To show $\int_0^\infty \left(\frac{\ln x}{1-x}\right)^2dx = \frac{2 π^2}{3}$

Question

Please help me, Wolfram alpha isn't showing steps to this integration, It's stuck in my head

Answer 1

Define

$$I:=\int_0^\infty \frac{\ln^2x}{(1-x)^2}\:dx =\int_0^1 \frac{\ln^2x}{(1-x)^2}\:dx + \int_1^\infty \frac{\ln^2x}{(1-x)^2}\:dx$$

The second integral transforms into the first one under the change of variable $u=1/x$. Therefore, $$I = 2\cdot\int_0^1 \frac{\ln^2x}{(1-x)^2}\:dx $$ Now, for $|x|<1$, we have $$\frac{1}{(1-x)^2}=1+2x+3x^2+\cdots = \sum_{n=0}^\infty \:(n+1)x^n$$ Therefore, $$I = 2\cdot \sum_{n=0}^\infty \:\left\{(n+1)\int_0^1 x^n\ln^2x\:dx\right\}$$ Note that for $b \geqslant 0$, $$ J(b):=\int_0^1 x^b \:dx=\frac{1}{b+1} \implies J''(b)=\frac{2}{(b+1)^3}=\int_0^1x^b\ln^2x\:dx $$ It follows that $$I = 2\cdot\sum_{n=0}^\infty \frac{2}{(n+1)^2} = 4\cdot \zeta(2) = \boxed{\frac{2\pi^2}{3}}$$ as desired. The problem is solved.