TITLE: Prove that $\int\limits_0^1 x^a(1-x)^{-1}\ln x \,dx = -\sum\limits_{n=1}^\infty \frac{1}{(n+a)^2}$
QUESTION [12 upvotes]: Prove that $$\int_0^1 x^a(1-x)^{-1}\ln x \,dx = -\sum_{n=1}^\infty \frac{1}{(n+a)^2}$$
I know that we have a product of $x^a$, $\displaystyle\sum_{n=0}^\infty x^n$, and $\displaystyle\sum_{n=0}^\infty \frac{(1-x)^n}{n}$, but it hasn't helped me so far. 
Any tips?
We are given that $a>-1$.

REPLY [2 votes]: $\newcommand{\+}{^{\dagger}}
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$\ds{\int_{0}^{1}x^{a}\pars{1 - x}^{-1}\ln\pars{x}\,\dd x
     = -\sum_{j = 1}^{\infty}{1 \over \pars{\,j + a\,}^{2}}}$

With $\ds{\quad x \equiv \expo{-t}\quad\iff\quad t = -\ln\pars{x}}$:
  \begin{align}
&\color{#00f}{\large\int_{0}^{1}x^{a}\pars{1 - x}^{-1}\ln\pars{x}\,\dd x}=
\int_{\infty}^{0}\expo{-at}\bracks{1 - \expo{-t}}^{-1}\pars{-t}
\pars{-\expo{-t}\,\dd t}
\\[3mm]&=
-\int_{0}^{\infty}t\expo{-\pars{a + 1}t}\sum_{j = 0}^{\infty}\expo{-jt}\,\dd t
=-\sum_{j = 0}^{\infty}\int_{0}^{\infty}t\expo{-\pars{j + a + 1}t}\,\dd t
\\[3mm]&=-\sum_{j = 0}^{\infty}{1 \over \pars{\,j + a + 1\,}^{2}}\
\overbrace{\int_{0}^{\infty}t\expo{-t}\,\dd t}^{\ds{1!\ = 1}}
=\color{#00f}{\large -\sum_{j = 1}^{\infty}{1 \over \pars{\,j + a\,}^{2}}}
\end{align}