How to approach $\sum _{k\ge 1}\frac{\left(-1\right)^k\:H_k}{\left(2k+1\right)^2}$

Question

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.

Answer 1

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

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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 ?.