TITLE: Prove $\int^\infty_0\frac x{e^x-1}dx=\frac{\pi^2}{6}$
QUESTION [21 upvotes]: I know that $$\int^\infty_0\frac x{e^x-1}dx=\frac{\pi^2}{6}$$ For substituting $u=2$ into $$\zeta(u)\Gamma(u)=\int^\infty_0\frac{x^{u-1}}{e^x-1}dx$$
However, I suspect that there is an easier proof, maybe by the use of complex analysis. I haven't learnt the zeta function yet. All I know is the above formula and $\zeta(2)=\frac{\pi^2}{6}$. Can we can use the above integral to find out some of the $\zeta$'s value?

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_{x\ =\ 0}^{x\ \to\ \infty}x\,\dd\bracks{\ln\pars{1 - \expo{-x}}}
=-\int_{0}^{\infty}\ln\pars{1 - \expo{-x}}\,\dd x
\\[3mm]&=\sum_{n = 1}^{\infty}{1 \over n}\int_{0}^{\infty}\expo{-nx}\,\dd x
=\sum_{n = 1}^{\infty}{1 \over n^{2}}\
\overbrace{\int_{0}^{\infty}\expo{-x}\,\dd x}^{\ds{=\ 1}}\
=\sum_{n = 1}^{\infty}{1 \over n^{2}}=\color{#00f}{\large{\pi^{2} \over 6}}
\end{align}