Show that $\dfrac{1}{e^x-1} = \dfrac{1}{x} - \dfrac{1}{2} + 2x \sum_{n=1}^{\infty} \dfrac{1}{(4n^2 \pi^2 +x^2)}$.

Question

In the fifth method of deriving the functional equation for Riemann's zeta function in Titchmarsh's book the following 'expansion' is assumed without proof : $$\dfrac{1}{e^x-1} = \dfrac{1}{x} - \dfrac{1}{2} + 2x \sum_{n=1}^{\infty} \dfrac{1}{4n^2 \pi^2 +x^2}.$$ I have never encountered this type of expansion and also WolframAlpha does not give a clue for the identity. How this equality is derived?

I have a second question regarding the integral just below this identity in the book

How to compute the integral $\int_0^{\infty} \dfrac{x^s}{4n^2 \pi^2 +x^2} dx$? Again using WolframAlpha I couldn't get any hint.

Answer 1

In this answer, it is shown that $$ \sum_{n=1}^\infty\frac1{n^2+u^2}=\frac\pi{2u}\coth(\pi u)-\frac1{2u^2}\tag1 $$ Therefore, $$ \begin{align} \sum_{n=1}^\infty\frac1{4n^2\pi^2+x^2} &=\frac1{4\pi^2}\sum_{n=1}^\infty\frac1{n^2+u^2}\tag{2a}\\ &=\frac1{4\pi^2}\left(\frac\pi{2u}\coth(\pi u)-\frac1{2u^2}\right)\tag{2b}\\ &=\frac1{4x}\coth\left(\frac{x}2\right)-\frac1{2x^2}\tag{2c}\\ &=\frac1{4x}\frac{e^x+1}{e^x-1}-\frac1{2x^2}\tag{2d}\\ &=\frac1{2x}\frac1{e^x-1}+\frac1{4x}-\frac1{2x^2}\tag{2d}\\ \end{align} $$ Explanation:
$\text{(2a):}$ $x=2\pi u$
$\text{(2b):}$ apply $(1)$
$\text{(2c):}$ undo the substitution from $\text{(2a)}$
$\text{(2d):}$ $\coth\left(\frac{x}2\right)=\frac{e^x+1}{e^x-1}$
$\text{(2e):}$ $\frac{e^x+1}{e^x-1}=\frac2{e^x-1}+1$

Solving $(2)$ for $\frac1{e^x-1}$ gives $$ \frac1{e^x-1}=\frac1x-\frac12+2x\sum_{n=1}^\infty\frac1{4n^2\pi^2+x^2}\tag3 $$ In this answer, it is shown that $$ \int_0^\infty\frac{x^n}{1+x^m}\,\mathrm{d}x=\frac{\pi}{m}\csc\left(\pi\frac{n+1}{m}\right)\tag4 $$ Therefore, $$ \begin{align} \int_0^\infty\frac{x^s}{4n^2\pi^2+x^2}\,\mathrm{d}x &=(2n\pi)^{s-1}\int_0^\infty\frac{x^s}{1+x^2}\,\mathrm{d}x\tag{5a}\\ &=(2n\pi)^{s-1}\frac\pi2\csc\left(\pi\frac{s+1}2\right)\tag{5b}\\ &=(2n\pi)^{s-1}\frac\pi2\sec\left(\frac{s\pi}2\right)\tag{5c} \end{align} $$ Explanation:
$\text{(5a):}$ substitute $x\mapsto2n\pi x$
$\text{(5b):}$ apply $(4)$
$\text{(5c):}$ $\csc(x+\pi/2)=\sec(x)$

Answer 2

This is an alternative answer to your first question. Let $n\ge 2$ be an integer and $x \ne 1$ be a real number. By partial fraction decomposition, we have \begin{align*} \frac{1}{{x^{2n} - 1}} & = \frac{1}{n}\frac{1}{{x^2 - 1}} + \frac{1}{n}\sum\limits_{k = 1}^{n - 1} {\frac{{x\cos \left( {\frac{{\pi k}}{n}} \right) - 1}}{{x^2 - 2x\cos \left( {\frac{{\pi k}}{n}} \right) + 1}}} \\ & = \frac{1}{n}\frac{1}{{x^2 - 1}} - \frac{{n - 1}}{{2n}} + \frac{1}{n}\sum\limits_{k = 1}^{n - 1} {\frac{{(x - 1)^2 + x - 1}}{{(x - 1)^2 + 4x\sin ^2 \left( {\frac{{\pi k}}{{2n}}} \right)}}} . \end{align*} Substituting $x = 1 + \frac{t}{2n}$ with $t\ne 0$ a real number, we find $$ \frac{1}{{\left( {1 + \frac{t}{{2n}}} \right)^{2n} - 1}} = \frac{1}{{t\left( {1 + \frac{t}{{4n}}} \right)}} - \frac{{n - 1}}{{2n}} + \sum\limits_{k = 1}^{n - 1} {\frac{{\frac{{t^2 }}{n} + 2t}}{{t^2 + 4\pi ^2 k^2 \left( {1 + \frac{t}{{2n}}} \right)\left( {\left( {\frac{{2n}}{{\pi k}}} \right)^2 \sin ^2 \left( {\frac{{\pi k}}{{2n}}} \right)} \right)}}} . $$ Passing to the limit $n\to+\infty$ yields the desired result: $$ \frac{1}{{{\rm e}^t - 1}} = \frac{1}{t} - \frac{1}{2} + \sum\limits_{k = 1}^{\infty} {\frac{{2t}}{{t^2 + 4\pi ^2 k^2 }}} . $$