TITLE: Compute $\int_0^{1/2}\frac{\left(\operatorname{Li}_2(x)\right)^2}{x}dx$ or $\sum_{n=1}^\infty \frac{H_n^{(2)}}{n^32^n}$
QUESTION [14 upvotes]: Prove that
I encountered this integral while working on the sum $\displaystyle \sum_{n=1}^\infty \frac{H_n^{(2)}}{n^32^n}$. Both of the integral and the sum were proposed by Cornel Valean:
The integral:

$$I=\int_0^{1/2}\frac{\left(\operatorname{Li}_2(x)\right)^2}{x}dx=\frac12\ln^32\zeta(2)-\frac78\ln^22\zeta(3)-\frac58\ln2\zeta(4)+\frac{27}{32}\zeta(5)+\frac78\zeta(2)\zeta(3)\\-\frac{7}{60}\ln^52-2\ln2\operatorname{Li}_4\left(\frac12\right)-2\operatorname{Li}_5\left(\frac12\right);$$

The sum:

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

This sum itself is related, through the Cauchy product of $\ln(1-x)\text{Li}_3(x)$, to the following sum:

$$\sum_{n=1}^\infty \frac{H_n^{(3)}}{n^22^n}=4\operatorname{Li}_5\left(\frac12\right)+3\ln2\operatorname{Li}_4\left(\frac12\right)-\frac{81}{64}\zeta(5)+\frac5{16}\ln2\zeta(4)\nonumber\\ -\frac78\zeta(2)\zeta(3)+\frac{7}{8}\ln^22\zeta(3)-\frac5{12}\ln^32\zeta(2)+\frac{11}{120}\ln^52$$

The main integral is very related to the integral $\int_0^1 \frac{\ln^3(1-x)\ln(1+x)}{x}dx$ which I managed to solve using three tough results of alternating series, so again I am looking for a different approach that does not use these results ( mentioned in the link) to compute $I$.
Here is how the two integrals are related:

$$\int_0^{1/2}\frac{\left(\operatorname{Li}_2(x)\right)^2}{x}dx\overset{IBP}{=}\operatorname{Li}_2\left(\frac12\right)\operatorname{Li}_3\left(\frac12\right)-\ln2\operatorname{Li}_4\left(\frac12\right)-\operatorname{Li}_5\left(\frac12\right)+\sum_{n=1}^\infty\frac{H_n^{(4)}}{n2^n}$$

From this identity, we have $\sum_{n=1}^\infty\frac{H_n^{(4)}}{n2^n}=-\frac16\int_0^1\frac{\ln^3(1-x)\ln(1+x)}{x}dx$
Then

$$\int_0^{1/2}\frac{\left(\operatorname{Li}_2(x)\right)^2}{x}dx=\operatorname{Li}_2\left(\frac12\right)\operatorname{Li}_3\left(\frac12\right)-\ln2\operatorname{Li}_4\left(\frac12\right)-\operatorname{Li}_5\left(\frac12\right)\\-\frac16\int_0^1\frac{\ln^3(1-x)\ln(1+x)}{x}dx$$

So, any elegant way to solve any of these two integrals?

REPLY [2 votes]: A solution by Cornel I. Valean to $\displaystyle \int_0^{1/2}\frac{(\operatorname{Li}_2(x))^2}{x}\textrm{d}x$ (based on a reduction strategy to weight $5$ alternating harmonic series by exploiting a Landen identity type in the series form)
We need the following identity presented in (Almost) Impossible Integrals, Sums, and Series (see page $285$, Sect. $4.11$), slightly rearranged,
$$\frac{1}{6}\sum_{n=1}^{\infty} (-1)^{n-1}\frac{x^{n-1}}{n}(H_n^3+3H_n H_n^{(2)}+2H_n^{(3)})=\frac{1}{x}\operatorname{Li}_4\left(\frac{x}{x+1}\right). \tag1$$
Integrating both sides of $(1)$ from $x=0$ to $x=1$, then letting the variable change $\displaystyle \frac{x}{x+1}\mapsto x$ and integrating by parts two times in the right-hand side and then rearranging, we get that
$$\int_0^{1/2}\frac{(\operatorname{Li}_2(x))^2}{x}\textrm{d}x$$
$$\small =-\frac{1}{12}\log^5(2)+\frac{1}{3}\log^3(2)\zeta(2)-\frac{7}{16}\log^2(2)\zeta(3)-\frac{5}{8}\log(2)\zeta(4)+\frac{7}{16}\zeta(2)\zeta(3)-\log(2)\operatorname{Li}_4\left(\frac{1}{2}\right)-\operatorname{Li}_5\left(\frac{1}{2}\right)$$
$$+\frac{1}{6}\sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_n^3}{n^2}+\frac{1}{2}\sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_n H_n^{(2)}}{n^2}+\frac{1}{3}\sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_n^{(3)}}{n^2}$$
$$=\frac{1}{2}\log^3(2)\zeta(2)-\frac{7}{8}\log^2(2)\zeta(3)-\frac{5}{8}\log(2)\zeta(4)+\frac{27}{32}\zeta(5)+\frac{7}{8}\zeta(2)\zeta(3)-\frac{7}{60}\log^5(2)\\-2\log(2)\operatorname{Li}_4\left(\frac{1}{2}\right)-2\operatorname{Li}_5\left(\frac{1}{2}\right),$$
where the remaining series are found and calculated in (Almost) Impossible Integrals, Sums, and Series, pages $311$-$312$.
End of story