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How can I find a closed form for

$$\int_0^{\infty}\log^n\left(\frac{e^x}{e^x-1}\right)dx, n\in\mathbb{N}$$

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  • $\begingroup$ Maple finds it fo concrete $n$:$2\,\zeta \left( 3 \right)$ if $n=2$, ${\frac {3703808}{255}}\,{\pi }^{16} $ if $n=15$ ans so on. $\endgroup$ – user64494 Nov 18 '13 at 6:25
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Substitute $u = -\log{(1-e^{-x})}$, then after a little algebra, you will find that $dx = -du/(e^u-1)$, and the integral becomes

$$\int_0^{\infty} du \frac{u^n e^{-u}}{1-e^{-u}}$$

This is a well known integral that has value

$$\Gamma(n+1) \zeta(n+1)$$

This may be shown by Taylor expanding the denominator to get

$$\sum_{k=0}^{\infty} \int_0^{\infty} du \, u^n \, e^{-(k+1) u} = n!\sum_{k=0}^{\infty}\frac{1}{(k+1)^{n+1}}$$

It should be noted in passing that restriction of $n$ to integers is not necessary and $n$ may be a real, or even a complex number so long as the integral converges.

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