Prove that $\int_0^\infty \frac{\ln x}{x^n-1}\,dx = \Bigl(\frac{\pi}{n\sin(\frac{\pi}{n})}\Bigr)^2$ This question inspired me to ask the following.
Prove that
$$I_n = \int_0^\infty \frac{\ln x}{x^n-1}\,dx = \left(\frac{\pi}{n\sin\left(\frac{\pi}{n}\right)}\right)^2,$$
for $\Re(n)>1$.
For some cases there is a nice specific form of $I_n$. For example
$$\begin{align}
I_2 & = \frac{\pi^2}{4} \\
I_3 & = \frac{4\pi^2}{27} \\
I_4 & = \frac{\pi^2}{8} \\
I_5 & = \frac{2\left(5+\sqrt 5\right)\pi^2}{125} \\
I_6 & = \frac{\pi^2}{9} \\
I_7 & = \frac{2\pi^2}{49\left(1-\sin\left(\frac{3\pi}{14}\right)\right)} \\
I_8 & = \frac{\left(2+\sqrt 2\right)\pi^2}{32}
\end{align}$$
 A: We can use a contour integral in the complex plane, as I showed here for the case $n=3$.  Now, however, we use
$$\oint_C dz \frac{\log^2{z}}{z^n-1}$$
where $C$ is the modified keyhole contour about the positive real axis of outer radius $R$ and inner radius $\epsilon$.  
Let's evaluate this integral over the contours.  As before, there are $8$ pieces to evaluate, as follows:
$$\int_{\epsilon}^{1-\epsilon} dx \frac{\log^2{x}}{x^n-1} + i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{\log^2{\left (1+\epsilon e^{i \phi}\right )}}{(1+\epsilon e^{i \phi})^n-1} \\ + \int_{1+\epsilon}^R   dx \frac{\log^2{x}}{x^n-1} + i R \int_0^{2 \pi} d\theta \, e^{i \theta} \frac{\log^2{\left (R e^{i \theta}\right )}}{R^n e^{i n \theta}-1} \\ + \int_R^{1+\epsilon} dx \frac{(\log{x}+i 2 \pi)^2}{x^n-1} + i \epsilon \int_{2 \pi}^{\pi} d\phi \, e^{i \phi} \frac{(\log{\left (1+\epsilon e^{i \phi}\right )}+i 2 \pi)^2}{(1+\epsilon e^{i \phi})^n-1} \\ + \int_{1-\epsilon}^{\epsilon} dx \frac{(\log{x}+i 2 \pi)^2}{x^n-1} + i \epsilon \int_{2 \pi}^0 d\phi \, e^{i \phi} \frac{\log^2{\left (\epsilon e^{i \phi}\right )}}{\epsilon^n e^{i n \phi}-1} $$ 
As $R \to \infty$, the fourth integral vanishes as $\log^2{R}/R^{n-1}$.  As $\epsilon \to 0$, the second integral vanishes as it is $O(\epsilon^n)$, while the eighth integral vanishes as $\epsilon \log^2{\epsilon}$.  This leaves the first, third, fifth, sixth and seventh integrals, which in the above limits, become
$$PV \int_0^{\infty} dx \frac{\log^2{x} - (\log{x}+i 2 \pi)^2}{x^n-1} + i \frac{4 \pi^3}{n}$$
The residue computation is a little more involved, because we now have $n-1$ poles at which we need to evaluate residues.  The contour integral is thus
$$i 2 \pi \sum_{k=1}^{n-1} \frac{-4 \pi^2 k^2/n^2}{n e^{i 2 (n-1) \pi k/n}} = -i \frac{8 \pi^3}{n^3} \sum_{k=1}^{n-1} k^2 \, e^{-i 2 (n-1) \pi k/n} $$
The sum is doable, but the algebra is a bit hideous.  The result is
$$\sum_{k=1}^{n-1} k^2 \, e^{-i 2 (n-1) \pi k/n} = \frac12 \left ( \frac{n}{\sin^2{\frac{\pi}{n}}} - n^2\right ) - i \frac12 n^2 \cot{\frac{\pi}{n}} $$
Equating real and imaginary parts of both equations for the contour integral yields
$$\int_0^{\infty} dx \frac{\log{x}}{x^n-1} = \frac{\pi^2}{n^2 \sin^2{\frac{\pi}{n}}} $$
$$PV \int_0^{\infty} dx \frac{1}{x^n-1} = -\frac{\pi}{n} \cot{\frac{\pi}{n}} $$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
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\begin{align}
\color{#66f}{\large
\overbrace{\int_{0}^{\infty}{\ln\pars{x} \over x^{n} - 1}\,\dd x}^{\ds{\color{#c00000}{x^{n}\ \mapsto\ x}}}}\ &=\
-\int_{0}^{\infty}{\ln\pars{x^{1/n}} \over 1 - x}\,{1 \over n}\,x^{1/n - 1}\,\dd x
=-\,{1 \over n^{2}}\int_{0}^{\infty}{x^{1/n - 1}\ln\pars{x} \over 1 - x}\,\dd x
\\[5mm]&=-\,{1 \over n^{2}}\lim_{\mu\ \to\ 0}\partiald{}{\mu}
\int_{0}^{\infty}{x^{\mu + 1/n - 1} - x^{1/n - 1} \over 1 - x}\,\dd x
\\[5mm]&=-\,{1 \over n^{2}}\lim_{\mu\ \to\ 0}\partiald{}{\mu}
\\&\bracks{
\int_{0}^{1}{x^{\mu + 1/n - 1} - x^{1/n - 1} \over 1 - x}\,\dd x
+\int_{1}^{0}{x^{-\mu - 1/n + 1} - x^{-1/n + 1} \over 1 - 1/x}\,
\pars{-\,{\dd x \over x^{2}}}}
\\[5mm]&=-\,{1 \over n^{2}}\lim_{\mu\ \to\ 0}\partiald{}{\mu}
\int_{0}^{1}{x^{\mu + 1/n - 1} - x^{-\mu - 1/n}\over 1 - x}\,\dd x
\\[5mm]&={1 \over n^{2}}\lim_{\mu\ \to\ 0}\partiald{}{\mu}\bracks{
\int_{0}^{1}{1 - x^{\mu + 1/n - 1} \over 1 - x}\,\dd x
-\int_{0}^{1}{1 - x^{-\mu - 1/n} \over 1 - x}\,\dd x}
\\[5mm]&={1 \over n^{2}}\lim_{\mu\ \to\ 0}\partiald{}{\mu}\bracks{
\Psi\pars{\mu + {1 \over n}} - \Psi\pars{-\mu - {1 \over n} + 1}}
\end{align}
where $\ds{\Psi\pars{z}}$ is the
Digamma Function $\color{#000}{\bf 6.3.1}$.

