TITLE: Computing $\int_0^\infty\mathrm{d} x\frac{x}{e^x+1}$ with contour integration
QUESTION [10 upvotes]: Let's set:
$$
\int_0^\infty\mathrm{d}x\frac{x}{e^x+1}=I.
$$
I would like to compute it using, presumably, the methods of complex analysis and contour integration.

REPLY [8 votes]: $\newcommand{\+}{^{\dagger}}
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$\large\tt\mbox{Just another one !!!}$

In the first step we integrate by parts:
  \begin{align}
&\color{#00f}{\large\int_{0}^{\infty}{x \over \expo{x} + 1}\,\dd x}
=\int_{0}^{\infty}\ln\pars{1 + \expo{-x}}\,\dd x=
\int_{0}^{\infty}\sum_{\ell = 1}^{\infty}{\pars{-1}^{\ell + 1} \over \ell}\,
\expo{-\ell x}\,\dd x
\\[3mm]&=\sum_{\ell = 1}^{\infty}{\pars{-1}^{\ell + 1} \over \ell}\
\overbrace{\int_{0}^{\infty}\expo{-\ell x}\,\dd x}^{\ds{1 \over \ell}}
=\sum_{\ell = 1}^{\infty}{\pars{-1}^{\ell + 1} \over \ell^{2}}
=\sum_{\ell = 0}^{\infty}
\bracks{{1 \over \pars{2\ell + 1}^{2}} - {1 \over \pars{2\ell + 2}^{2}}}
\\[3mm]&=\bracks{\sum_{\ell = 1}^{\infty}{1 \over \ell^{2}}
-\sum_{\ell = 1}^{\infty}{1 \over \pars{2\ell}^{2}}}
-{1 \over 4}\sum_{\ell = 1}^{\infty}{1 \over \ell^{2}}
=\half\ \underbrace{\sum_{\ell = 1}^{\infty}{1 \over \ell^{2}}}
_{\ds{\zeta\pars{2} = {\pi^{2} \over 6}}} =\ \color{#00f}{\large{\pi^{2} \over 12}}
\end{align}

$\zeta\pars{z}$ is the
Riemann Zeta Function .