Evaluate $I(p)=\int _{0}^{\infty} \frac{\sin^2 (px)}{x(e^{2x}-1)}dx,~p\neq0.$

Question

The integral I cannot evaluate is $I(p)=\displaystyle \int \limits_{0}^{\infty} \dfrac{\sin^2 (px)}{x(e^{2x}-1)}dx \;\; , p\neq0.$

I must evaluate the integral using these two following statements :

(1) $\displaystyle \frac{dI(p)}{dp} = \frac{1}{2} \int \limits_{0}^{\infty} \bigg(\sum_{k=1}^{\infty}e^{-kx}\sin(px) \bigg)dx .$

(2) $\displaystyle \coth (x) = 1/x +\sum_{n=1}^{\infty} \frac{2x}{x^2 +n^2 \pi^2 } $.

The result is $ \displaystyle I(p) =\ln \frac{\sinh(\pi p )}{\pi p}$ , but I need to understand the method and the statement (1). Thanks for replies .

Answer 1

The key behind $\bf (1)$ is that we can interchange the integral and the differential operator when suitable conditions are met. Here, we differentiate w.r.t. $p$. In more detail we have \begin{align*} I(p):=\int_0^\infty\frac{\sin^2(px)}x\frac{{\rm d}x}{e^{2x}-1}\implies\frac{\rm d}{{\rm d}p}I(p)&=\int_0^\infty\frac\partial{\partial p}\frac{\sin^2(px)}x\frac{{\rm d}x}{e^{2x}-1}\\ &=\int_0^\infty\frac{2\sin(px)\cos(px)x}x\frac{{\rm d}x}{e^{2x}-1}\\ &=\int_0^\infty\sin(2px)\frac{{\rm d}x}{e^{2x}-1}&&;~2x\mapsto x\\ &=\frac12\int_0^\infty\frac{\sin(px)}{e^x-1}{\rm d}x \end{align*} Now utilize the geometric series $\sum_{k\ge0}e^{-kx}=\frac1{1-e^{-x}}$. Note that we do not use the one given by $\sum_{k\ge0}e^{kx}=\frac1{1-e^x}$ by simple convergence considerations. Therefore, we continue to arrive at $\bf(1)$ $$I'(p)=\frac12\int_0^\infty\frac{\sin(px)}{e^x-1}{\rm d}x=\frac12\int_0^\infty\sin(px)\frac{e^{-x}}{1-e^{-x}}{\rm d}x=\frac12\int_0^\infty\sin(px)\left[\sum_{k\ge1}e^{-kx}\right]{\rm d}x$$ Interchange the order of summation and integration to conclude \begin{align*} I'(p)=\frac12\int_0^\infty\left[\sum_{k\le1}e^{-kx}\sin(px)\right]&=\frac12\sum_{k\ge1}\int_0^\infty e^{-kx}\sin(px){\rm d}x\\ &=\frac12\sum_{k\ge1}\left[\frac{e^{-kx}}{k^2+p^2}(k\sin(px)+p\cos(px))\right]_0^\infty\\ &=\frac14\sum_{k\ge1}\frac{2p}{k^2+p^2} \end{align*} Now, using $\bf(2)$ for $x=\pi p$ we deduce the identity $$\coth(\pi p)=\frac1{\pi p}+\sum_{n\ge1}\frac{2\pi p}{(\pi p)^2+n^2\pi^2}\implies\pi\coth(\pi p)-\frac1p=\sum_{n\ge1}\frac{2p}{p^2+n^2}$$ Thus $$I'(p)=\frac14\sum_{k\ge1}\frac{2p}{k^2+p^2}=\frac14\left[\pi\coth(\pi p)-\frac1p\right]$$ Integrating back yields $$I(p)=\int I'(p){\rm d}p\implies I(p)=\frac14\int\pi\coth(\pi p)-\frac1p{\rm d} p=\frac14\log(\sinh(\pi p))-\log p+c$$ Now, note that $I(0)=0$ and $\lim_{p\to0}\frac{\sinh(\pi p)}p=\pi$ by L'Hospital. Therefore, $c=\frac14\log\pi$.

$$\therefore~I(p)=\int_0^\infty\frac{\sin^2(px)}x\frac{{\rm d}x}{e^{2x}-1}=\frac14\log\left(\frac{\sinh(\pi p)}{\pi p}\right)$$

