A generalized integral

Let $m, n \in \mathbb{N}_{n \geq 2}$. Prove that:

$$\int_{0}^{1}\frac{1-x^{n-1}}{\left ( 1-x \right )\left ( 1-x^n \right )}\left ( -\ln x \right )^{m-1}\, {\rm d}x= \left ( 1-\frac{1}{n^m} \right )\Gamma(m)\zeta(m)$$

Solution

We note that:

\begin{align*}
\int_{0}^{1}\frac{1-x^{n-1}}{\left ( 1-x \right )\left ( 1-x^n \right )}\left ( -\ln x \right )^{m-1}\,{\rm d}x &=\int_{0}^{1}\left ( \frac{1}{1-x} - \frac{x^{n-1}}{1-x^n} \right )\left ( - \ln x \right )^{m-1}\, {\rm d}x \\
 &= \int_{0}^{1}\frac{\left ( -\ln x \right )^{m-1}}{1-x}\, {\rm d}x - \int_{0}^{1}\frac{x^{n-1}}{1-x^n} \left ( -\ln x \right )^{m-1}\, {\rm d}x \qquad (1)
\end{align*}

However:

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

and similarly:

\begin{align*}
\int_{0}^{1}\frac{x^{n-1}}{1-x^n}\left ( -\ln x  \right )^{m-1}\, {\rm d}x &\overset{u=x^n}{=\! =\! =\!} \frac{1}{n^m}\int_{0}^{1}\frac{\left ( -\ln u \right )^{m-1}}{1-u} \, {\rm d}u\\
 &= \frac{1}{n^m}\Gamma (m)\zeta(m)
\end{align*}

Returning back to $(1)$ the result follows.