Improper integral of $\frac{x}{e^x-1}$ This integral came up in an exercise on the estimation of the specific heat of a 1-D solid and is probably a standard integral, possibly one that can be solved by contour integration:
\begin{equation}
\int_0^{+\infty} \frac{x}{e^x-1} dx
\end{equation} 
I have some rudimentary knowledge of contour integrals, but I can't come up with a proper path, also because of the many singularities along the imaginary axis. Any suggestion?

Make a change or variables, $e^{-x}=u$, and write the integral as 
$$\int_0^\infty \frac{x e^{-x}}{1 - e^{-x}}dx$$
Now substituting gives
$$-\int_0^1 \frac{\log u}{1 - u}du $$
This is a known integral that evaluates to $\dfrac{\pi^2}{6}$. 
If you want to prove it, you can use the dilogarithm. Let $1-u=x$, so that
$$-\int_0^1 \frac{\log(1 - x)}{x}dx$$
Now, since we're working on $(0,1)$ it is legitimate to use
$$\frac{{ - \log \left( {1 - x} \right)}}{x} = \sum_{n = 1}^\infty  {\frac{{{x^{n - 1}}}}{n}} $$
Integrating termwise gives
$$\int_0^t \frac{-\log(1 - x)}{x} dx = \sum_{n = 1}^\infty \frac{t^n}{n^2}  = \mathrm{Li}_2 (t)$$
Evaluating at $t=1$ gives
$$ -\int_0^1 \frac{\log(1 - x)}{x} dx = \sum_{n = 1}^\infty \frac{1}{n^2}  = \frac{\pi^2}{6}$$
