the Fourier serier of ln(sin t) given in (2.121)} # " Z π4 ∞ X cos(2nt) = ln(2) (16t − 4π)dt + 2 (16t − 4π) − ln(2) − dt n 0 0 n=1 Z π4 Z π4 ∞ X 1 = − ln(2) (16t − 4π)dt − 2 (16t − 4π) cos(2nt)dt n 0 0 n=1    π ∞ X π2 1 4 cos(2nt) 2π sin(2nt) 8t sin(2nt) 4 = − ln(2) − −2 + − 2 n n2 n n 0 n=1 " #  ∞ nπ X 1 4 cos 4 2 = 3 ln(2)ζ(2) − 2 − 2 2 n n n n=1  ∞ ∞ nπ X cos X 1 2 +8 = 3 ln(2)ζ(2) − 8 3 n n3 n=1 n=1 ( ) ∞ ∞  nπ  X X note that an cos = 0 − a2 + 0 + a4 + · · · = (−1)n a2n 2 n=1 n=1 Z

π 4

302

Chapter 4. Harmonic Series

= 3 ln(2)ζ(2) − 8 = 3 ln(2)ζ(2) −

∞ ∞ X X 1 (−1)n + 8 3 3 (2n) n n=1 n=1

∞ ∞ X X 1 (−1)n + 8 3 3 n n n=1 n=1

= 3 ln(2)ζ(2) − Li3 (−1) + 8ζ(3) 3 = 3 ln(2)ζ(2) + ζ(3) + 8ζ(3) 4 35 = 3 ln(2)ζ(2) + ζ(3). 4

4.9.3

P∞

n=1

4n Hn (2n) n3 n

Show that   ∞ X 4n Hn 1 1  = −8 Li + ζ(4) + 8 ln2 (2)ζ(2) − ln4 (2). 4 2n n3 2 3 n=1 n

Solution

Set z =

(4.232)

x in (2.50), ∞ √ 1 X 4n xn  2. arcsin2 ( x) = 2 n=1 2n n n

Multiply through by − ln(1−x) then integrate using − x

R1 0

xn−1 ln(1 − x)dx =

√ Z 1 ∞ X 4n Hn arcsin2 ( x) ln(1 − x)  dx 2n n3 = 2 x 0 n=1 n Z π2 √ x=sin θ = −8 x2 cot x ln(cos x)dx.

Hn n ,

(4.233)

0

To compute this integral, recall the Fourier series of cot x ln(cos x) given in (2.139): cot x ln(cos x) =

∞ X

(−1)n

Z 0

n=1

1

 1 − t n−1 t dt sin(2nx), 1+t

0 < x < π.

On multiplying both sides by x2 then integrating from x = 0 to π/2, we have Z 0

π 2

x2 cot x ln(cos x)dx

4.9. Harmonic Series with Binomial Coefficient in the Denominator

=

∞ X

(−1)n

1

Z 0

n=1

303

  Z π  2 2  1 − t n−1 x sin(2nx)dx t dt    1+t |0 {z } 

IBP

=

∞ X

(−1)n

1

Z 0

n=1

 cos(nπ) 3ζ(2) cos(nπ) 1 − t n−1 − t dt 1+t 4n3 4n  π sin(nπ) 1 − 3+ 4n 4n2

{note that cos(nπ) = (−1)n and sin(nπ) = 0 for integer n}   Z 1 ∞ X (−1)n 3ζ(2)(−1)n 1 1 − t n−1 t dt − − + 0 = (−1)n 4n3 4n 4n3 0 1+t n=1 {change the order of integration and summation} ! Z ∞ X 1 1 1−t tn 3ζ(2)tn (−t)n = − − dt 4 0 t(1 + t) n=1 n3 n n3  Z  1 1 1 2 = − (Li3 (t) + 3ζ(2) ln(1 − t) − Li3 (−t)) dt 4 0 t 1+t Z Z 1 1 Li3 (t) − Li3 (−t) 1 1 Li3 (t) − Li3 (−t) dt − dt = 4 0 t 2 0 1+t | | {z } {z } I1

