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?