[itex]\int_0^\infty \frac{u}{e^u - 1}[/itex] I know that this integral is [itex]\frac{\pi^2}{6}[/itex], just from having seen it before, but I'm not really sure if I can evaluate it directly to show that. I know that: [itex] \zeta(x) = \frac{1}{\Gamma(x)} \int_0^\infty \frac{u^{x-1}}{e^u -1} du [/itex] Does the value [itex]\frac{\pi^2}{6}[/itex] come from using other methods of showing the result for [itex]\zeta(2)[/itex] and solving the equation, or is that integral another way of evaluating [itex]\zeta(2)[/itex]?