Alternating harmonic sum $\sum_{k\geq 1}\frac{(-1)^k}{k^3}H_k$

Question

How to analytically prove

$$\sum_{k\geq 1}\frac{(-1)^k}{k^3}H_k=-\frac{11\pi^4}{360}+\frac{\ln^42-\pi^2\ln^22}{12}+2\mathrm{Li}_4\left(\frac12\right)+\frac{7\ln 2}{4}\zeta(3) $$

As O.L answer

where $$H_k = \sum_{n\geq 1}^{k}\frac{1}{n}.$$

Addition

So far I developed the following

$$\sum_{k\geq 1} \frac{H_k}{k^2} \, x^{k} = \text{Li}_3(x)-\, \text{Li}_3(1-x)+\, \log(1-x) \text{Li}_2(1-x) +\frac{1}{2}\log(x) \log^2(1-x)+\zeta(3)$$

where $\text{Li}_3(x)$ is the trilogarithm .

For the derivation see http://www.mathhelpboards.com/f10/interesting-logarithm-integral-5301/

Update

A frined on another site gave the following answer

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{k\ \geq\ 1}{\pars{-1}^{k} \over k^{3}}H_{k} & = \sum_{k = 1}^{\infty}\pars{-1}^{k}H_{k}\ \overbrace{\bracks{{1 \over 2}\int_{0}^{1}\ln^{2}\pars{x}x^{k - 1}\,\dd x}} ^{\ds{1 \over k^{3}}} \\[5mm] & = {1 \over 2}\int_{0}^{1}\ln^{2}\pars{x} \bracks{\sum_{k = 1}^{\infty}H_{k}\pars{-x}^{k}}\,{\dd x \over x} \\[5mm] & = {1 \over 2}\int_{0}^{1}\ln^{2}\pars{x} \bracks{-\,{\ln\pars{1 + x} \over 1 + x}}\,{\dd x \over x} = -\,{1 \over 2}\int_{0}^{1} {\ln^{2}\pars{x}\ln\pars{1 + x} \over \pars{1 + x}x}\,\dd x \\[5mm] & = {1 \over 2}\int_{0}^{1}{\ln^{2}\pars{x}\ln\pars{1 + x} \over 1 + x}\,\dd x - {1 \over 2}\int_{0}^{1}{\ln^{2}\pars{x}\ln\pars{1 + x} \over x}\,\dd x \\[1cm] & = {1 \over 6}\int_{0}^{1}{3\ln^{2}\pars{x}\ln\pars{1 + x} - 3\ln\pars{x}\ln^{2}\pars{1 + x} \over 1 + x}\,\dd x \\[5mm] & + {1 \over 2}\int_{0}^{1}{\ln\pars{x}\ln^{2}\pars{1 + x} \over 1 + x}\,\dd x + {1 \over 2}\int_{0}^{-1}\mrm{Li}_{2}'\pars{x}\ln^{2}\pars{-x}\,\dd x \\[1cm] & = {1 \over 6}\int_{0}^{1}{\ln^{3}\pars{x} \over 1 + x}\,\dd x - {1 \over 6}\int_{0}^{1}{\ln^{3}\pars{1 + x} \over 1 + x}\,\dd x \\[5mm] &- {1 \over 6}\int_{0}^{1}\ln^{3}\pars{x \over 1 + x}\,{\dd x \over 1 + x} - {1 \over 6}\int_{0}^{1}{\ln^{3}\pars{1 + x} \over x}\,\dd x \\[5mm] & - \int_{0}^{-1}\mrm{Li}_{3}'\pars{x}\ln\pars{-x}\,\dd x \\[1cm] & = -\,{1 \over 6}\int_{0}^{-1}{\ln^{3}\pars{-x} \over 1 - x}\,\dd x - {1 \over 24}\,\ln^{4}\pars{2} - {1 \over 6}\int_{0}^{1/2}{\ln^{3}\pars{x} \over 1 - x}\,\dd x \\[5mm] & +{1 \over 6}\int_{1}^{2}{\ln^{3}\pars{x} \over 1 - x}\,\dd x\ +\ \underbrace{\quad\int_{0}^{-1}\mrm{Li}_{4}'\pars{x}\,\dd x\quad} _{\ds{= \,\mrm{Li}_{4}\pars{-1} = -\,{7 \over 720}\,\pi^{4}}}\label{1}\tag{1} \end{align}

