A formula for $\int\limits_0^\infty (\frac{x}{e^x-1})^n dx$

Question

The Stirling numbers of the first kind $\begin{bmatrix} n \\ k \end{bmatrix}$ are defined by $\sum\limits_{k=0}^n \begin{bmatrix} n \\ k \end{bmatrix}x^k:=\prod\limits_{k=0}^{n-1}(x+k)$ with $n\in\mathbb{N}_0$ .

Which proof exists for

$$\frac{1}{n}\int\limits_0^\infty \left(\frac{x}{e^x-1}\right)^n dx=\sum\limits_{k=1}^n (-1)^{k-1}\begin{bmatrix} n \\ n-k+1 \end{bmatrix}\zeta(k+1)\quad,\quad n\in\mathbb{N}$$

?

I am also looking for literature where a proof is written, so that it isn’t necessary to write one down here.

Answer 1

Your identity is just a consequence of integration by parts, together with the identity: $$ \int_{0}^{+\infty}\frac{x^k}{e^x-1}\,dx = k!\cdot\zeta(k+1)\tag{1} $$ that simply follows by expanding $\frac{1}{e^x-1}$ as $e^{-x}+e^{-2x}+e^{-3x}+\ldots$
So the whole point is just to compute $$ \frac{d^{n-1}}{dx^{n-1}}\frac{1}{(e^x-1)^n} \tag{2}$$ by recalling, for instance, that by stars and bars we have: $$ \frac{1}{(1-x)^{m+1}}=\sum_{n\geq 0}\binom{m+n}{n}x^n.\tag{3} $$

Answer 2

A more general formula holds. Take a look at formula (10) in this paper by Adamchik for $p=n+1$ and $a=n$. The hypergeometric function on the l.h.s. can be written as the integral given just before formula (16). By letting $x=\ln(1/t)$ one can easily obtain your integral.