# Show that : int_0^infty e^(-x)/(x^2)(1/(1-e^(-x))-1/x-1/2)^2dx=7/36-ln A+(zeta(3))(2pi^2)

Show that :
${\int }_{0}^{\mathrm{\infty }}\frac{{\text{e}}^{-x}}{{x}^{2}}{\left(\frac{1}{1-{\text{e}}^{-x}}-\frac{1}{x}-\frac{1}{2}\right)}^{2}\phantom{\rule{thinmathspace}{0ex}}\text{d}x=\frac{7}{36}-\mathrm{ln}A+\frac{\zeta \left(3\right)}{2{\pi }^{2}}$
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Define
$f\left(s\right)={\int }_{0}^{\mathrm{\infty }}{x}^{s-1}\frac{{e}^{-x}}{{x}^{2}}{\left(\frac{1}{1-{e}^{-x}}-\frac{1}{x}-\frac{1}{2}\right)}^{2}dx.$
This defines an analytic function on the domain $\mathrm{Re}\left(s\right)>0$ and the problem is to evaluate f(1).
We have
$f\left(s\right)={\int }_{0}^{\mathrm{\infty }}{x}^{s-3}{e}^{-x}\left(\frac{1}{\left(1-{e}^{-x}{\right)}^{2}}+\frac{1}{{x}^{2}}+\frac{1}{4}-\frac{2}{x\left(1-{e}^{-x}\right)}-\frac{1}{1-{e}^{-x}}+\frac{1}{x}\right)dx$

For $\mathrm{Re}\left(s\right)>4$, integrating by parts on the first term gives
$f\left(s\right)={\int }_{0}^{\mathrm{\infty }}\left(\frac{\left(s-3\right){x}^{s-4}}{{e}^{x}-1}+{x}^{s-5}{e}^{-x}+\frac{{x}^{s-3}{e}^{-x}}{4}-\frac{2{x}^{s-4}}{{e}^{x}-1}-\frac{{x}^{s-3}}{{e}^{x}-1}+{x}^{s-4}{e}^{-x}\right)dx.$
$={\int }_{0}^{\mathrm{\infty }}\left(\frac{\left(s-5\right){x}^{s-4}}{{e}^{x}-1}+{x}^{s-5}{e}^{-x}+\frac{{x}^{s-3}{e}^{-x}}{4}-\frac{{x}^{s-3}}{{e}^{x}-1}+{x}^{s-4}{e}^{-x}\right)dx.$
Again assuming Re(s)>4, this gives us
$f\left(s\right)=\left(s-5\right)\mathrm{\Gamma }\left(s-3\right)\zeta \left(s-3\right)+\mathrm{\Gamma }\left(s-4\right)+\frac{1}{4}\mathrm{\Gamma }\left(s-2\right)-\mathrm{\Gamma }\left(s-2\right)\zeta \left(s-2\right)+\mathrm{\Gamma }\left(s-3\right),$
We may write
$f\left(s\right)=\frac{\left(s-4\right)\left(s-5\right)\zeta \left(s-3\right)+1+\frac{1}{4}\left(s-4\right)\left(s-3\right)-\left(s-4\right)\left(s-3\right)\zeta \left(s-2\right)+s-4}{\left(s-4\right)\left(s-3\right)\left(s-2\right)\left(s-1\right)}\mathrm{\Gamma }\left(s\right)$
and so
$f\left(1\right)=\underset{s\to 1}{lim}\left(\frac{\left(s-5\right)\zeta \left(s-3\right)}{\left(s-3\right)\left(s-2\right)\left(s-1\right)}-\frac{\zeta \left(s-2\right)+\frac{1}{12}}{\left(s-2\right)\left(s-1\right)}+\frac{\frac{1}{3}\left(s-4\right)\left(s-3\right)+s-3}{\left(s-4\right)\left(s-3\right)\left(s-2\right)\left(s-1\right)}\right)$
$=\underset{s\to 1}{lim}\left(\frac{\left(s-5\right)\zeta \left(s-3\right)}{\left(s-3\right)\left(s-2\right)\left(s-1\right)}-\frac{\zeta \left(s-2\right)+\frac{1}{12}}{\left(s-2\right)\left(s-1\right)}+\frac{1}{3\left(s-4\right)\left(s-2\right)}\right)$
$=-2{\zeta }^{\prime }\left(-2\right)+{\zeta }^{\prime }\left(-1\right)+\frac{1}{9}$
$=\frac{\zeta \left(3\right)}{2{\pi }^{2}}+\frac{1}{12}-\mathrm{ln}A+\frac{1}{9}$
$=\frac{7}{36}-\mathrm{ln}A+\frac{\zeta \left(3\right)}{2{\pi }^{2}}.$
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