The remaining integrals are evaluated by successive integration by parts. Namely,

\begin{align} \int{\ln^{3}\pars{\pm x} \over 1 - x}\,\dd x & = -\ln\pars{1 - x}\ln^{3}\pars{\pm x} - 3\int\mrm{Li}_{2}'\pars{x}\ln^{2}\pars{\pm x}\,\dd x \\[5mm] & = -\ln\pars{1 - x}\ln^{3}\pars{\pm x} - 3\,\mrm{Li}_{2}\pars{x}\ln^{2}\pars{\pm x} + 6\int\mrm{Li}_{3}'\pars{x}\ln\pars{\pm x}\,\dd x \\[1cm] & = -\ln\pars{1 - x}\ln^{3}\pars{\pm x} - 3\,\mrm{Li}_{2}\pars{x}\ln^{2}\pars{\pm x} + 6\,\mrm{Li}_{3}\pars{x}\ln\pars{\pm x} \\[5mm] & - 6\int\mrm{Li}_{4}'\pars{x}\,\dd x \\[1cm] & = -\ln\pars{1 - x}\ln^{3}\pars{\pm x} - 3\,\mrm{Li}_{2}\pars{x}\ln^{2}\pars{\pm x} + 6\,\mrm{Li}_{3}\pars{x}\ln\pars{\pm x} \\[5mm] & - 6\,\mrm{Li}_{4}\pars{x}\label{2}\tag{2} \end{align}

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With \eqref{1} and \eqref{2}: $$\bbox[15px,#ffe,border:1px dotted navy]{\ds{ \sum_{k\ \geq\ 1}{\pars{-1}^{k} \over k^{3}}H_{k} = -\,{11 \over 360}\,\pi^{4} - {1 \over 12}\ln^{2}\pars{2}\pi^{2} + {1 \over 12}\,\ln^{4}\pars{2} + 2\,\mrm{Li}_{4}\pars{1 \over 2} + {7 \over 4}\,\ln\pars{2}\zeta\pars{3}}} $$

Answer 2

Let us first recall that harmonic numbers have generating function \begin{align} \sum_{k=1}^{\infty}H_kx^k=-\frac{\ln(1-x)}{1-x}, \end{align} and therefore \begin{align} S=\sum_{k=1}^{\infty}\frac{(-1)^k}{k^3}H_k&=\frac{1}{2}\sum_{k=1}^{\infty}(-1)^kH_k\int_0^{\infty}e^{-kx}x^2dx=\\ &=-\frac{1}{2}\int_0^{\infty}\frac{\ln(1+e^{-x})}{1+e^{-x}}x^2dx. \end{align} Mathematica knows how to evaluate the last integral in terms of zeta values and polylogarithms. Its answer is $$S=-\frac{11\pi^4}{360}+\frac{\ln^42-\pi^2\ln^22}{12}+2\mathrm{Li}_4\left(\frac12\right)+\frac{7\ln 2}{4}\zeta(3).$$ It is unlikely that it can be simplified further: Wolfram Alpha proposes alternative expressions for $\mathrm{Li}_{2}\left(\frac12\right)$ and $\mathrm{Li}_{3}\left(\frac12\right)$ in terms of elementary functions and zeta values, but does not suggest anything simpler for $\mathrm{Li}_{4}\left(\frac12\right)$.

Answer 3

Related problems: (I), (II), (III). Your sum is a special case of the following general case which I derived an integral representation for it

$$ A(p,q) =\sum_{k=1}^{\infty} \frac{(-1)^{k+1}H^{(p)}_k}{k^q} = \frac{\left( -1 \right) ^{q}}{\Gamma(q)}\int _{0}^{1}\!{\frac { \left( \ln\left( u \right)\right)^{q-1}{\mathrm{Li}_{p}(-u)} }{ u\left( 1+u \right) }}{du}. $$

where $ \mathrm{Li}_{p}(z) $ is the polylogarithm function. So, letting $p=1$ and $q=3$ in the above formula gives an integral representation for your sum

$$ A(1,3) =\sum_{k=1}^{\infty} \frac{(-1)^{k+1}H^{(1)}_k}{k^3} = \frac{\left( -1 \right) ^{3}}{\Gamma(3)}\int _{0}^{1}\!{\frac { \left( \ln\left( u\right) \right)^{3-1}{\mathrm{Li}_{1}(-u)} }{ u (1+u) }}{du}.$$

$$ \implies A(1,3) =\frac{1}{2}\int _{0}^{1}\!{\frac { \left( \ln \left( u \right) \right) ^{2} \ln \left( 1+u \right) }{u\left(1+u\right)}}{du} \sim 0.8592471579. $$

See here for related techniques.

Note:

1) $$ \mathrm{Li}_{1}(-u)=-\ln(1+u). $$