TITLE: Integral ${\large\int}_0^1\ln(1-x)\ln(1+x)\ln^2x\,dx$
QUESTION [33 upvotes]: This problem was posted at I&S a week ago, and no attempts to solve it have been posted there yet. It looks very alluring, so I decided to repost it here:

Prove:
  $$\int_0^1\ln(1-x)\ln(1+x)\ln^2x\,dx=24-\frac{4\pi^2}3-\frac{11\pi^4}{720}-12\ln2\\+2\ln^22-\frac16\ln^42+\pi ^2\ln2+\frac{\pi^2}6\ln^22-4\operatorname{Li}_4\!\left(\tfrac12\right)-\frac{35}4\zeta(3)+\frac72\zeta(3)\ln2.$$

I found a paper where some similar integrals are evaluated: J. A. M. Vermaseren, Harmonic sums, Mellin transforms and Integrals, Int. J. Mod. Phys. A, 14 (1999), 2037-2076, DOI: 10.1142/S0217751X99001032
, but it's not quite easy to read for me. Maybe it could be of some help for this problem.

REPLY [4 votes]: using an identity proved by Cornel Ioan Valean and it can be found in his book " Almost impossible integrals, sums, and series":
$$\ln(1-x)\ln(1+x)=-\sum_{n=1}^\infty\left(\frac{H_{2n}-H_n}{n}+\frac1{2n^2}\right)x^{2n} $$
multiply both sides by $\ln^2x$ then integrate
\begin{align}
I&=\int_0^1\ln(1-x)\ln(1+x)\ln^2\ dx=-\sum_{n=1}^\infty\left(\frac{H_{2n}-H_n}{n}+\frac1{2n^2}\right)\int_0^1x^{2n}\ln^2x\ dx\\
&=-\sum_{n=1}^\infty\left(\frac{H_{2n}-H_n}{n}+\frac1{2n^2}\right)\left(\frac2{(2n+1)^3}\right)\\
&=-2\sum_{n=1}^\infty\frac{H_{2n}}{n(2n+1)^3}+2\sum_{n=1}^\infty\frac{H_n}{n(2n+1)^3}-\sum_{n=1}^\infty\frac{1}{n^2(2n+1)^3}\\
&=-2\sum_{n=1}^\infty\frac{H_n}{n(n+1)^3}-2\sum_{n=1}^\infty\frac{(-1)^nH_n}{n(n+1)^3}+2\sum_{n=1}^\infty\frac{H_n}{n(2n+1)^3}-\sum_{n=1}^\infty\frac{1}{n^2(2n+1)^3}\\
&=-2S_1-2S_2+2S_3-S_4
\end{align}
The first sum:
\begin{align}
S_1=\sum_{n=1}^\infty\frac{H_n}{n(n+1)^3}&=\sum_{n=1}^\infty\frac{H_n}{n(n+1)}-\sum_{n=1}^\infty\frac{H_n}{(n+1)^2}-\sum_{n=1}^\infty\frac{H_n}{(n+1)^3}\\
&=\zeta(2)-\zeta(3)-\frac14\zeta(4)
\end{align}
The second sum:
\begin{align}
S_2&=\sum_{n=1}^\infty\frac{(-1)^nH_n}{n(n+1)^3}=\sum_{n=1}^\infty\frac{(-1)^nH_n}{n(n+1)}-\sum_{n=1}^\infty\frac{(-1)^nH_n}{(n+1)^2}-\sum_{n=1}^\infty\frac{(-1)^nH_n}{(n+1)^3}\\
&=\sum_{n=1}^\infty\frac{(-1)^nH_n}{n(n+1)}-\operatorname{Li}_3(-1)+\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^2}-\operatorname{Li}_4(-1)+\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}\\
&=\left(\ln^22-\frac12\zeta(2)\right)+\frac34\zeta(3)+\frac78\zeta(4)+\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^2}+\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}
\end{align}
using the generating function$$\sum_{n=1}^\infty\frac{x^nH_n}{n^2}=\frac12\ln x\ln^2(1-x)+\ln(1-x)\operatorname{Li}_2(1-x)+\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x)+\zeta(3)$$
we get$$\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^2}=-\frac58\zeta(3)$$
I was able here to prove$$\begin{align}
\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}=2\operatorname{Li_4}\left(\frac12\right)-\frac{11}4\zeta(4)+\frac74\ln2\zeta(3)-\frac12\ln^22\zeta(2)+\frac{1}{12}\ln^42
\end{align}$$
therefore$$S_2=2\operatorname{Li_4}\left(\frac12\right)-\frac{15}8\zeta(4)+\frac18\zeta(3)-\frac12\zeta(2)+\frac74\ln2\zeta(3)-\frac12\ln^22\zeta(2)+\frac{1}{12}\ln^42+\ln^22$$
The third sum:
using the following identity $$\sum_{n=1}^\infty\frac{H_n}{n(n+k)}=\frac{H_k^2}{2k}+\frac{H_k^{(2)}}{2k}-\frac{H_k}{k^2}+\frac{\zeta(2)}{k}$$
differentiate both sides with respect to $k$ twice then let $k=1/2$
\begin{align}
S_3=\sum_{n=1}^\infty\frac{H_n}{n(2n+1)^3}&=\frac18\sum_{n=1}^\infty\frac{H_n}{n(n+1/2)^3}\\
&=\frac72\ln2\zeta(3)+3\ln2\zeta(2)-\frac{45}{16}\zeta(4)-\frac72\zeta(3)+2\ln^22
\end{align}
The fouth sum:
\begin{align}
S_4=\sum_{n=1}^\infty\frac1{n^2(2n+1)^3}&=\sum_{n=1}^\infty\left(\frac1{n^2}+\frac{8}{(2n+1)^2}+\frac{4}{(2n+1)^3}-6\left(\frac1n-\frac{1}{n+1/2}\right)\right)\\
&=\zeta(2)+(6\zeta(2)-8)+\left(\frac72\zeta(3)-4\right)-6H_{1/2}\\
&=\frac72\zeta(3)+7\zeta(2)+12\ln2-24
\end{align}
combining $S_1$, $S_2$, $S_3$ and $S_4$ , we have
$$I=24-8\zeta(2)-\frac{35}{4}\zeta(3)-\frac{11}{8}\zeta(4)-12\ln2+2\ln^22-\frac16\ln^42+6\ln2\zeta(2)+\ln^22\zeta(2)+\frac72\ln2\zeta(3)-4\operatorname{Li}_4\left(\frac12\right)$$