3 + ζ(2) 4

For I1 ,

1

Z |0

I2

Z 1 3 ln(1 − t) ln(1 − t) dt − ζ(2) dt . t 2 1+t {z } {z } |0 I3

I4

7 15 I1 = Li4 (1) − Li4 (−1) = ζ(4) + ζ(4) = ζ(4). 8 8

For I2 , Z 1 ln(1 + t) Li2 (t) ln(1 + t) Li2 (−t) dt + dt t t 0 0 Z ∞ X 1 (−1)n 1 n−1 1 7 t Li2 (t)dt − Li22 (−t) 0 = ln(2)ζ(3) + 4 n 2 0 n=1

7 I2 = ln(2)ζ(3) − 4 IBP

Z

1

{recall the result from (3.103)}   ∞ X 7 (−1)n ζ(2) Hn 5 = ln(2)ζ(3) + − 2 − ζ(4) 4 n n n 16 n=1 =

∞ X 7 5 (−1)n Hn 5 ln(2)ζ(3) − ζ(4) − − ζ(4) 3 4 4 n 16 n=1

304

Chapter 4. Harmonic Series

{recall the result from (4.143)}   25 1 1 4 1 − ζ(4) + ln2 (2)ζ(2) − ln (2). = −2 Li4 2 16 2 12 For I3 , I3 = − Li2 (1) = −ζ(2). For I4 , Z

1

I4 = 0

Z 1 ∞ X ln(y) 1 dy = y n−1 ln(y)dy n 2 0 2−y 0 n=1   ∞ X 1 1 2 1 1 =− = − Li2 = ln (2) − ζ(2). 2 n n 2 2 2 2 n=1 ln(1 − t) 1−x=y dt = 1+t

Z

1

Combining all four integrals reveals π 2

Z 0

  1 4 1 1 x cot x ln(cos x)dx = Li4 ln (2). − ζ(4) − ln2 (2)ζ(2) + 2 8 24 2

Plugging this integral in (4.233) completes the solution.

P∞

4.9.4

(2)

n=1

4n Hn (2n) n2 n

Show that   ∞ (2) X 4n Hn 1 1  2 = 8 Li4 − ζ(4) + 4 ln2 (2)ζ(2) + ln4 (2). 2n n 2 3 n=1 n Solution

Multiply both sides of (4.54): (2)

Hn Hn + n2 n by

4n n(2n n)

(4.234)

ζ(2) − = n

Z

1

xn−1 ln(x) ln(1 − x)dx

0

then consider the summation, ∞ ∞ ∞ (2) X X 4n Hn X 4n Hn 4n 1    + − ζ(2) 2n n3 2n 2n n2 n2 n=1 n n=1 n n=1 n ! Z 1 ∞ ln(x) ln(1 − x) X 4n xn  = dx 2n n x 0 n=1 n √ {replace z by x in (2.49) to get the sum}

(4.235)

4.9. Harmonic Series with Binomial Coefficient in the Denominator

305

 √ √  ln(x) ln(1 − x) 2 x arcsin x √ dx x 1−x 0 Z π2 √ x=sin θ = 16 θ ln(sin θ) ln(cos θ)dθ.

Z =

1

0

Let θ →

π 2

Z

− θ using cos( π2 − θ) = sin θ and sin( π2 − θ) = cos θ, π 2

Z θ ln(sin θ) ln(cos θ)dθ =

0

π 2

0 π 2

Z

π = 2 ( Z add

2 Z π2

ln(cos θ) ln(sin θ)dθ − 0

0

π 2

 − θ ln(cos θ) ln(sin θ)dθ θ ln(cos θ) ln(sin θ)dθ. )

θ ln(sin θ) ln(cos θ)dθ to both sides then divide by 2

0

Z π π 2 ln(sin θ) ln(cos θ)dθ 4 0 {substitute the result from (3.109)} 3 15 = ln2 (2)ζ(2) − ζ(4). 4 32 =

(4.236)

Therefore, ∞ ∞ ∞ (2) X X 4n 1 15 4n Hn X 4n Hn    2 = 12 ln2 (2)ζ(2) − ζ(4). + − ζ(2) 2n n3 2n 2n 2 n n 2 n=1 n n=1 n n=1 n

We have ζ(2)

 2 ∞ X π 4n 1 15  = ζ(2) = ζ(4) 2n n2 2 2 n=1 n

follows from (2.50) on setting z = 1. On collecting this result along with (4.232), the solution is finalized. A different solution may be found in [28, p. 334]

