Evaluate integral

Dec 2015
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Earth
\(\displaystyle I= \int_{0}^{\infty} \dfrac{\sin^2 (px)}{x(e^{2x}-1)}dx \: \) , p-parameter .
 
Dec 2015
1,076
166
Earth
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 ?
 
Last edited:
Dec 2015
1,076
166
Earth
The result is a typo. it must be log instead of ln .
\(\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*}

\).

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

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

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

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