Infinite Series $\sum_{n=1}^\infty\frac{H_n}{n^32^n}$

Question

I'm trying to find a closed form for the following sum $$\sum_{n=1}^\infty\frac{H_n}{n^3\,2^n},$$ where $H_n=\displaystyle\sum_{k=1}^n\frac{1}{k}$ is a harmonic number.

Could you help me with it?

Answer 1

In the same spirit as Robert Israel's answer and continuing Raymond Manzoni's answer (both of them deserve the credit because of inspiring my answer) we have $$ \sum_{n=1}^\infty \frac{H_nx^n}{n^2}=\zeta(3)+\frac{1}{2}\ln x\ln^2(1-x)+\ln(1-x)\operatorname{Li}_2(1-x)+\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x). $$ Dividing equation above by $x$ and then integrating yields \begin{align} \sum_{n=1}^\infty \frac{H_nx^n}{n^3}=&\zeta(3)\ln x+\frac12\color{red}{\int\frac{\ln x\ln^2(1-x)}{x}\ dx}+\color{blue}{\int\frac{\ln(1-x)\operatorname{Li}_2(1-x)}x\ dx}\\&+\operatorname{Li}_4(x)-\color{green}{\int\frac{\operatorname{Li}_3(1-x)}x\ dx}.\tag1 \end{align} Using IBP to evaluate the green integral by setting $u=\operatorname{Li}_3(1-x)$ and $dv=\frac1x\ dx$, we obtain \begin{align} \color{green}{\int\frac{\operatorname{Li}_3(1-x)}x\ dx}&=\operatorname{Li}_3(1-x)\ln x+\int\frac{\ln x\operatorname{Li}_2(1-x)}{1-x}\ dx\qquad x\mapsto1-x\\ &=\operatorname{Li}_3(1-x)\ln x-\color{blue}{\int\frac{\ln (1-x)\operatorname{Li}_2(x)}{x}\ dx}.\tag2 \end{align} Using Euler's reflection formula for dilogarithm $$ \operatorname{Li}_2(x)+\operatorname{Li}_2(1-x)=\frac{\pi^2}6-\ln x\ln(1-x), $$ then combining the blue integral in $(1)$ and $(2)$ yields $$ \frac{\pi^2}6\int\frac{\ln (1-x)}{x}\ dx-\color{red}{\int\frac{\ln x\ln^2(1-x)}{x}\ dx}=-\frac{\pi^2}6\operatorname{Li}_2(x)-\color{red}{\int\frac{\ln x\ln^2(1-x)}{x}\ dx}. $$ Setting $x\mapsto1-x$ and using the identity $H_{n+1}-H_n=\frac1{n+1}$, the red integral becomes \begin{align} \color{red}{\int\frac{\ln x\ln^2(1-x)}{x}\ dx}&=-\int\frac{\ln (1-x)\ln^2 x}{1-x}\ dx\\ &=\int\sum_{n=1}^\infty H_n x^n\ln^2x\ dx\\ &=\sum_{n=1}^\infty H_n \int x^n\ln^2x\ dx\\ &=\sum_{n=1}^\infty H_n \frac{\partial^2}{\partial n^2}\left[\int x^n\ dx\right]\\ &=\sum_{n=1}^\infty H_n \frac{\partial^2}{\partial