A difficult logarithmic integral and its relation to alternating Euler Sums

Different approach to compute our main sum $\displaystyle\sum_{n=1}^\infty(-1)^n\frac{H_nH_n^{(3)}}{n}$.

From here we have

$$\int_0^1\frac{\ln^ax\ln\left(\frac{1+x}{2}\right)}{1-x}=(-1)^aa!\sum_{n=1}^\infty\frac{(-1)^nH_n^{a+1}}{n}\tag{1}$$ Using the identity

$$\ln^2(1+x)=2\sum_{n=1}^\infty\frac{H_n}{n+1}(-x)^{n+1}=2\sum_{n=1}^\infty(-1)^n\left(\frac{H_n}{n}-\frac{1}{n^2}\right)x^n\tag{2}$$

Multiply both sides of (2) by $\frac{\ln^2x}{1-x}$ then integrate from $x=0$ to $1$ we have

\begin{align} I&=\int_0^1\frac{\ln^2x\ln^2(1+x)}{1-x}\ dx=2\sum_{n=1}^\infty(-1)^n\left(\frac{H_n}{n}-\frac{1}{n^2}\right)\int_0^1\frac{x^n\ln^2x}{1-x}\ dx\\ &=2\sum_{n=1}^\infty(-1)^n\left(\frac{H_n}{n}-\frac{1}{n^2}\right)\left(2\zeta(3)-2H_n^{(3)}\right)\\ &=4\zeta(3)\underbrace{\sum_{n=1}^\infty(-1)^n\left(\frac{H_n}{n}-\frac{1}{n^2}\right)}_{\text{use (2) where}\ x=1}+4\sum_{n=1}^\infty(-1)^n\frac{H_n^{(3)}}{n^2}-4\sum_{n=1}^\infty(-1)^n\frac{H_nH_n^{(3)}}{n}\\ &=2\ln^22\zeta(3)+4\sum_{n=1}^\infty(-1)^n\frac{H_n^{(3)}}{n^2}-4\sum_{n=1}^\infty(-1)^n\frac{H_nH_n^{(3)}}{n}\tag{3} \end{align}

On the other hand:

\begin{align} I&=\small{\int_0^1\frac{\ln^2x\ln^2(1+x)}{1-x}\ dx\overset{x\mapsto 1-x}=\int_0^1\frac{\ln^2(1-x)\ln^2(2-x)}{x}=\int_0^1\frac{\ln^2(1-x)}{x}\left(\ln2+\ln\left(1-\frac x2\right)\right)^2\ dx}\\ &=\small{\ln^22\int_0^1\frac{\ln^2(1-x)}{x}\ dx+2\ln2\underbrace{\int_0^1\frac{\ln^2(1-x)}{x}\ln\left(1-\frac x2\right)\ dx}_{x\mapsto 1-x}+\underbrace{\int_0^1\frac{\ln^2(1-x)}{x}\ln^2\left(1-\frac x2\right)\ dx}_{\text{use (2)}}}\\ &=\small{2\ln^22\zeta(3)+2\ln2\underbrace{\int_0^1\frac{\ln^2x}{1-x}\ln\left(\frac{1+x}{2}\right)\ dx}_{\text{use (1)}}+2\sum_{n=1}^\infty\frac1{2^n}\left(\frac{H_n}{n}-\frac1{n^2}\right)\int_0^1x^{n-1}\ln^2(1-x)\ dx}\\ &=2\ln^22\zeta(3)+4\ln2\sum_{n=1}^\infty(-1)^n\frac{H_n^{(3)}}{n}+2\sum_{n=1}^\infty\frac1{2^n}\left(\frac{H_n}{n}-\frac1{n^2}\right)\left(\frac{H_n^2+H_n^{(2)}}{n}\right)\\ &=\small{2\ln^22\zeta(3)+4\ln2\sum_{n=1}^\infty(-1)^n\frac{H_n^{(3)}}{n}+2\sum_{n=1}^\infty\frac{H_n^3}{n^22^n}+2\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^22^n}-2\sum_{n=1}^\infty\frac{H_n^2}{n^32^n}-2\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^32^n}}\quad \quad \quad \quad \text{(4)} \end{align}

