Integral with logarithm and exponential

Let $k$ be a positive integral. Evaluate the integral:

$$\int_{0}^{\infty}\frac{e^x-1}{e^x+1}\log^k\left(\frac{e^x+1}{e^x-1}\right)\,{\rm d}x$$

(Ovidiu Furdui)

Solution

We have successively:

\begin{align*}
\int_{0}^{\infty} \frac{e^x-1}{e^x+1} \log^k \left ( \frac{e^x+1}{e^x-1} \right )\, {\rm d}x &\overset{\begin{subarray}{c}y=\frac{e^x-1}{e^x+1} \\
{\rm d}x= - \frac{2}{y^2-1} \, {\rm d}y \end{subarray}}{=\! =\! =\! =\! =\! =\! =\! =\!  =\! =\!}2(-1)^k \int_{0}^{1} \frac{y}{1-y^2} \log^k y \, {\rm d}y \\
 &= 2(-1)^k \int_{0}^{1}y \log^k y  \sum_{m=0}^{\infty} y^{2m} \, {\rm d}y\\
 &= 2(-1)^k \sum_{m=0}^{\infty} \int_{0}^{1}y^{2m+1} \log^k y \, {\rm d}y\\
 &= 2k! \sum_{m=0}^{\infty} \frac{1}{\left ( 2m+2 \right )^{k+1}}  \\
 &= 2k! \sum_{m=0}^{\infty} \frac{1}{2^{k+1} \left ( m+1 \right )^{k+1}}\\
 &= \frac{k!}{2^k} \sum_{m=0}^{\infty} \frac{1}{\left ( m+1 \right )^{k+1}} \\
 &=\frac{k!}{2^k} \sum_{m=1}^{\infty} \frac{1}{m^{k+1}} \\
 &=\frac{k!}{2^k} \zeta(k+1)
\end{align*}