Calculate $\int_{0}^{1} \frac{\mathrm{ln} \mathit{x}}{1-x}dx$ Please help me with this integral
$$\int_{0}^{1} \frac{\mathrm{ln} \mathit{x}}{1-x}dx$$
If it is useful, the previous demand is to prove that $\int_{0}^{1} \frac{\mathrm{ln} \mathit{x}}{1-x^a}dx$ is convergent for all $a>0$
 A: The integral of interest is 
$$\int_0^1 \frac{\log(x)}{1-x}\,dx=-\text{Li}_2(1)=-\frac{\pi^2}{6}$$
where $\text{Li}_2(x)$ is the dilogarithm function.

ALTERNATIVE DEVELOPMENT:
Note that we can expand $\log(x)$ as $\log(x)=-\sum_{n=1}^{\infty}\frac{(1-x)^n}{n}$.  Then, we have
$$\begin{align}
\int_0^1 \frac{\log(x)}{1-x} \,dx&=\lim_{\epsilon \to 0^+}\int_\epsilon^1 \frac{\log(x)}{1-x} \,dx \tag 1\\\\
&=-\lim_{\epsilon \to 0^+}\sum_{n=1}^{\infty} \frac1n \int_\epsilon^1 (1-x)^{n-1}\,dx \tag 2\\\\
&=-\lim_{\epsilon \to 0^+}\sum_{n=1}^{\infty} \frac{(1-\epsilon)^n}{n^2} \tag 3\\\\
&=-\sum_{n=1}^{\infty}\frac1{n^2} \tag 4\\\\
&=-\frac{\pi^2}{6}
\end{align}$$
where the legitimacy of interchanging the integral and series in going from $(1)$ to $(2)$ and the legitimacy of interchanging the limit with the series in going from $(3)$ to $(4)$ is provided by uniform convergence.


We could rely on Lebesgue integration and the Dominated Convergence Theorem rather than Riemann integration and Uniform Convergence.  We proceed as follows.

First, we can write
$$\begin{align}
\int_0^1 \sum_{n=1}^N \frac{(1-x)^n}{n}\,dx&=\int_0^1 \int_0^1 \left(\frac{1-(xy)^N}{1-xy}\right)\,dy\,dx\\\\
\end{align}$$
Next, inasmuch as $\left|\int_0^1 \left(\frac{1-(xy)^N}{1-xy}\right)\,dy \right|\le -\frac{\log(1-x)}{x}$ for $x\in (0,1)$, and $\int_0^1 \left|-\frac{\log(1-x)}{x}\right|\,dx$ is Lebesgue integrable, the Dominated Convergence Theorem guarantees that 
$$\begin{align}
-\frac{\pi^2}{6}&=-\lim_{N\to \infty}\sum_{n=1}^N \frac1n \int_0^1  (1-x)^{n-1}\,dx\\\\
&=-\lim_{N\to \infty}\int_0^1 \sum_{n=1}^N \frac{(1-x)^n}{n}\,dx\\\\
&=-\int_0^1 \sum_{n=1}^\infty \frac{(1-x)^n}{n}\,dx\\\\
&=\int_0^1 \frac{\log(1-x)}{x}\,dx
\end{align}$$
as expected!
A: By setting $x=e^{-t}$ we get:
$$ \int_{0}^{1}\frac{\log(x)}{1-x}\,dx = -\int_{0}^{+\infty}\frac{t e^{-t}}{1-e^{-t}}\,dt = -\sum_{n\geq 0}\int_{0}^{+\infty}te^{-(n+1)t}\,dt=-\zeta(2). $$
In a similar way,
$$ \int_{0}^{1}\frac{\log(x)\,dx }{1-x^{\alpha}}=-\sum_{n\geq 1}\frac{1}{(n+\alpha)^2}=-\psi'(\alpha+1)$$
for any $\alpha>0$.
