integration of a series in $x\in(0,1)$ Prove that if $p>0$,
$$ \int_0^1 \frac{x^{p-1}}{1-x}\log\left(\frac{1}{x}\right)dx=\frac{1}{p^2}+\frac{1}{(p+1)^2}+\dots$$
 A: Set $x=e^{-t}$. Then you have:
$$ I = \int_{0}^{1}\frac{- x^{p-1}\log x}{1-x}dx = \int_{0}^{+\infty}\frac{t e^{-pt}}{1-e^{-t}}dt=\int_{0}^{+\infty}\sum_{n=p}^{+\infty}t e^{-nt}\,dt,$$
but since:
$$\int_{0}^{+\infty}t e^{-nt}\,dt = \frac{1}{n^2}\int_{0}^{+\infty}te^{-t}\,dt=\frac{1}{n^2},$$
your claim follows.
A: Thanks for the comments, I will type out both the answers here for anyone else that still may be interested in this question.
\begin{equation}\begin{aligned}
\int_0^1\frac{x^{p-1}}{1-x}\log\left(\frac{1}{x}\right)dx&=\int_0^1 -x^{p-1}\log(x)\left(\sum_{n\geq 0}x^n\right) dx\\
&=\sum_{n\geq 0}\int_0^1 -\log(x)x^{n+p-1} dx\\
&=\sum_{n\geq 0} \left[-\log(x)\frac{x^{n+p}}{n+p}\right]^1_0+\int_0^1 \frac{x^{n+p-1}}{n+p} dx\\
&=\sum_{n\geq 0} \left[\frac{x^{n+p}}{(n+p)^2}\right]^1_0=\sum_{n\geq 0}\frac{1}{(n+p)^2}
\end{aligned}\end{equation}
The third line holds using l'hopitals.
alternatively, as suggested, one could substitute $x=e^{-t}$ in the second line
yielding the same result. 
from the second step, after subtitution we have,
\begin{equation}\begin{aligned}
&=\sum_{n\geq 0}\int_0^{\infty} -\log(e^{-t})e^{-t(n+p)} dt=\sum_{n\geq 0}\int_0^{\infty} te^{-t(n+p)}\\
&=\sum_{n\geq 0} \left[-t\frac{e^{-t(n+p)}}{n+p}\right]^{\infty}_0+\int_0^{\infty} \frac{e^{-t(n+p)}}{n+p} dx\\
&=\sum_{n\geq 0} \left[\frac{e^{-t(n+p)}}{(n+p)^2}\right]^{\infty}_0=\sum_{n\geq 0}\frac{1}{(n+p)^2}
\end{aligned}\end{equation}