4.9.5

P∞

n=1

2 4n Hn 2n 2 ( )n n

Show that   ∞ X 4n Hn2 1 81  = −24 Li + ζ(4) + 12 ln2 (2)ζ(2) − ln4 (2). (4.237) 4 2n n2 2 2 n=1 n

306

Chapter 4. Harmonic Series

Solution

Multiply both sides of (2.71): (2)

1

Z

xn−1 ln2 (1 − x)dx =

0

by

n

4

1

n (2n n)

Hn2 + Hn n

then take the summation,

! Z 1 2 ∞ ∞ ∞ (2) X ln (1 − x) X 4n xn 4n Hn2 X 4n Hn   2 =  dx 2n n2 + 2n 2n n n x 0 n=1 n n=1 n n=1 n √ {replace z by x in (2.49) to get the sum}  √ √  Z 1 2 x arcsin x ln (1 − x) = 2 √ dx x 1−x 0 Z π2 √ x=sin u = 16 u ln2 (cos u)du 0 u= π 2 −x

=

Z

π 2

16

π 2

0

Z

π 2

 − x ln2 (sin x)dx Z

2

ln (sin x)dx − 16

= 8π 0

π 2

x ln2 (sin x)dx,

0

where π 2

ln2 (t) arcsin t √ dt 1 − t2 0 0 Z 1 ln(t) arcsin2 t IBP dt = − t 0 {recall the series expansion of arcsin2 t in (2.50)} Z 1 ∞ ∞ 1 X 4n 1 1 X 4n 1 2n−1   4. =− t ln(t)dt = 2 n=1 2n n2 0 8 n=1 2n n n n Z

x ln2 (sin x)dx

sin x=t

Z

1

=

For the latter sum, set x = 1 in (2.59): (2) ∞ ∞ ∞ (2) X X 4n Hn−1 x2n 2 4n Hn x2n X 4n x2n    arcsin4 x = = − 2n 2n 2n n4 , 3 n2 n2 n=1 n n=1 n n=1 n

we obtain

∞ ∞ (2) X X 4n Hn 15 4n 1   = − ζ(4). 2n n4 2n 2 n 4 n=1 n n=1 n

(4.238)

4.9. Harmonic Series with Binomial Coefficient in the Denominator

307

Then, we have π 2

Z

x ln2 (sin x)dx =

0

∞ (2) 15 1 X 4n Hn  − ζ(4). 2 8 n=1 2n n 32 n

Substitute this integral back then rearrange the terms, Z π2 ∞ ∞ (2) X X 4n Hn2 15 4n Hn 2   ln (sin x)dx − 3 = ζ(4) + 8π . 2n n2 2n 2 n2 0 n=1 n n=1 n The solution ends on gathering the results from (3.108) and (4.234).

P∞

4.9.6

n=1

4n H2n (2n) n3 n

Show that   ∞ X 4n H2n 1 65 5  = −20 Li4 + ζ(4)+8 ln2 (2)ζ(2)− ln4 (2). (4.239) 2n 3 n 2 8 6 n=1 n Solution

Differentiate both sides of (4.230): n Z 1  1 x 1  = dx (1 + x)2 n 2n 0 x n

with respect to n, d d 1 = dn n 2n dn n

Z 0

1

1 x



x (1 + x)2

n dx

{use differentiation under the integral sign theorem given in (2.78)}  n Z 1 x 1 ∂ dx = (1 + x)2 0 x ∂n   n Z 1 1 x x = ln dx. (1 + x)2 (1 + x)2 0 x Let’s find the derivative of n 12n : By the definition of the central binomial coefficient: (n)   a Γ(a + 1) = , b Γ(b + 1)Γ(a − b + 1) we have 1 n

 2n = n

1 Γ2 (n + 1) · . n Γ(2n + 1)

308

Chapter 4. Harmonic Series

Use Γ(n + 1) = nΓ(n) given in (1.32), 1 n

 2n = n

1 n2 Γ2 (n) Γ2 (n) · = . n 2nΓ(2n) 2Γ(2n)

Differentiate both sides, d Γ2 (n) d 1  = dn n 2n dn 2Γ(2n) n {use Γ0 (n) = Γ(n)ψ(n) given in (1.167)} 2Γ(2n)Γ2 (n)ψ(n) − 2Γ(2n)Γ2 (n)ψ(2n) 2Γ2 (2n) Γ2 (n) = (ψ(n) − ψ(2n)) Γ(2n) {use ψ(n + 1) = Hn − γ given in (1.169)} 2 = (Hn−1 − γ − H2n−1 + γ) 2n n n   1 write Hn−1 = Hn − n   1 1 2  = Hn − − H2n + n 2n n 2n n 2H2n 1 2Hn = 2n − 2n − 2 2n . n n n n n n