n^2}\left[\frac {x^{n+1}}{n+1}\right]\\ &=\sum_{n=1}^\infty H_n \left[\frac{x^{n+1}\ln^2x}{n+1}-2\frac{x^{n+1}\ln x}{(n+1)^2}+2\frac{x^{n+1}}{(n+1)^3}\right]\\ &=\ln^2x\sum_{n=1}^\infty\frac{H_n x^{n+1}}{n+1}-2\ln x\sum_{n=1}^\infty\frac{H_n x^{n+1}}{(n+1)^2}+2\sum_{n=1}^\infty\frac{H_n x^{n+1}}{(n+1)^3}\\ &=\frac12\ln^2x\ln^2(1-x)-2\ln x\left[\sum_{n=1}^\infty\frac{H_{n+1} x^{n+1}}{(n+1)^2}-\sum_{n=1}^\infty\frac{x^{n+1}}{(n+1)^3}\right]\\&+2\left[\sum_{n=1}^\infty\frac{H_{n+1} x^{n+1}}{(n+1)^3}-\sum_{n=1}^\infty\frac{x^{n+1}}{(n+1)^4}\right]\\ &=\frac12\ln^2x\ln^2(1-x)-2\ln x\left[\sum_{n=1}^\infty\frac{H_{n} x^{n}}{n^2}-\sum_{n=1}^\infty\frac{x^{n}}{n^3}\right]\\&+2\left[\sum_{n=1}^\infty\frac{H_{n} x^{n}}{n^3}-\sum_{n=1}^\infty\frac{x^{n}}{n^4}\right]\\ &=\frac12\ln^2x\ln^2(1-x)-2\ln x\left[\sum_{n=1}^\infty\frac{H_{n} x^{n}}{n^2}-\operatorname{Li}_3(x)\right]\\&+2\left[\sum_{n=1}^\infty\frac{H_{n} x^{n}}{n^3}-\operatorname{Li}_4(x)\right]. \end{align} Putting all together, we have \begin{align} \sum_{n=1}^\infty \frac{H_nx^n}{n^3}=&\frac12\zeta(3)\ln x-\frac18\ln^2x\ln^2(1-x)+\frac12\ln x\left[\sum_{n=1}^\infty\frac{H_{n} x^{n}}{n^2}-\operatorname{Li}_3(x)\right]\\&+\operatorname{Li}_4(x)-\frac{\pi^2}{12}\operatorname{Li}_2(x)-\frac12\operatorname{Li}_3(1-x)\ln x+C.\tag3 \end{align} Setting $x=1$ to obtain the constant of integration, \begin{align} \sum_{n=1}^\infty \frac{H_n}{n^3}&=\operatorname{Li}_4(1)-\frac{\pi^2}{12}\operatorname{Li}_2(1)+C\\ \frac{\pi^4}{72}&=\frac{\pi^4}{90}-\frac{\pi^4}{72}+C\\ C&=\frac{\pi^4}{60}. \end{align} Thus \begin{align} \sum_{n=1}^\infty \frac{H_nx^n}{n^3}=&\frac12\zeta(3)\ln x-\frac18\ln^2x\ln^2(1-x)+\frac12\ln x\left[\sum_{n=1}^\infty\frac{H_{n} x^{n}}{n^2}-\operatorname{Li}_3(x)\right]\\&+\operatorname{Li}_4(x)-\frac{\pi^2}{12}\operatorname{Li}_2(x)-\frac12\operatorname{Li}_3(1-x)\ln x+\frac{\pi^4}{60}.\tag4 \end{align} Finally, setting $x=\frac12$, we obtain \begin{align} \sum_{n=1}^\infty \frac{H_n}{2^nn^3}=\color{purple}{\frac{\pi^4}{720}+\frac{\ln^42}{24}-\frac{\ln2}8\zeta(3)+\operatorname{Li}_4\left(\frac12\right)}, \end{align} which matches Cleo's answer.

