# Evaluate integral

#### idontknow

$$\displaystyle I= \int_{0}^{\infty} \dfrac{\sin^2 (px)}{x(e^{2x}-1)}dx \:$$ , p-parameter .

#### idontknow

What the book suggests is : $$\displaystyle dI/dp = \frac{1}{2} \int_{0}^{\infty} [\sum_{k=1}^{\infty}e^{-kx}\sin(px) ]dx$$.
Now use $$\displaystyle \coth (x) = 1/x +\sum_{n=1}^{\infty} \frac{2x}{x^2 +n^2 \pi^2 }$$ to get $$\displaystyle I_p =\ln \frac{\sinh(\pi p )}{\pi p}$$.
Anyone knows where these come from ? how to apply them and evaluate the integral ?

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#### idontknow

$$\displaystyle I'(p) =\frac12\int_0^\infty\frac{\sin(px)}{e^x-1}{\rm d}x=\frac12\int_0^\infty\frac{\sin(px)}{e^x-1}{\rm d}x=\frac12\int_0^\infty\sin(px)\frac{e^{-x}}{1-e^{-x}}{\rm d}x=\frac12\int_0^\infty\sin(px)\left[\sum_{k\ge1}e^{-kx}\right]{\rm d}x$$.

\displaystyle \begin{align*} I'(p)=\frac12\int_0^\infty\left[\sum_{k\ge1}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*}.

$$\displaystyle \coth(\pi p)=\frac1{\pi p}+\sum_{n\ge1}\frac{2\pi p}{(\pi p)^2+n^2\pi^2}\implies\pi\coth(\pi p)-\frac1p=\sum_{n\ge1}\frac{2p}{p^2+n^2} .$$

$$\displaystyle I'(p)=\frac14\sum_{k\ge1}\frac{2p}{k^2+p^2}=\frac14\left[\pi\coth(\pi p)-\frac1p\right]$$ ; $$\displaystyle \; \; I(p)=\int I'(p){\rm d}p\implies I(p)=\frac14\int\pi\coth(\pi p)-\frac1p{\rm d} p=\frac14\log(\sinh(\pi p))-\log p+c$$.

$$\displaystyle I(0)=0 ; \; \; \lim_{p\to0}\frac{\sinh(\pi p)}p=\pi ; \; \; c=\frac14\log\pi .$$

$$\displaystyle I(p)=\frac14\log\left(\frac{\sinh(\pi p)}{\pi p}\right) .$$

topsquark
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