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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?

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    Generally speaking, $\int_0^\infty\frac{x^n}{e^x-1}dx=n!\cdot\zeta(n+1)$2013-11-04

2 Answers 2

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The integrand can be expressed as a geometric series with first term $te^{-t}$ and common ratio $e^{-t}$. Integrate term-by-term (after justifying it, of course) and see if you don't recognize the result as $\zeta(2)$.

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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...

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    @Eric: I didn't want to bring it up. You're right, of course. ;)2011-09-14