---

References :

$[1]\ $ Harmonic number

$[2]\ $ Polylogarithm

Answer 2

$$\sum_{n=1}^\infty\frac{H_n}{n^3\,2^n}=\frac{\pi^4}{720}+\frac{\ln^42}{24}-\frac{\ln2}8\zeta(3)+\operatorname{Li}_4\left(\frac12\right).$$

Answer 3

Note: Please note the top voted answer by @Tunk-Fey is regrettably not correct. Contrary to his claim his final expression (4) when evaluated at $x=\frac{1}{2}$ does not match @Cleo's answer but differs by $\frac{\pi^4}{120}$ from the correct identity: \begin{align*} \sum_{n=1}^\infty \frac{H_n}{n^32^n}&=-\frac{1}{8}\ln 2\zeta(3)+\frac{1}{24}\ln^4(2)+\frac{\pi^4}{720}+ \operatorname{Li}_4\left(\frac{1}{2}\right)\\ &\stackrel{.}{=}0.55824 \end{align*} A rather detailed analysis of the deviation from the correct result is provided in this answer.

Nevertheless it was a pleasure to review his answer which contains nice and instructive aspects. Here I provide a solution in a similar spirit which hopefully overcomes the problems of his answer.

Raymond Manzoni's has nicely demonstrated that for $|x|<1$ \begin{align*} \sum_{n=1}^\infty \frac{H_nx^n}{n^2}&=\zeta(3)+\frac{1}{2}\ln x\ln^2(1-x)+\ln(1-x)\operatorname{Li}_2(1-x)\\ &\qquad+\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x) \end{align*}

This result is our starting point.

\begin{align*} \sum_{n=1}^\infty \frac{H_nx^n}{n^3}&=\int\sum_{n=1}^\infty \frac{H_nx^{n-1}}{n^2}dx\\ &=\zeta(3)\ln(x)+\frac{1}{2}\int\frac{1}{x}\ln x\ln^2(1-x)dx+\int\frac{\ln(1-x)}{x}\operatorname{Li}_2(1-x)dx\\ &\qquad+\int\frac{1}{x}\operatorname{Li}_3(x)dx-\int\frac{1}{x}\operatorname{Li}_3(1-x)dx+C\tag{1}\\ \end{align*}

At first we consider $\int\frac{1}{x}\operatorname{Li}_3(1-x)dx$. Integration by parts with $u=\frac{1}{x}$ and $dv=\operatorname{Li}_3(1-x)dx$ gives

\begin{align*} \int\frac{1}{x}\operatorname{Li}_3(1-x)dx&=\ln x\operatorname{Li}_3(1-x)+\int\frac{\ln x}{1-x}\operatorname{Li}_2(1-x)dx\\ &=\ln x\operatorname{Li}_3(1-x)+\frac{1}{2}\operatorname{Li}_2^2(1-x)+C \end{align*} Once again integration by parts on the RHS with $u=\frac{\ln x}{1-x}$ and $dv=\operatorname{Li}_2(1-x)dx$ gives \begin{align*} \int\frac{\ln x}{1-x}\operatorname{Li}_2(1-x)dx&=\operatorname{Li}_2^2(1-x) -\int\frac{\ln x}{1-x}\operatorname{Li}_2(1-x)dx\\ \Longrightarrow\int\frac{\ln x}{1-x}\operatorname{Li}_2(1-x)dx&=\frac{1}{2}\operatorname{Li}_2^2(1-x)+C \end{align*}

It follows \begin{align*} \int\frac{1}{x}\operatorname{Li}_3(1-x)dx&=\operatorname{Li}_3(1-x)\ln x+\frac{1}{2}\operatorname{Li}_2^2(1-x)+C \end{align*}

and we obtain substituting this result in (1) and noting that \begin{align*} \int\frac{1}{x}\operatorname{Li}_3(x)dx=\operatorname{Li}_4(x)+C \end{align*}

\begin{align*} \sum_{n=1}^\infty \frac{H_nx^n}{n^3}&=\zeta(3)\ln x+\frac{1}{2}\int\frac{1}{x}\ln x\ln^2(1-x)dx+\int\frac{\ln(1-x)}{x}\operatorname{Li}_2(1-x)dx\\ &\qquad+\operatorname{Li}_4(x)-\left(\operatorname{Li}_3(1-x)\ln x+\frac{1}{2}\operatorname{Li}_2^2(1-x)\right)+C\tag{2}\\ \end{align*}