From (3) and (4) we conclude that

$$\sum_{n=1}^\infty(-1)^n\frac{H_nH_n^{(3)}}{n}=\\ \small{\sum_{n=1}^\infty(-1)^n\frac{H_n^{(3)}}{n^2}-\ln2\sum_{n=1}^\infty(-1)^n\frac{H_n^{(3)}}{n}-\frac12\sum_{n=1}^\infty\frac{H_n^3}{n^22^n}-\frac12\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^22^n} +\frac12\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^32^n}+\frac12\sum_{n=1}^\infty\frac{H_n^2}{n^32^n}}\tag{5}$$

We have the following results:

$$S_1=\sum_{n=1}^\infty(-1)^n\frac{H_n^{(3)}}{n^2}=\frac{21}{32}\zeta(5)-\frac34\zeta(2)\zeta(3)$$

$$S_2=\sum_{n=1}^\infty(-1)^n\frac{H_n^{(3)}}{n}=\frac34\ln2\zeta(3)-\frac{19}{16}\zeta(4)$$

$$S_3=\sum_{n=1}^\infty\frac{H_n^3}{n^22^n}=-14\operatorname{Li}_5\left(\frac12\right)-9\ln2\operatorname{Li}_4\left(\frac12\right)+\frac{279}{16}\zeta(5)-\frac{25}{4}\ln2\zeta(4)-\frac78\zeta(2)\zeta(3)\\-\frac74\ln^22\zeta(3)+\frac{13}{12}\ln^32\zeta(2)-\frac{31}{120}\ln^52$$

$$S_4=\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^22^n}=2\operatorname{Li}_5\left(\frac12\right)+\ln2\operatorname{Li}_4\left(\frac12\right)-\frac{31}{32}\zeta(5)+\frac{1}{8}\ln2\zeta(4)+\frac18\zeta(2)\zeta(3)\\-\frac{1}{12}\ln^32\zeta(2)+\frac{1}{40}\ln^52$$

$$S_5=\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^32^n}=-2\operatorname{Li}_5\left(\frac12\right)-3\ln2\operatorname{Li}_4\left(\frac12\right)+\frac{23}{64}\zeta(5)-\frac1{16}\ln2\zeta(4)+\frac{23}{16}\zeta(2)\zeta(3)\\-\frac{23}{16}\ln^22\zeta(3)+\frac7{12}\ln^32\zeta(2)-\frac{13}{120}\ln^52$$

$$S_6=\sum_{n=1}^\infty\frac{H_n^2}{n^32^n}=-2\operatorname{Li}_5\left(\frac12\right)-\ln2\operatorname{Li}_4\left(\frac12\right)+\frac{279}{64}\zeta(5)-\frac{37}{16}\ln2\zeta(4)-\frac{9}{16}\zeta(2)\zeta(3)\\+\frac{7}{16}\ln^22\zeta(3)+\frac1{12}\ln^32\zeta(2)-\frac{1}{40}\ln^52$$

By substituting these results in (5) we get

$$\sum_{n=1}^\infty(-1)^n\frac{H_nH_n^{(3)}}{n}=4 \operatorname{Li}_5\left(\frac{1}{2}\right)+2\ln2\operatorname{Li}_4\left(\frac{1}{2}\right)-\frac{167}{32}\zeta(5)+\frac{49}{16}\ln2\zeta(4)-\frac{3}{8}\ln^22\zeta(3)\\-\frac{1}{6}\ln^32\zeta(2)+\frac{1}{16}\zeta(2)\zeta(3)+\frac{1}{20}\ln^52$$

NOTE:

$S_1$ and $S_2$ can be found here, $S_3$ and $S_4$ can be found here and $S_5$ and $S_6$ can be found here.