How can i evaluate $\int _0^1\frac{\ln ^2\left(x\right)\ln \left(1-x\right)\ln \left(1+x\right)}{x}\:dx$

Question

I want to evaluate $$\int _0^1\frac{\ln ^2\left(x\right)\ln \left(1-x\right)\ln \left(1+x\right)}{x}\:dx$$ I tried integration by parts and shape it in a way that i could expand either $\ln$ terms. $$-\int _0^1\frac{\ln ^3\left(x\right)\ln \left(1-x\right)}{1+x}dx+\int _0^1\frac{\ln ^3\left(x\right)\ln \left(1+x\right)}{1-x}\:dx$$ After this i tried expand the terms but i still couldnt go through, any different approaches are welcome.

Answer 1

To evaluate this you can make use of the following identity $$\ln \left(1-x\right)\ln \left(1+x\right)=-\sum _{k=1}^{\infty }x^{2k}\frac{H_{2k}-H_k}{k}-\frac{1}{2}\sum _{k=1}^{\infty }\frac{x^{2k}}{k^2}$$ Resuming on your integral, $$\int _0^1\frac{\ln \left(1-x\right)\ln ^2\left(x\right)\ln \left(1+x\right)}{x}\:dx$$ $$=-\sum _{k=1}^{\infty }\frac{H_{2k}-H_k}{k}\int _0^1x^{2k-1}\ln ^2\left(x\right)\:dx-\frac{1}{2}\sum _{k=1}^{\infty }\frac{1}{k^2}\int _0^1x^{2k-1}\ln ^2\left(x\right)\:dx$$

$$=-\frac{1}{4}\sum _{k=1}^{\infty }\frac{H_{2k}}{k^4}+\frac{1}{4}\sum _{k=1}^{\infty }\frac{H_k}{k^4}-\frac{1}{8}\sum _{k=1}^{\infty }\frac{1}{k^5}$$ $$=-\frac{7}{4}\sum _{k=1}^{\infty }\frac{H_k}{k^4}-2\sum _{k=1}^{\infty }\frac{\left(-1\right)^kH_k}{k^4}-\frac{1}{8}\zeta \left(5\right)$$ $$=-\frac{21}{4}\zeta \left(5\right)+\frac{7}{4}\zeta \left(2\right)\zeta \left(3\right)-\zeta \left(2\right)\zeta \left(3\right)+\frac{59}{16}\zeta \left(5\right)-\frac{1}{8}\zeta \left(5\right)$$ $$=\frac{3}{4}\zeta \left(2\right)\zeta \left(3\right)-\frac{27}{16}\zeta \left(5\right)$$ Those sums are evaluated here.