Sum $S=\sum _{k=1}^{\infty } \frac{(-1)^k H_k}{k^3}?$

Question

We know that $$\sum _{k=1}^{\infty } \frac{H_k}{k^3} = \frac{\pi^4}{72}.$$ Is there a closed form for the sum $$S=\sum _{k=1}^{\infty } \frac{(-1)^k H_k}{k^3}?$$ Mathematica doesn't give anything resembling a closed form and I have no idea if one exists.

Answer 1

Since: $$ \sum_{k\geq 1} H_k x^k = \frac{-\log(1-x)}{1-x} $$ we have: $$ \sum_{k\geq 1} \frac{H_k}{k+1} x^{k+1} = \frac{1}{2}\log^2(1-x), $$

$$ \sum_{k\geq 1} \frac{(-1)^{k+1}H_k}{k+1} x^{k} = \frac{1}{2x}\log^2(1+x), $$ and since $\int_{0}^{1} x^k\log(x) = -\frac{1}{(k+1)^2}$, it follows that: $$ S=\sum_{k\geq 1} \frac{(-1)^{k}H_k}{(k+1)^3}=\int_{0}^{1}\frac{\log^2(1+x)\log x}{2x}\,dx \tag{1}$$ for the last integral, Mathematica returns: $$ S = \frac{\pi ^4}{48}-\frac{\log^4 2}{12}+\frac{\pi^2\log^2 2}{12} -2 \operatorname{Li}_4\left(\frac{1}{2}\right)-\frac{7}{4}\zeta(3)\log 2 \tag{2}$$ and since: $$ \sum_{k\geq 1}\frac{(-1)^{k}\frac{1}{k+1}}{(k+1)^3}=-1+\frac{7\pi^4}{720}\tag{3} $$ it follows that:

$$ \sum_{k\geq 1}\frac{(-1)^k H_k}{k^3}=-\frac{11\pi ^4}{360}+\frac{\log^4 2}{12}-\frac{\pi^2\log^2 2}{12} +2 \operatorname{Li}_4\left(\frac{1}{2}\right)+\frac{7}{4}\zeta(3)\log 2.\tag{4}$$