Evaluate $\int_0^1 \frac{\ln (1 - x) \ln (1 + x)}{x} \, dx$

Solution 1:

Similarly to this post. We have: $$4ab=(a+b)^2-(a-b)^2$$ Thus we can take $a=\ln(1-x)$ and $b=\ln(1+x)$ to get: $$I=\int_0^1 \frac{\ln(1-x)\ln(1+x)}{x}dx=\frac14 \int_0^1 \frac{\ln^2(1-x^2)}{x}dx-\frac14 \int_0^1 \frac{\ln^2\left(\frac{1-x}{1+x}\right)}{x}dx$$ By setting $x^2 =t$ in the first integral and $\frac{1-x}{1+x}=t$ in the second one we get: $$I=\frac18 \int_0^1 \frac{\ln^2(1-t)}{t}dt -\frac12 \int_0^1 \frac{\ln^2 t}{1-t^2}dt=\frac18 \int_0^1 \frac{\ln^2t}{1-t}dt-\frac12 \int_0^1 \frac{\ln^2 t}{1-t^2}dt$$ $$=\frac18\sum_{n\ge 0}\int_0^1 t^n \ln^2 tdt-\frac12 \sum_{n\ge 0}\int_0^1 t^{2n}\ln^2 tdt$$ Using the following fact: $$\int_0^1 t^a dt=\frac{1}{a+1}\Rightarrow \int_0^1 t^a \ln^k tdt=\frac{d^k}{da^k} \left(\frac{1}{a+1}\right)=\frac{(-1)^k k!}{(a+1)^{k+1}}$$ $$\Rightarrow I=\frac18\sum_{n\ge 0}\frac{2}{(n+1)^3} -\frac12 \sum_{n\ge 0}\frac{2}{(2n+1)^3}=\frac28 \zeta(3) -\frac78 \zeta(3)=-\frac58\zeta(3)$$

Solution 2:

we can use the generating function : $$\sum_{n=1}^\infty\frac{H_n}{n^2}x^n=\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x)+\ln(1-x)\operatorname{Li}_2(1-x)+\frac12\ln x\ln^2(1-x)+\zeta(3)$$

which can be directly derived from $\ \displaystyle\sum_{n=1}^\infty\frac{H_n}{n}x^n=\frac12\ln^2(1-x)+\operatorname{Li}_2(x)$

I can provide the proof if needed. and I think it was proved here on some post.