TITLE: Show $\int_{0}^{1} \frac{\ln x}{1-x}dx$=$\sum_{1}^{\infty}\frac{1}{n^2}$ and converges
QUESTION [6 upvotes]: I found this question

a) show that the follow integral converges: $\int_{0}^{1} \frac{\ln x}{1-x}dx $
b) $\int_{0}^{1} \frac{\ln x}{1-x}dx$=$\sum_{1}^{\infty}\frac{1}{n^2}$

for the first part I try with comparison test because $f(x):=\frac{\ln x}{1-x} \le0$ in $ [0,1]$
and I know from L'Hôpital's rule that $\lim_{x\to1} \frac{\ln x}{1-x}=-1 $
thanks ahead

REPLY [5 votes]: a) At $x=0$
$$
\frac{\ln x}{1-x}\sim \ln x
$$
and
$$
\int \ln x\,dx = x(\ln x - 1) + C;
$$
but
$$
\lim_{x\to0} x(\ln x - 1) = 0
$$
At $x=1$
$$
\frac{\ln x}{1 - x}\sim -1,
$$
as you finded. So, this integral converges.
b) Expand $1/(1-x)$ into series:
$$
\frac{1}{1-x} = \sum_{n=0}^\infty x^n.
$$
So,
$$
I = \int_0^1 \frac{\ln x}{1 - x}dx = \sum_{n=0}^\infty \underbrace{\int\limits_0^1 x^n\ln x\,dx}_{J_n}
$$
Let's evaluate $J_n$:
$$
J_n \overset{x=e^{-u}}{=} \int_0^\infty e^{-(n+1)u}u\,du
\overset{z=(n+1)u}{=} \frac{1}{(n+1)^2} \int_0^\infty e^{-z}z\,dz = \frac{1}{(n+1)^2}
$$
and
$$
I = \sum_{n=0}^\infty \frac{1}{(n+1)^2} = \sum_{k=1}^\infty \frac{1}{k^2}
$$