Evaluate $\sum_{n\geq1} \frac{(-1)^{n+1}H_n^2}{(n+1)^2}$.

Question

I am looking for a closed for $$\sum_{n\geq1} \frac{(-1)^{n+1}H_n^2}{(n+1)^2}.$$ I believe there is a closed form for the sum as we have seen in [1] which poses as, presumably, a more difficult sum of higher order. Any hint would be much obliged; thank you! [1]: How to find ${\large\int}_0^1\frac{\ln^3(1+x)\ln x}x\mathrm dx$

Answer 1

Besides the solution suggested in the comment, here is another idea:

Recall the generating function (see relation 10)

$$\small{\sum_{n=1}^\infty\frac{ H_n^{2}}{n+1}x^{n}=\frac{6\operatorname{Li}_3(1-x)-3\operatorname{Li}_2(1-x)\ln(1-x)-\ln^3(1-x)-3\zeta(2)\ln(1-x)-6\zeta(3)}{3x}}$$

where if we replace $x$ by $-x$ then $\int_0^1$ we have

$$\small{\sum_{n=1}^\infty\frac{ (-1)^nH_n^{2}}{(n+1)^2}=-\int_0^1\frac{6\operatorname{Li}_3(1+x)-3\operatorname{Li}_2(1+x)\ln(1+x)-\ln^3(1+x)-3\zeta(2)\ln(1+x)-6\zeta(3)}{3x}}dx$$

$$=2\int_0^1 \frac{\zeta(3)-\text{Li}_3(1+x)}{x}dx+\int_0^1\frac{\text{Li}_2(1+x)\ln(1+x)}{x}dx$$

$$+\frac13\int_0^1\frac{\ln^3(1+x)}{x}dx+\zeta(2)\int_0^1\frac{\ln(1+x)}{x}dx$$

$$=2I_1+I_2+\frac13I_3+\zeta(2)I_4$$

Starting with the simplest integral

$$I_4=-\text{Li}_2(-x)|_0^1=\frac12\zeta(2)$$

By integration by parts we have

$$I_1=\int_0^1\frac{\ln x\text{Li}_2(1+x)}{1+x}dx$$

write $\ln x=\ln x+i \pi-i \pi=\ln(-x)-i \pi$

$$\Longrightarrow I_1=\int_0^1\frac{\ln(-x)\text{Li}_2(1+x)}{1+x}dx-i\pi\int_0^1\frac{\text{Li}_2(1+x)}{1+x}dx$$

$$=-\frac12\text{Li}_2^2(1+x)|_0^1-i\pi\text{Li}_3(1+x)|_0^1$$

$$=-\frac12(\text{Li}_2^2(2)-\zeta^2(2))-i\pi(\text{Li}_3(2)-\zeta(3))$$

$$=\frac{i\pi}{8}\zeta(3)-\frac{25}{16}\zeta(4)$$

By Landen's identity $\text{Li}_2(1-x)=\zeta(2)-\ln x\ln(1-x)-\text{Li}_2(x)$ with replacing $x$ by $-x$ we have

$$I_2=\zeta(2)\int_0^1\frac{\ln(1+x)}{x}dx-\int_0^1\frac{\ln(-x)\ln^2(1+x)}{x}dx-\int_0^1\frac{\text{Li}_2(-x)\ln(1+x)}{x}dx$$

$$=\zeta(2)I_4-\int_0^1\frac{\ln x\ln^2(1+x)}{x}dx-i\pi \int_0^1\frac{\ln^2(1+x)}{x}dx+\frac12\text{Li}_2^2(-x)|_0^1$$

where

$$\int_0^1\frac{\ln x\ln^2(1+x)}{x}dx=-\int_0^1\frac{\ln^2x\ln(1+x)}{1+x}dx$$

$$=-4\operatorname{Li_4}\left(\frac12\right)+\frac{15}4\zeta(4)-\frac72\ln2\zeta(3)+\ln^22\zeta(2)-\frac{1}{6}\ln^42$$

the last integral is already calculated in this solution ( see the integral $I$).

and

$$\int_0^1\frac{\ln^2(1+x)}{x}dx=\frac14\zeta(3)$$

follows from this generalization which also gives

$$I_3=\int_0^1\frac{\ln^3(1+x)}{x}dx=6\zeta(4)-\frac{21}{4}\ln2\zeta(3)+\frac32\ln^22\zeta(2)-\frac14\ln^42-6\text{Li}_4\left(\frac12\right)$$

Collecting all the pieces we have

$$\sum_{n=1}^\infty\frac{ (-1)^nH_n^{2}}{(n+1)^2}=2\operatorname{Li}_4\left(\frac12\right)-\frac{33}{16}\zeta(4)+\frac{7}{4}\ln2\zeta(3)-\frac{1}{2}\ln^22\zeta(2)+\frac1{12}\ln^42$$