<s>[INST] 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 $a>-1$.
 [/INST] $$
\begin{align}
&\int_0^1x^a(1-x)^{-1}\log(x)\,\mathrm{d}x\\
&=-\frac{\mathrm{d}}{\mathrm{d}a}\int_0^1\frac{1-x^a}{1-x}\,\mathrm{d}x\\
&=-\frac{\mathrm{d}}{\mathrm{d}a}\int_0^1\left[\left(1-x^a\right)+\left(x-x^{a+1}\right)+\left(x^2-x^{a+2}\right)+\dots\right]\,\mathrm{d}x\\
&=-\frac{\mathrm{d}}{\mathrm{d}a}\left[\left(1-\frac1{a+1}\right)+\left(\frac12-\frac1{a+2}\right)+\left(\frac13-\frac1{a+3}\right)+\dots\right]\\
&=-\left[\frac1{(a+1)^2}+\frac1{(a+2)^2}+\frac1{(a+3)^2}+\dots\right]
\end{align}
$$

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
$$
\begin{align}
\int_0^1x^a(1-x)^{-1}\log(x)\,\mathrm{d}x
&=\int_0^1\sum_{k=0}^\infty x^{a+k}\log(x)\,\mathrm{d}x\\
&=-\sum_{k=0}^\infty\frac1{(a+k+1)^2}\\
&=-\sum_{k=1}^\infty\frac1{(a+k)^2}
\end{align}
$$
</s>