=

Therefore, we have 2Hn 2H2n 1    2n − 2n − 2 2n = n n n n n n

Z 0

1

1 ln x

Now multiply both sides of (4.240) by



4n 2n2

x (1 + x)2



x (1 + x)2

n

then consider the summation,

∞ ∞ ∞ X 4n Hn X 4n H2n 1 X 4n 1   3 −  4 2n n3 − 2n n 2 n=1 2n n n n=1 n n=1 n | {z } S

 n   4x  X Z ∞ (1+x)2 1 11 x   dx = ln 2 2 0 x (1 + x)2 n n=1     Z 1 1 1 x 4x = ln Li2 dx 2 0 x (1 + x)2 (1 + x)2 

dx.

(4.240)

4.9. Harmonic Series with Binomial Coefficient in the Denominator 1



   1 2 2(x − 1) 1−x ln (x) + 2 Li2 (−x) ln dx 2 x(1 + x) 1+x 0 Z Z 1 1 ln2 (x) ln(1 + x) 1 1 ln2 (x) ln(1 − x) 5 dx − dx = − ζ(4) + 4 2 0 x 2 0 x | {z } | {z }

5 1 = − ζ(4) − 4 2

IBP

309

Z

I1

ln(1 − x) Li2 (−x) ln(1 + x) Li2 (−x) dx −2 dx x x {z } {z } |0

Z

1

|0

I3

− |0 1

−4 |0

1

Z

+2

Z

I2

1

Z

I4

2

1

2

ln (x) ln(1 − x) ln (x) ln(1 + x) dx + dx 1+x 1+x {z } |0 {z } Z

I5

I6

1

ln(1 − x) Li2 (−x) ln(1 + x) Li2 (−x) dx +4 dx . 1+x 1+x |0 {z } {z } Z

I7

I8

For I1 , expand ln(1 − x) in series, I1 = −

Z ∞ ∞ X X 1 1 n−1 2 1 = −2ζ(4). x ln (x)dx = −2 n 0 n4 n=1 n=1

For I2 , expand ln(1 + x) in series, I2 = −

Z ∞ ∞ X X −1)n 1 n−1 2 (−1)n 7 = ζ(4). x ln (x)dx = −2 4 n n 4 0 n=1 n=1

For I3 , expand Li2 (−x) in series, Z ∞ ∞ X X (−1)n 1 n−1 (−1)n Hn I3 = x ln(1 − x)dx = − . 2 n n3 0 n=1 n=1 For I4 ,

5 1 I4 = − Li22 (−1) = − ζ(4). 2 16

I5 is given in (3.131).

For I6 , set a = 2 in (3.89), I6 =

∞ X 7 (−1)n Hn ζ(4) + 2 . 4 n3 n=1

310

Chapter 4. Harmonic Series

For I7 , expand

Li2 (−x) 1+x

in series given in (2.3),

I7 = −

∞ X

(−1)

n

(2) Hn−1

Z

xn−1 ln(1 − x)dx

0

n=1

=

=

1

(2) ∞ X (−1)n Hn−1 Hn n n=1

∞ ∞ (2) X (−1)n Hn Hn X (−1)n Hn − . n n3 n=1 n=1

For I8 , apply integration by parts, Z 1 2 1 1 ln3 (1 + x) I8 = − ln (2)ζ(2) + dx 4 2 0 x {substitute the result from (3.39)} 21 1 1 = 3ζ(4) − ln(2)ζ(3) + ln2 (2)ζ(2) − ln4 (2) − 3 Li4 8 2 8

  1 . 2

For S, multiply both sides of (4.54): (2)

Hn Hn + 2 n n by

4n n(2n n)

ζ(2) = n

Z

1

xn−1 ln(x) ln(1 − x)dx

0

then take the summation, ∞ ∞ ∞ (2) X X 4n Hn X 4n Hn 4n 1    + − ζ(2) 2n n3 2n 2n n2 2 n n=1 n n=1 n n=1 n ! Z 1 ∞ ln(x) ln(1 − x) X 4n xn  dx = 2n n x 0 n=1 n

