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

I was playing around with trying to prove the following alternating Euler sum: $$\sum_{n = 1}^\infty \frac{(-1)^n H_n}{n^2} = -\frac{5}{8} \zeta (3).$$ Here $$H_n$$ is the Harmonic number.

At least two different proofs for this result that I could find can be seen here and here. Embarking on an alternative proof I did the following. From the integral representation for the Harmonic number of $$H_n = \int_0^1 \frac{1 - x^n}{1 - x} \, dx$$ I rewrote the alternating Euler sum as \begin{align} \sum_{n = 1}^\infty \frac{(-1)^n H_n}{n^2} &= \int_0^1 \frac{1}{1 - x} \left [\sum_{n = 1}^\infty \frac{(-1)^n}{n^2} - \sum_{n = 1}^\infty \frac{(-1)^n x^n}{n^2} \right ]\\ &= \int_0^1 \frac{-\pi^2/12 - \operatorname{Li}_2 (-x)}{1 - x} \, dx\\ &= \int_0^1 \frac{\ln (1 - x) \ln (1 + x)}{x} \, dx, \end{align} where integration by parts has been used. I expect this last integral can be knocked over with relative easy but I have been going around in circles now for quite some time without any success.

So my question is, how can this last integral be evaluated (a real method is preferred) that does not rely on the alternating Euler sum I started out with?

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: $$\int_0^1 t^a dt=\frac{1}{a+1}\Rightarrow \int_0^1 t^a \ln^k tdt=\frac{d^k}{da^k} \left(\frac{1}{a+1}\right)=\frac{(-1)^k k!}{(a+1)^{k+1}}$$ $$\Rightarrow I=\frac18\sum_{n\ge 0}\frac{2}{(n+1)^3} -\frac12 \sum_{n\ge 0}\frac{2}{(2n+1)^3}=\frac28 \zeta(3) -\frac78 \zeta(3)=-\frac58\zeta(3)$$
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 $$\ \displaystyle\sum_{n=1}^\infty\frac{H_n}{n}x^n=\frac12\ln^2(1-x)+\operatorname{Li}_2(x)$$