Integration of $\int_{0}^{1} x^{a}(1-x)^{-1}\log (x) dx $ I need to compute the following integral for $a>-1$,
$$\int_{0}^{1} x^{a}(1-x)^{-1}\log (x) dx $$
My attempt:
By change of variable $x=1+t$:
\begin{align*}
I &= \int_{0}^{1} x^{a}(1-x)^{-1}\log (x) dx= \int_{0}^{1} \frac{(1+t)^a}{t} \log(1+t) dt \\
&= \sum_1^{\infty} \frac{(-1)^{n+1}}{n}  \int_{0}^{1} \frac{(1+t)^a t^n}{t}dt=\sum_1^{\infty} \frac{(-1)^{n+1}}{n}  \int_{0}^{1} (1+t)^a t^{n-1}dt
\end{align*}
 A: Expanding $(1-x)^{-1}$:
$$\begin{align}
\int_{0}^{1} x^{a}(1-x)^{-1}\log x\,dx&=\sum_{n=0}^\infty\int_0^1x^{n+a}\log x\,dx\\
&=\sum_{n=0}^\infty\Biggl(\frac{x^{n+a+1}}{n+a+1}\,\log x-\frac{x^{n+a+1}}{(n+a+1)^2}\,\Biggr|_0^1\Biggr)\\
&=-\sum_{n=0}^\infty\frac{1}{(n+a+1)^2}.
\end{align}$$
Since $a>-1$ (or otherwise the integral would not converge), we have $n+a+1\ne0$ for all $n$.
A: As you note, put $x=1-t$
 , $dx=-dt$
 $$\int_{0}^{1}x^{a}\left(1-x\right)^{-1}\log\left(x\right)dx=\sum_{n\geq1}\frac{1}{n}\int_{0}^{1}\left(1-t\right)^{a}t^{n-1}dt=\sum_{n\geq1}\frac{B\left(a+1,n\right)}{n}$$
 where $B\left(a,b\right)$
  is Beta function. If $a$
  is an integer, can be rewritten as$$\sum_{n\geq1}\frac{B\left(a+1,n\right)}{n}=\frac{\pi^{2}}{6}-\sum_{n=1}^{a}\frac{1}{n^{2}}.$$ You can find a proof of this fact here partial sum of Basel problem related to series involving Beta function.
A: Alternately, notice that the integral is the derivative with regard to a of $\displaystyle\int_0^1x^a(1-x)^b~dx$, which evaluates to $B\big(a+1,b+1\big)$. But since $\Gamma(0)$ is undefined, we'll have to ultimately evaluate the expression by taking its limit as $b\to-1$. However, that limit is nothing more than minus the derivative of $\psi_0(t)$ at $t=2$, namely $-\psi_1(2)=1-\zeta(2)$.
