Integral

[Tolaso J Kos]

Evaluate the following: \( \displaystyle \int_0^\infty \ln^n \left( \frac{e^x}{e^x-1} \right) \, dx \,\,\,\, n\in \mathbb{N}\) although we have no problem if \( n \) is real or complex as long as the integral converges.

[Tolaso J Kos]

Successively we have:
$$\begin{aligned}
\int_{0}^{\infty}\ln^n \left ( \frac{e^x}{e^x-1} \right )\,dx &\overset{u=-\ln \left ( 1-e^{-x} \right )}{=\! =\! =\! =\! =\! =\! =\! =\! =\! }\int_{0}^{\infty}\frac{u^n e^{-u}}{1-e^{-u}}\,du \\
&= \sum_{k=0}^{\infty}\int_{0}^{\infty}u^n e^{-(k+1)u}\,du=n! \sum_{k=0}^{\infty}\frac{1}{\left ( k+1 \right )^{n+1}}\\
&= \Gamma (n+1)\zeta (n+1)
\end{aligned}$$