\begin{align}\color{#66f}{\large
\int_{0}^{\infty}{\ln\pars{x} \over x^{n} - 1}\,\dd x}
&={1 \over n^{2}}\bracks{\Psi'\pars{1 \over n} + \Psi'\pars{-\,{1 \over n} + 1}}
\end{align}

With the Euler Reflection Formula $\color{#000}{\bf 6.4.7}$
$$
\Psi'\pars{1 - z} + \Psi'\pars{z} = - \pi\,\totald{\cot\pars{\pi z}}{z}
=\pi^{2}\csc^{2}\pars{\pi z}
$$

we'll find
  $$
\color{#66f}{\large
\int_{0}^{\infty}{\ln\pars{x} \over x^{n} - 1}\,\dd x
={\pi^{2} \over n^{2}}\,\csc^{2}\pars{\pi \over n}}
$$

A: \begin{align}
\int_0^\infty \frac{\ln x}{x^n-1}\,dx
 =& \>\frac1n\int_0^\infty  \int_0^\infty \frac{1}{(1+y)(x^n+y)} dy \ dx \\
=&\>\frac{\pi }{n^2}\csc\frac\pi n \int_0^\infty\frac{y^{\frac1n-1}}{1+y}dy
=\frac{\pi^2}{n^2}\csc^2\frac\pi n
\end{align}
A: $$
\begin{align}
\int_0^\infty\frac{\log(x)}{x^n-1}\mathrm{d}x
&=\int_{-\infty}^\infty\frac{x}{e^{nx}-1}e^x\,\mathrm{d}x\\
&=\int_0^\infty\frac{x}{e^{nx}-1}e^x\,\mathrm{d}x+\int_0^\infty\frac{x}{1-e^{-nx}}e^{-x}\,\mathrm{d}x\\
&=\int_0^\infty x(e^{(1-n)x}+e^{(1-2n)x}+e^{(1-3n)x}+\dots)\,\mathrm{d}x\\
&+\int_0^\infty x(e^{-x}+e^{(-1-n)x}+e^{(-1-2n)x}+\dots)\,\mathrm{d}x\\
&=\frac1{(n-1)^2}+\frac1{(2n-1)^2}+\frac1{(3n-1)^2}+\dots\\
&+1+\frac1{(n+1)^2}+\frac1{(2n+1)^2}+\frac1{(3n+1)^2}+\dots\\
&=\frac1{n^2}\sum_{k\in\mathbb{Z}}\frac1{\left(k+\frac1n\right)^2}\\
&=\frac{\pi^2}{n^2}\csc^2\left(\frac\pi{n}\right)
\end{align}
$$
where the last step uses the derivative of $(7)$ from this answer
