How to evaluate this integral to get pi^2/6:

[hb1547]

\int_0^\infty \frac{u}{e^u - 1}

I know that this integral is \frac{\pi^2}{6}, just from having seen it before, but I'm not really sure if I can evaluate it directly to show that.

I know that:

\zeta(x) = \frac{1}{\Gamma(x)} \int_0^\infty \frac{u^{x-1}}{e^u -1} du

Does the value \frac{\pi^2}{6} come from using other methods of showing the result for \zeta(2) and solving the equation, or is that integral another way of evaluating \zeta(2)?

 

[sunjin09]

\int_0^\infty \frac{u}{e^u - 1}

I know that this integral is \frac{\pi^2}{6}, just from having seen it before, but I'm not really sure if I can evaluate it directly to show that.

I know that:

\zeta(x) = \frac{1}{\Gamma(x)} \int_0^\infty \frac{u^{x-1}}{e^u -1} du

Does the value \frac{\pi^2}{6} come from using other methods of showing the result for \zeta(2) and solving the equation, or is that integral another way of evaluating \zeta(2)?

never mind ... my complex variable technique is rusty ...

 

[hb1547]

Anyone else have any input?