Show $\int_{0}^{1} \frac{\ln x}{1-x}dx$=$\sum_{1}^{\infty}\frac{1}{n^2}$ and converges

Question

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

Answer 1

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 $$ I = \sum_{n=0}^\infty \frac{1}{(n+1)^2} = \sum_{k=1}^\infty \frac{1}{k^2} $$