TITLE: Evaluate the integral $\int_0^\infty \frac{t}{e^t-1}\mathrm dt$
QUESTION [5 upvotes]: An integral related to the zeta function at the point $2$ is given by
$$\zeta(2) = \int\nolimits_0^\infty \dfrac{t}{e^t - 1}\mathrm dt$$
How to calculate this integral?

REPLY [5 votes]: Somewhat equivalent to Gerry's answer: let $t=-\log(1-u)$, giving the integral
$$-\int_0^1 \frac{\log(1-u)}{u}\mathrm du$$
Expand the logarithm as a series, swap summation and integration, and then you should be able to see something familiar...