How to approach $\sum _{k\ge 1}\frac{\left(-1\right)^k\:H_k}{\left(2k+1\right)^2}$ I am currently trying to find a way to evaluate
$$\sum _{k\ge 1}\frac{\left(-1\right)^k\:H_k}{\left(2k+1\right)^2}$$
but i dont have any hope into accomplishing it, i am also not sure if it has a closed form since programs i've used cant find it.
 A: Some progress:
$$S=\sum_{k=1}^{\infty} \frac{(-1)^k H_k}{(2k+1)^2}=
\int_{0}^{\infty}\sum_{k=0}^{\infty} (-1)^k H_k x e^{-(2k+1)x} dx$$
Next use
https://en.wikipedia.org/wiki/Harmonic_number
$$\sum_{n=1}^{\infty} H_n z^n=-\frac{\ln(1-z)}{1-z}$$
Then take $e^{-x}=t$
$$S=-\int_{0}^{\infty} x e^{-x} \frac{\ln(1+e^{-2x})}{1+e^{-2x}} dx=-\int_{0}^1 \log t \frac{\ln(1+t^2)}{1+t^2} dt$$
Let $t=\tan u$.
$$\implies S=2\int_{0}^{\pi/4} \ln \tan u \ln \cos u ~du$$
I may get back
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[10px,#ffd]{\sum _{k\ \geq\ 1}{\pars{-1}^k\, H_{k} \over
\pars{2k + 1}^{2}}} =
\sum _{k\ =\ 1}^{\infty}{\pars{-1}^k\, H_{k}\
\overbrace{\bracks{-\int_{0}^{1}\ln\pars{x}x^{2k}\,\dd x}}
^{\ds{1 \over \pars{2k + 1}^{2}}}}
\\[5mm] = &\
-\int_{0}^{1}\ln\pars{x}\sum _{k\ =\ 1}^{\infty}H_{k}\
\pars{-x^{2}}^{k}\,\dd x =
-\int_{0}^{1}\ln\pars{x}\bracks{-\,{\ln\pars{1 + x^{2}} \over
1 + x^{2}}}\,\dd x
\\[5mm] = &\
\Re\int_{0}^{1}{\color{red}{2\ln\pars{x}\ln\pars{1 + x \ic}} \over
1 + x^{2}}\,\dd x
\\[2mm] &\ \mbox{The above}\ "\color{red}{double\ \ln\ product}"\
\mbox{is rewritten by means of the identity}
\\ &\ 2ab = a^{2} + b^{2} - \pars{a - b}^{2}. \mbox{Namely,}
\end{align}
\begin{align}
&\bbox[10px,#ffd]{\sum _{k\ \geq\ 1}{\pars{-1}^k\, H_{k} \over
\pars{2k + 1}^{2}}} =
\\[5mm] = &\
\int_{0}^{1}{\ln^{2}\pars{x} \over 1 + x^{2}}\,\dd x +
\Re\int_{0}^{1}{\ln^{2}\pars{1 + x\ic} \over 1 + x^{2}}\,\dd x -
\Re\int_{0}^{1}\ln^{2}\pars{x \over 1 + x\ic}\,
{\dd x  \over 1 + x^{2}}
\\[5mm] = &\
-\Im\int_{0}^{1}{\ln^{2}\pars{x} \over \ic - x}\,\dd x + \bracks{%
{1 \over 2}\,\Im\int_{1}^{1 + \ic}{\ln^{2}\pars{x} \over 2 - x}\,\dd x
-
\,{1 \over 2}\,\Im\int_{1}^{1 + \ic}{\ln^{2}\pars{x} \over -x}\,\dd x}
\\[2mm]  &\
-\,{1 \over 2}\,\Im\int_{0}^{1/2 - \ic/2}{\ln^{2}\pars{x} \over -\ic/2 - x}
\,\dd x
\end{align}
These integrals are of the form ( they are evaluated by performing two times an integration by parts ):
$$
\int{\ln^{2}\pars{x} \over a - x}\,\dd x =
2\,\mrm{Li}_{3}\pars{x \over a} -
2\ln\pars{x}\mrm{Li}_{2}\pars{x \over a} -
\ln^{2}\pars{x}\ln\pars{a - x \over a}
$$
Can you take from here ?.