{make use of (2.49) for the sum}  √ √  ln(x) ln(1 − x) 2 x arcsin x √ = dx x 1−x 0 Z π2 √ x=sin θ = 16 θ ln(sin θ) ln(cos θ)dθ Z

1

0

{recall the result from (4.236)} 15 = 12 ln2 (2)ζ(2) − ζ(4). 2

(4.241)

4.9. Harmonic Series with Binomial Coefficient in the Denominator

311

By setting z = 1 in (2.50), we find ∞ X π2 4n 1  2n n2 = 2 = 3ζ(2). n=1 n

Substitute this result in (4.241), ∞ ∞ (2) X X 4n Hn 4n Hn   = − + 12 ln2 (2)ζ(2). 2n n3 2n 2 n n=1 n n=1 n

(4.242)

Adding (4.242) and (4.238) yields S :=

∞ ∞ X X 4n 1 4n Hn 15 2   = 12 ln (2)ζ(2) − ζ(4) − 2n n4 2n n3 . 4 n=1 n n=1 n

Collecting all integrals (I1 to I8 ) along with S yields   ∞ X 4n H2n 67 21 1  − ζ(4) + ln(2)ζ(3) − 7 ln2 (2)ζ(2) = 8 Li 4 2n 3 n 2 8 2 n=1 n ∞ ∞ ∞ (2) X X 1 (−1)n Hn Hn (−1)n Hn 3 X 4n Hn  3. + ln4 (2) + 4 −4 + 3 n n3 2 n=1 2n n n n=1 n=1

Substituting the results from (4.146), (4.143), and (4.232) completes the solution. P∞ Remark: Usually the two Mathematica commands for approximating n=1 f (n): NSum[f(n),{n,1,Infinity}] NSum[f(n),{n,1,Infinity},WorkingPrecision->10] don’t give the right approximation for series involving the binomial coefficient due to the slow convergence. The following replacement works fine and with high accuracy: major=Normal@Series[f(n),{n,Infinity,12}]; majorsum=Sum[major,{n,Infinity}]; majorsum+NSum[f(n)-major,{n,1,Infinity},NSumTerms->20, WorkingPrecision->20,Method->"WynnEpsilon"] On reaching the end of the book, I would like to say that there are still a wide range of results about the harmonic series left to be discovered by the reader by employing and manipulating the identities provided in the second chapter. Even though the book has presented different solutions for several problems, there are still more paths to take to reach the same results, since the realm of harmonic series is full of hidden secrets and magic.

Table of Mathematica Commands R{expression} J{expression} sin(x) sin(x) tan(x) sec(x) csc(x) cot(x) arcsin(x) arccos(x) arctan(x)  a b

Γ(x) ζ(x) β(x) η(x) ln(x) Lia (x) Hx (a) Hx Hx ψ(x) ψ (a) (x) limx→a f (x) limx→a f (x, y)

y→b da dxa f (x) ∂5 ∂x2 ∂y 3 f (x, y) 5 limx→a ∂x∂2 ∂y3 f (x, y) y→b

ComplexExpand[Re[expression]] ComplexExpand[Im[expression]] Sin[x] Cos[x] Tan[x] Sec[x] Csc[x] Cot[x] ArcSin[x] ArcCos[x] ArcTan[x] Binomial[a,b] Gamma[x] Zeta[x] DirichletBeta[x] DirichletEta[x] Log[x] PolyLog[a,x] HarmonicNumber[x] HarmonicNumber[x,a] Log[2]-(-1)ˆx LerchPhi[-1,1,x+1] PolyGamma[0,x] PolyGamma[a,x] Limit[f(x),{x->a}] Limit[f(x,y),{x,y}->{a,b}] D[f(x),{x,a}] D[f(x,y),{x,2},{y,3}] Normal[Series[D[f(x,y),{x,2},{y,3}],{x,a,0} ,{y,b,0}]]//FullSimplify//Expand 312

Table of Mathematica Commands

Rb Rab a

f (x)dx

Integrate[f(x),{x,a,b}]

f (x) ≈

NIntegrate[f(x),{x,a,b}] (or) NIntegrate[f(x),{x,a,b},WorkingPrecision ->12] Sum[f(n),{n,1,Infinity}] NSum[f(n),{n,1,Infinity}] (or) NSum[f(n),{n,1,Infinity},WorkingPrecision ->12] (or) major=Normal@Series[f(n),{n,Infinity,12}]; majorsum=Sum[major,{n,Infinity}]; majorsum+NSum[f(n)-major,{n,1,Infinity}, NSumTerms->20,WorkingPrecision->20,Method ->"WynnEpsilon"] EulerGamma Catalan E Pi I