WolframAlpha agrees on this for $p=1$ suggesting that in your given solution is a typo.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \mrm{I}\pars{p} & = \bbox[5px,#ffd]{\left.\int_{0}^{\infty}{\sin^{2}\pars{px} \over x\pars{\expo{2x} - 1}} \,\dd x\,\right\vert_{\large\ p\ \not=\ 0}} = \int_{0}^{\infty}{\sin^{2}\pars{px/2} \over \expo{x} - 1}\,{\dd x \over x} \\[5mm] & = {1 \over 2}\int_{0}^{\infty}{1 - \cos\pars{px} \over \expo{x} - 1}\,{\dd x \over x} = {1 \over 2}\,\Re\int_{0}^{\infty}{1 + \ic px - \expo{\ic px} \over \expo{x} - 1}\,{\dd x \over x} \\[5mm] & = {1 \over 2}\,\Re\int_{0}^{\infty}{\expo{-x} + \ic px\expo{-x} - \expo{-\pars{1 - \ic p}x} \over 1 - \expo{-x}}\,{\dd x \over x} \\[5mm] & \stackrel{\large\,\,\,\,\,\,\, t\ =\ \expo{-x}}{=}\,\,\, -\,{1 \over 2}\,\Re\int_{0}^{1}{1 - \ic p\ln\pars{t} - t^{-\ic p} \over 1 - t}\,{\dd t \over \ln\pars{t}} \\[5mm] & = -\,{1 \over 2}\,\Re\int_{0}^{1}{1 - \ic p\ln\pars{t} - t^{-\ic p} \over 1 - t}\,\pars{-\int_{0}^{\infty}t^{\,\xi}\,\dd\xi}\dd t \\[5mm] & = {1 \over 2}\,\Re\int_{0}^{\infty}\int_{0}^{1}{t^{\xi} - \ic pt^{\xi}\ln\pars{t} - t^{\xi - \ic p} \over 1 - t}\,\dd t\,\dd\xi \\[5mm] & = {1 \over 2}\,\Re\int_{0}^{\infty}\bracks{% \int_{0}^{1}{1 - t^{\xi - \ic p} \over 1 - t}\,\dd t - \int_{0}^{1}{1 - t^{\xi} \over 1 - t}\,\dd t - \ic p\int_{0}^{1}{t^{\xi}\ln\pars{t} \over 1 - t}\,\dd t}\dd\xi \\[5mm] & = {1 \over 2}\,\Re\int_{0}^{\infty}\bracks{% \Psi\pars{\xi - \ic p + 1} - \Psi\pars{\xi + 1}}\dd\xi \\[5mm] & = {1 \over 2}\,\Re\lim_{\xi \to \infty} \ln\pars{\Gamma\pars{\xi + 1 - \ic p} \over \Gamma\pars{\xi + 1}} - {1 \over 2}\,\Re\ln\pars{\Gamma\pars{1 - \ic p} \over \Gamma\pars{1}} \\[5mm] & =\ \overbrace{{1 \over 2}\lim_{\xi \to \infty} \ln\pars{\verts{\Gamma\pars{\xi + 1 - \ic p} \over \Gamma\pars{\xi + 1}}}}^{\ds{=\ 0\,,\ \mbox{See}\ \color{red}{below}}}\ -\ {1 \over 4}\ln\pars{\Gamma\pars{1 - \ic p}\Gamma\pars{1 + \ic p}} \\[5mm] & = -\,{1 \over 4} \ln\pars{\Gamma\pars{1 - \ic p}\ic p\Gamma\pars{\ic p}} = -\,{1 \over 4} \ln\pars{\ic p\,{\pi \over \sin\pars{\pi\ic p}}} \\[5mm] & = \bbx{\large{{1 \over 4}\ln\pars{\sinh\pars{\pi p} \over \pi p}}} \\ & \end{align}

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$\color{red}{\mbox{Note that}}$ \begin{align} &\verts{\Gamma\pars{\xi + 1 - \ic p} \over \Gamma\pars{\xi + 1}} = \verts{\pars{\xi - \ic p}! \over \xi!} \,\,\,\stackrel{\mrm{as}\ \xi\ \to\ \infty}{\sim}\,\,\, \verts{\root{2\pi}\pars{\xi - \ic p}^{\xi - \ic p + 1/2} \expo{-\xi + \ic p} \over \root{2\pi}\xi^{\xi + 1/2}\expo{-\xi}} \\[5mm] & \stackrel{\mrm{as}\ \xi\ \to\ \infty}{\sim}\,\,\, \verts{\xi^{\xi - \ic p + 1/2}\pars{1 - \ic p/\xi}^{\xi} \over \xi^{\xi + 1/2}} \,\,\,\stackrel{\mrm{as}\ \xi\ \to\ \infty}{\sim}\,\,\, \verts{\xi^{-\ic p}\expo{-\ic p}} = \verts{\expo{-\ic p\ln\pars{\xi}}\expo{-\ic p}} = {\Large 1} \end{align}