TITLE: Integral $I=\int_0^1\frac{\ln x}{x^n-1}dx$
QUESTION [6 upvotes]: Hi I am trying to obtain a closed form for$$
I_n=\int_0^1\frac{\ln x}{x^n-1}dx, \quad n\geq 1.
$$
This integral is quite nice and generates many other known closed form results such as 
$$
\int_0^1\frac{\ln x}{x^2-1} dx=\frac{\pi^2}{8}, \quad \int_0^1\frac{\ln x}{x-1} dx=\frac{\pi^2}{6}.
$$
In these cases I use residue methods, but am unsure how to generalize as in this case of $I_n$.
Thank you

REPLY [2 votes]: $$
\begin{align}
\int_0^1\frac{\log(x)}{x^n-1}\mathrm{d}x
&=\int_0^\infty\frac{u}{1-e^{-nu}}e^{-u}\,\mathrm{d}u\\
&=\int_0^\infty u(e^{-u}+e^{-(n+1)u}+e^{-(2n+1)u}+\dots)\,\mathrm{d}u\\
&=1+\frac1{(n+1)^2}+\frac1{(2n+1)^2}+\dots\\
&=\frac1{n^2}\left(\frac1{(0+1/n)^2}+\frac1{(1+1/n)^2}+\frac1{(2+1/n)^2}+\dots\right)\\[4pt]
&=\frac1{n^2}\psi^{\,\prime}\left(\frac1n\right)
\end{align}
$$
The order of integration and summation can be interchanged using Tonelli's Theorem since everything is positive in the last integral.