Integral $I=\int_0^1\frac{\ln x}{x^n-1}dx$ 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
 A: We have
$$I_n=\int_0^1\frac{\ln x}{x^n-1}dx=-\sum_{k=0}^\infty\int_0^1x^{nk}\ln x dx=\sum_{k=0}^\infty\frac1{(nk+1)^2}$$
and for $n=1$ and $n=2$ the two sums are known.
A: (Edited for rigour, credit for this goes to RandomVariable).
$$I(\alpha)=-\int_0^b\frac{x^{\alpha}dx}{1-x^n}$$
$$I(\alpha)=-\left(\frac{b^{\alpha+n+1}}{ \alpha+n+1}+\frac{b^{\alpha+2n+1}}{ \alpha+2n+1}+\cdots\right)$$
Now, as $I'(\alpha)=\int_0^b\frac{\log(x) x^{\alpha}dx}{x^n-1}$, or rather, $J(\alpha)= \lim_{b \to 1}I'(\alpha)$ is something the value of which we want to find,
$$I'(\alpha)=-\log(b)\left(\frac{b^{\alpha+n+1}}{ \alpha+n+1}+\frac{b^{\alpha+2n+1}}{ \alpha+2n+1}+\cdots\right)+\left(\frac{b^{\alpha+n+1}}{( \alpha+n+1)^2}+\frac{b^{\alpha+2n+1}}{( \alpha+2n+1)^2}+\cdots\right)$$
So 
$$J(\alpha)=\lim_{b \to 1}I'(\alpha)=\left(\frac{1}{( \alpha+n+1)^2}+\frac{1}{( \alpha+2n+1)^2}+\cdots\right).$$
Then set $\alpha=0$ and we have accordance with the other answer:
$$J(1)=\left(\frac{1}{(n+1)^2}+\frac{1}{(2n+1)^2}+\cdots\right).$$
A: $\newcommand{\+}{^{\dagger}}
 \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\down}{\downarrow}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
 \newcommand{\wt}[1]{\widetilde{#1}}$
$$
\Psi\pars{s + 1} + \gamma = \int_{0}^{1}{1 - t^{s} \over 1 - t}
\quad\imp\quad
\Psi'\pars{1 \over n} = -\int_{0}^{1}{t^{1/n - 1}\ln\pars{t} \over 1 - t}\,\dd t
$$

With $\ds{t = x^{n}}$:
  $$
\Psi'\pars{1 \over n}=
-\int_{0}^{1}{\pars{x^{n}}^{1/n - 1}\ln\pars{x^{n}} \over 1 - x^{n}}\,nx^{n - 1}
\,\dd x=n^{2}\int_{0}^{1}{\ln\pars{x} \over x^{n}- 1}\,\dd x
$$

$$\color{#00f}{\large%
\int_{0}^{1}{\ln\pars{x} \over x^{n}- 1}\,\dd x
= {1 \over n^{2}}\,\Psi'\pars{1 \over n}}
$$
Note that $\ds{\Psi'\pars{1} = {\pi^{2} \over 6}}$ and
$\ds{\Psi'\pars{\half} = {\pi^{2} \over 2}}$.

Some details are given in my previous answer.

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