Prove that $\int\limits_0^1 x^a(1-x)^{-1}\ln x \,dx = -\sum\limits_{n=1}^\infty \frac{1}{(n+a)^2}$ 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 $a>-1$.
 A: We first define the function 
$$
f(a,b)= \int_0^b \frac{x^a}{1-x}\,dx, \,\,\,b\in (0,1).
$$
Using the fact that
$$
\frac{1}{1-x}=\sum_{k\geq 0}x^k,
$$
we obtain 
$$
f(a,b)= \int_0^b x^a\sum_{k\geq 0}x^k \,dx= \sum_{k\geq 0} \int_0^b x^{k+a}=\sum_{k\ge 0}\frac{b^{k+a+1}}{k+a+1}=\sum_{k\ge 1}\frac{b^{k+a}}{k+a},
$$
and thus
$$
\frac{\partial}{\partial a}f(a,b)= \int_0^b \frac{x^a}{1-x}\ln x\,\,dx
=-\sum_{k\ge 1}\frac{b^{k+a}}{(k+a)^2}.
$$
We are allowed to differentiate the series and interchange $\partial/\partial a$ with the summation, since the differentiated series converges locally uniformly in $a$, when $a\in (0,\infty)$.
Now we are allowed to take the limits as $b\searrow 1$, and obtain the identity.
Justification of 
$
\,\,\,\displaystyle
\lim_{b\searrow 1}\sum_{k=1}^\infty\frac{b^{k+a}}{(k+a)^2}=\sum_{k=1}^\infty\frac{1}{(k+a)^2}.
$
It suffices to show that, for every $\varepsilon>0$, there exists a $b_0\in(0,1)$, such that
$b\in (b_0,1)$ implies that 
$$
\sum_{k=1}^\infty\frac{1-b^{k+a}}{(k+a)^2}<\varepsilon.
$$
Since $\,\,\displaystyle\sum_{k=0}^\infty\frac{1}{(k+a)^2}\,\,$ converges, 
there exists an $n_0$, such that
$$\sum_{k=n_0+1}^\infty\frac{1}{(k+a)^2}<\frac{\varepsilon}{2}.$$
Clearly the function
$$
f_{n_0}(b)=\sum_{k=1}^{n_0}\frac{b^{k+a}}{(k+a)^2},
$$
is continuous in $b\in\mathbb R$, as it is a polynomial, and hence there is a $b_0\in(0,1)$, such that
$$
b\in (b_0,1) \quad\Longrightarrow\quad \frac{\varepsilon}{2}>\lvert f_{n_0}(b)-f_{n_0}(1)\rvert=\sum_{k=1}^{n_0}\frac{1-b^{k+a}}{(k+a)^2}
$$
Altogether, if $b\in(b_0,1)$, then
$$
\sum_{k=1}^\infty\frac{1-b^{k+a}}{(k+a)^2}=\sum_{k=1}^{n_0}\frac{1-b^{k+a}}{(k+a)^2}+\sum_{k=n_0+1}^\infty\frac{1-b^{k+a}}{(k+a)^2}<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon.
$$
A: $$
\begin{align}
&\int_0^1x^a(1-x)^{-1}\log(x)\,\mathrm{d}x\\
&=-\frac{\mathrm{d}}{\mathrm{d}a}\int_0^1\frac{1-x^a}{1-x}\,\mathrm{d}x\\
&=-\frac{\mathrm{d}}{\mathrm{d}a}\int_0^1\left[\left(1-x^a\right)+\left(x-x^{a+1}\right)+\left(x^2-x^{a+2}\right)+\dots\right]\,\mathrm{d}x\\
&=-\frac{\mathrm{d}}{\mathrm{d}a}\left[\left(1-\frac1{a+1}\right)+\left(\frac12-\frac1{a+2}\right)+\left(\frac13-\frac1{a+3}\right)+\dots\right]\\
&=-\left[\frac1{(a+1)^2}+\frac1{(a+2)^2}+\frac1{(a+3)^2}+\dots\right]
\end{align}
$$

Differentiation under the Integral
$$
\begin{align}
-\frac{\mathrm{d}}{\mathrm{d}a}\int_0^1\frac{1-x^a}{1-x}\,\mathrm{d}x
&=\lim_{h\to0}\int_0^1\frac{x^{a+h}-x^a}{h(1-x)}\,\mathrm{d}x\\
&=\lim_{h\to0}\int_0^1\frac{x^a}{1-x}\frac{x^h-1}{h}\,\mathrm{d}x
\end{align}
$$
Pointwise $\frac{x^h-1}{h}\to\log(x)$ and on $(0,1)$ we have $\left|\frac{x^h-1}{h}\right|\le|\log(x)|$. Dominated Convergence finishes things off.

Convergence of the Integral of the Sum
For $a\ge0$, note that on $[0,1]$, $\frac{1-x^a}{1-x}\le\max(a,1)$. Therefore,
$$
\begin{align}
\left|\int_0^1\frac{1-x^a}{1-x}\,\mathrm{d}x-\sum_{k=0}^{n-1}\int_0^1(1-x^a)x^k\,\mathrm{d}x\right|
&=\left|\int_0^1\frac{1-x^a}{1-x}x^n\,\mathrm{d}x\right|\\
&\le\frac{\max(a,1)}{n+1}
\end{align}
$$
For $-1\lt a\lt0$, the term on the right becomes
$$
\begin{align}
\left|\int_0^1\frac{1-x^a}{1-x}x^n\,\mathrm{d}x\right|
&=\left|\int_0^1\frac{x^{-a}-1}{1-x}x^{n+a}\,\mathrm{d}x\right|\\
&\le\frac1{n+a+1}
\end{align}
$$

An Alternate Approach
Note that integration by parts gives us
$$
\begin{align}
\int_0^1x^a\log(x)\,\mathrm{d}x
&=\frac1{a+1}\int_0^1\log(x)\,\mathrm{d}x^{a+1}\\
&=-\frac1{a+1}\int_0^1x^a\,\mathrm{d}x\\
&=-\frac1{(a+1)^2}
\end{align}
$$
Then using the sum of a geometric series, we have
$$
\begin{align}
\int_0^1x^a(1-x)^{-1}\log(x)\,\mathrm{d}x
&=\int_0^1\sum_{k=0}^\infty x^{a+k}\log(x)\,\mathrm{d}x\\
&=-\sum_{k=0}^\infty\frac1{(a+k+1)^2}\\
&=-\sum_{k=1}^\infty\frac1{(a+k)^2}
\end{align}
$$
A: $\newcommand{\+}{^{\dagger}}
 \newcommand{\angles}[1]{\left\langle #1 \right\rangle}
 \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\down}{\downarrow}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left( #1 \right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
 \newcommand{\wt}[1]{\widetilde{#1}}$
$\ds{\int_{0}^{1}x^{a}\pars{1 - x}^{-1}\ln\pars{x}\,\dd x
     = -\sum_{j = 1}^{\infty}{1 \over \pars{\,j + a\,}^{2}}}$

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}

A: Let $$F(a)= \int^1_0 \frac{x^a}{1-x}\,dx$$
Then knowing that 
$$\frac{1}{1-x}=\sum_{i\geq 0}x^j$$
we have 
$$F(a)= \int^1_0 x^a\sum_{i\geq 0}x^j \,dx= \sum_{i\geq 0} \int^1_0 x^{j+a}=\sum_{j=1}\frac{1}{j+a}$$
$$F'(a)= \int^1_0 \frac{x^a}{1-x}\ln x\,\,dx=-\sum_{j=1}\frac{1}{(j+a)^2}$$