P∞ f (n) Pn=1 ∞ n=1 f (n) ≈

γ G e π √

313

−1

References 1. Amrik, S.N.: Sum of series involving central binomial coefficient and harmonic number (2019). https://arxiv.org/pdf/1806.03998 2. Au, K.C.: Linear relations between logarithmic integrals of high weight and some closed-form evaluations (2019). https://arxiv.org/pdf/1910.12113.pdf 3. Balendran, S.: Mathematics Stack Exchange (2020). https://math.stackexchange.com/q/3959007 4. Bastien, G.: Elementary methods for evaluating Jordan’s sums (2013). https://arxiv.org/pdf/1301.7 662.pdf 5. Betal, K.C.: Mathematics Stack Exchange (2019). https://math.stackexchange.com/q/3260868 6. Bonar, D.D., Khoury, M.J.: Real Infinite Series, Classroom Resource Materials. The Mathematical Association of America, Washington, DC (2006) 7. Krantz, S.G.: Real Analysis And Foundations. Chapman & Hall/CRC Press, Washington, DC (2005) 8. Boyadzhi, K.N.: Power series with skew-harmonic numbers, dilogarithms, and double integrals (2017). https://arxiv.org/ftp/arxiv/papers/1701/1701.04484.pdf 9. D’Aurizio, J: Mathematics Stack Exchange (2016). https://math.stackexchange.com/q/1786846 10. Dutta, R.: Evaluation of a cubic Euler sum (2016). https://www.researchgate.net/publication/31170 3645 11. Flajolet, P., Salvy, B.: Euler Sums and Contour Integral Representations. Exp. Math. 7, 15–35 (1998) 12. Hintze, W.: Mathematics Stack Exchange (2020). https://math.stackexchange.com/q/3581958 13. Ian: Mathematics Stack Exchange (2015). https://math.stackexchange.com/q/1335512 14. Johnson, R.: Mathematics Stack Exchange (2012).https://math.stackexchange.com/q/116212 15. Johnson, R.: Mathematics Stack Exchange (2013).https://math.stackexchange.com/q/469785 16. Lewin, L.: Polylogarithms and Associated Functions. North-Holland, New York (1981) 17. Mathematics Stack Exchange (2020). https://math.stackexchange.com/q/3538399 18. Mathematics Stack Exchange (2020). https://math.stackexchange.com/q/3522967 19. Rao, S.: A formula of S. Ramanujan. Journal of Number Theory. 25, 1–19 (1987) 20. Song, M.: Mathematics Stack Exchange (2019). https://math.stackexchange.com/q/3325928 21. Song, M.: Mathematics Stack Exchange (2019). https://math.stackexchange.com/q/3337600 22. Vălean, C.I.: A master theorem of series and an evaluation of a cubic harmonic series, (2017). https://www.researchgate.net/publication/317143428 23. Vălean, C.I.: A new powerful strategy of calculating a class of alternating Euler sums (2019). https://www.researchgate.net/publication/333999069 24. Vălean, C.I.: A new proof for a classical quadratic harmonic series (2016). https://www.researchgate .net/publication/305699445 314

