TITLE: How to find $ \int_0^\infty \dfrac x{1+e^x}\ dx$
QUESTION [8 upvotes]: How to find
$$
\int_0^\infty \dfrac x{1+e^x}\ dx=\ ...?
$$
I don't know where should I start with. The correct answer from my textbook is $\frac{\pi^2}{12}$.
This is my homework with 10 questions but I can only answer 9 questions, this one I'm stuck. Can anyone help me? Thanks.

REPLY [9 votes]: $\newcommand{\+}{^{\dagger}}
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\begin{align}
&\color{#00f}{\large\int_{0}^{\infty}{x \over \expo{x} + 1}\,\dd x}
=\int_{0}^{\infty}x\pars{{1 \over \expo{x} + 1} - {1 \over \expo{x} - 1}}
\,\dd  x + \int_{0}^{\infty}{x \over \expo{x} - 1}\,\dd x
\\[3mm]&=-2\int_{0}^{\infty}{x \over \expo{2x} - 1}\,\dd x
 + \int_{0}^{\infty}{x \over \expo{x} - 1}\,\dd x
=\half\int_{0}^{\infty}{x\expo{-x} \over 1 - \expo{-x}}\,\dd x
\\[3mm]&=-\,\half\int_{0}^{\infty}\ln\pars{1 - \expo{-x}}\,\dd x
=\half\int_{0}^{\infty}\sum_{n = 1}^{\infty}{\expo{-nx} \over n}\,\dd x
=\half\sum_{n = 1}^{\infty}{1 \over n}\
\overbrace{\int_{0}^{\infty}\expo{-nx}\,\dd x}^{\ds{=\ {1 \over n}}}
\\[3mm]&=\half\
\overbrace{\sum_{n = 1}^{\infty}{1 \over n^{2}}}
^{\ds{\zeta\pars{2} = {\pi^{2} \over 6}}}
=\color{#00f}{\large{\pi^{2} \over 12}}
\end{align}

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