Improper integral with log.

[Tolaso J Kos]

Let $n \in \mathbb{N}$. Find a closed form for the integral:

$$\int_0^\infty \ln^n \left( \frac{e^x}{e^x-1} \right) \, {\rm d}x$$

[r9m]

\begin{align*}\int_0^{\infty} \log^n \left(\frac{e^{x}}{1-e^{x}}\right)\,dx&= (-1)^{n}\int_0^{\infty} \log^n (1-e^{-x})\,dx\\&= (-1)^{n}\int_0^1 \frac{\log^n (1-x)}{x}\,dx \qquad \text{ change of variable } e^{-x} \mapsto x \\&= (-1)^{n}\int_0^1 \frac{\log^n x}{1-x}\,dx = n!\zeta(n+1)\end{align*}

[Tolaso J Kos]

\begin{align*}\int_0^{\infty} \log^n \left(\frac{e^{x}}{1-e^{x}}\right)\,dx&= (-1)^{n}\int_0^{\infty} \log^n (1-e^{-x})\,dx\\&= (-1)^{n}\int_0^1 \frac{\log^n (1-x)}{x}\,dx \qquad \text{ change of variable } e^{-x} \mapsto x \\&= (-1)^{n}\int_0^1 \frac{\log^n x}{1-x}\,dx = n!\zeta(n+1)\end{align*}

Thank you r9m and welcome aboard!!

Allow me to supplement the last calculation of yours where you calculate the integral $ \displaystyle \int_0^1 \frac{\ln^n x}{1-x}\, {\rm d}x$

We have that:

$$\begin{align*}
\int_{0}^{1}\frac{\ln^n x}{1-x}\, {\rm d}x &= \int_{0}^{1}\ln^n x \sum_{k=0}^{\infty}x^k \, {\rm d}x\\
&=\sum_{k=0}^{\infty}\int_{0}^{1}x^k \ln^n x \, {\rm d}x \\
&=(-1)^n n!\sum_{k=0}^{\infty} \frac{1}{\left ( k+1 \right )^{n+1}}\\
&= (-1)^n n! \zeta(n+1)
\end{align*}$$

The integral $\int_0^1 x^k \ln^n x \, {\rm d}x$ is evaluated by applying parts $k$ times.