References

315

25. Vălean, C.I.: A note on two elementary logarithmic integrals, (2020). https://www.researchgate.net /publication/342703290 26. Vălean, C.I.: A simple strategy of calculating two alternating harmonic series generalizations (2019). https://www.researchgate.net/publication/333339284 27. Vălean, C.I.: A new perspective on the evaluation of a logarithmic integral, (2020). https://www.rese archgate.net/publication/339024876 28. Vălean, C.I.: Almost (Impossible) Integrals, Sums, and Series. Springer, New York (2019) 29. Vălean, C.I.: An easy approach to two classical Euler sums, (2020). https://www.researchgate.net/p ublication/339290253 30. Vălean, C.I.: Mathematics Stack Exchange (2021). https://math.stackexchange.com/q/3997602 31. Vălean, C.I.: Mathematics Stack Exchange (2019). https://math.stackexchange.com/q/3259984 32. Vălean, C.I.: Mathematics Stack Exchange (2019). https://math.stackexchange.com/q/3261717 33. Vălean, C.I.: Mathematics Stack Exchange (2018). https://math.stackexchange.com/q/3293328 34. Vălean, C.I.: Mathematics Stack Exchange (2018). https://math.stackexchange.com/q/3305999 35. Weisstein, E.W.: Analytic Continuation. https://mathworld.wolfram.com/AnalyticContinuation.html 36. Weisstein, E.W.: Catalan’s Constant. http://mathworld.wolfram.com/CatalansConstant.html 37. Weisstein, E.W.: Euler–Mascheroni Constant. https://mathworld.wolfram.com/Euler-MascheroniCo nstant.html 38. Weisstein, E.W.: Gamma Function. https://mathworld.wolfram.com/GammaFunction.html 39. Weisstein, E.W.: Gaussian Integral. https://mathworld.wolfram.com/GaussianIntegral.html 40. Weisstein, E.W.: Legendre Duplication Formula. https://mathworld.wolfram.com/LegendreDuplicat ionFormula.html 41. WikiBooks: Series developments. https://de.wikibooks.org/wiki/Formelsammlung_Mathematik: _Reihenentwicklungen#Arkusfunktionen 42. Wikipedia: Beta function. https://en.wikipedia.org/wiki/Beta_function 43. Wikipedia: Differentiation under the integral sign. https://en.wikipedia.org/wiki/Leibniz_integral_r ule

Index A Abel’s summation, 105, 110, 204, 275 Analytic continuation, 165 C Cauchy product, 23, 58, 73, 99, 100, 103, 104 Central binomial coefficient, 17, 83, 299, 302, 305, 307 Constant Catalan’s, 54, 177, 178, 238, 240, 266, 267, 272 Euler–Mascheroni, 49, 55 E Euler’s definition of gamma, 13 F Finite product, 6 Formula Digamma reflection, 47 Dilogarithm inversion, 36 Dilogarithm reflection, 34, 61, 66, 152 Euler’s, 8, 119, 158 Euler’s product, 123 Euler’s reflection, 14, 47, 138, 146, 166 Geometric series, 20, 23, 26, 90, 158, 254 Geometric sries, 7 Legendre duplication, 16 Polylogarithm inversion, 39, 144 Polylogarithm symmetry, 33 Geometric series, 263 Function Beta, 17, 95, 161, 166, 299 Digamma, 47, 160 Dirichlet beta, 26 Dirichlet eta, 25 Gamma, 10 Generating, 57, 59, 61, 63, 68, 70, 72, 75, 78, 80, 82, 86, 88, 280 Polygamma, 50

Polylogarithm, 30 Riemann zeta, 19, 22 Functional equation, 10 I Identity Algebraic, 167, 170, 182 Beta symmetry, 18 Beta–Gamma, 17 Digamma–Harmonic Number, 48, 155, 308 Landen’s dilogarithm, 35, 38, 66 Landen’s trilogarithm, 37, 63 Trigonometric, 117, 118, 305 Infinite product, 13, 124 Integral Caussian, 12 Divergent, 32 Double, 21, 224 Generalized improper logarithmic, 145 Generalized improper polylogarithmic, 139 Generalized inverse hyperbolic tangent, 156 Generalized inverse tangent, 144 Generalized logarithmic, 141 Generalized polylogarithmic, 140 Integration by parts, 10, 20, 69, 153, 154, 175, 187, 303, 309 L L’Hopital’s rule, 34, 126, 164 N Number Complex, 8 Harmonic, 40, 154, 156 Rational harmonic, 78, 202, 285 Skew harmonic, 44, 73, 194, 201, 277 316

Index

317

P Partial fraction decomposition, 100, 101, 142, 145, 174, 183 S Series Divergent, 44 Fourier, 112, 302 Generalized alternating harmonic, 198 Generalized alternating rational harmonic, 202 Generalized alternating skew harmonic, 201 Generalized rational harmonic, 202 Generalized skew harmonic, 194

Taylor, 8, 23, 74, 89, 291 Telescoping, 49 Stirling’s approximation, 11, 13 Sum Double, 7, 192, 196, 224 Generalized Euler, 204 Triple, 190, 195 T Theorem Binomial, 133 Differentiation under the integral sign, 51, 97, 138, 146, 205, 307 Lebesgue’s dominated convergence, 15