Sum $\sum^\infty_{n=1}\frac{(-1)^nH_n}{(2n+1)^2}$

I would like to seek your assistance in computing the sum $$\sum^\infty_{n=1}\frac{(-1)^nH_n}{(2n+1)^2}$$ I am stumped by this sum. I have tried summing the residues of $\displaystyle f(z)=\frac{\pi\csc(\pi z)(\gamma+\psi(-z))}{(2z+1)^2}$, unfortunately the sum disappears when I add the residues up. Another idea that came to my mind would be to use $$\sum^\infty_{n=1}(-1)^nH_nx^{2n}=-\frac{\ln(1+x^2)}{1+x^2}$$ and integrate once, divide by $x$ then integrate again. However, it seems that performing these successive integrations would turn out to be disastrous. Therefore, I would like to know if any other methods can be employed to crack this sum. I am particularly interested in finding out if the contour integration method still applies for these sort of sums, and in the case that it is still viable, what then would be the appropriate kernel to use?

Thank you for your help.

Solution 1:

Since $$\int_{0}^{1}x^{2n}\log x\,dx = -\frac{1}{(2n+1)^2}$$ we have $$ S = \sum_{n=1}^{+\infty}\frac{(-1)^n H_n}{(2n+1)^2}=\int_{0}^{1}\frac{\log(1+x^2)\log x}{1+x^2}\,dx $$ for which Mathematica gives: $$\frac{1}{192} \left(-3 \pi ^3-192 K \log 2-10 i \pi ^2 \log 2-12 \pi\log^2 2+2 i \left(4 \log^3 2-192 \text{Li}_3\left(\frac{1+i}{2}\right)+105 \zeta(3)\right)\right).$$ Thanks to @gammatester, it looks that the last formula follows setting $x=i$ in the line after $(608)$ in http://www.pi314.net/eng/hypergse13.php#x15-134002r658.

Solution 2:

Using @Jack D'Aurizio's idea, there is a simplification of his closed-form. $$ \sum^\infty_{n=1}\frac{(-1)^nH_n}{(2n+1)^2} = 2\,\Im\left[\operatorname{Li}_3\left(\frac{1+i}2\right)\right]-\frac{\pi^3}{64}-\frac{\pi}{16}\ln^2 2 - G \ln 2, $$ where $G$ is Catalan's contant, and $\operatorname{Li_3}$ is the trilogarithm function.