The next step is to calculate $\int\frac{1}{x}\ln x\ln^2(1-x)dx$. We use Euler's reflection formula \begin{align*} \operatorname{Li}_2(x)+\operatorname{Li}_2(1-x)=\frac{\pi^2}{6}-\ln x\ln(1-x) \end{align*} to split the integral into parts which can either be directly calculated or which can be transformed to the remaining integral. We obtain using the reflection formula

\begin{align*} \int&\frac{1}{x}\ln x\ln^2(1-x)dx\\ &=\int\frac{\ln(1-x)}{x}\left(\frac{\pi^2}{6}-\operatorname{Li}_2(x)-\operatorname{Li}_2(1-x)\right)\\ &=-\frac{\pi^2}{6}\operatorname{Li}_2(x)-\int\frac{\ln(1-x)}{x}\operatorname{Li}_2(x)dx -\int\frac{\ln(1-x)}{x}\operatorname{Li}_2(1-x)dx\\ &=-\frac{\pi^2}{6}\operatorname{Li}_2(x)+\frac{1}{2}\operatorname{Li}_2^2(x)dx -\int\frac{\ln(1-x)}{x}\operatorname{Li}_2(1-x)dx \end{align*}

Putting this result into (2) we get

\begin{align*} \sum_{n=1}^\infty \frac{H_nx^n}{n^3}&=\zeta(3)\ln x +\frac{1}{2}\left(-\frac{\pi^2}{6}\operatorname{Li}_2(x)+\frac{1}{2}\operatorname{Li}_2^2(x) -\int\frac{\ln(1-x)}{x}\operatorname{Li}_2(1-x)dx\right)\\ &\qquad+\int\frac{\ln(1-x)}{x}\operatorname{Li}_2(1-x)dx\\ &\qquad+\operatorname{Li}_4(x)-\left(\operatorname{Li}_3(1-x)\ln x+\frac{1}{2}\operatorname{Li}_2^2(1-x)\right)+C\\ &=\zeta(3)\ln x-\frac{\pi^2}{12}\operatorname{Li}_2(x)+\frac{1}{4}\operatorname{Li}_2^2(x) -\frac{1}{2}\operatorname{Li}_2^2(1-x)\\ &\qquad-\operatorname{Li}_3(1-x)\ln x+\operatorname{Li}_4(x)\\ &\qquad+\frac{1}{2}\int\frac{\ln(1-x)}{x}\operatorname{Li}_2(1-x)dx+C\tag{3}\\ \end{align*}

The most complex and cumbersome part is the remaining integral in (3). With the help of Wolfram Alpha a rather lengthy result is provided. After some simplifications we obtain \begin{align*} \int&\frac{\ln(1-x)}{x}\operatorname{Li}_2{(1-x)}dx\\ &=-\frac{1}{2}\ln^2(1-x)\ln^2x+\ln(1-x)\ln^3x-\frac{1}{4}\ln^4x\\ &\qquad-\operatorname{Li}_2(1-x)\left(\ln^2(1-x)-\ln(1-x)\ln x\right)+\operatorname{Li}_2(x)\ln^2 x\\ &\qquad-\operatorname{Li}_2\left(1-\frac{1}{x}\right)\left(\ln^2(1-x)-2\ln(1-x)\ln x+\ln^2 x\right)+\frac{1}{2}\operatorname{Li}_2^2(1-x)\\ &\qquad+2\left(\operatorname{Li}_3\left(1-\frac{1}{x}\right)\left(\ln(1-x)-\ln x\right)+\operatorname{Li}_3(1-x)\ln(1-x) -\operatorname{Li}_3(x)\ln x\right)\\ &\qquad-2\left(\operatorname{Li}_4(1-x)+\operatorname{Li}_4\left(1-\frac{1}{x}\right)-\operatorname{Li}_4(x)\right)+C\\ \end{align*}

Finally substituting this expression into (3) and doing some more simplifications we obtain

\begin{align*} \sum_{n=1}^\infty \frac{H_nx^n}{n^3}&=\zeta(3)\ln x-\frac{\pi^2}{12}\operatorname{Li}_2(x)+\frac{1}{4}\operatorname{Li}_2^2(x) -\frac{1}{2}\operatorname{Li}_2^2(1-x)\\ &\quad-\operatorname{Li}_3(1-x)\ln x+\operatorname{Li}_4(x)\\ &\quad+\frac{1}{2}\left(-\frac{1}{2}\ln^2(1-x)\ln^2x+\ln(1-x)\ln^3x-\frac{1}{4}\ln^4x\right.\\ &\quad\quad-\operatorname{Li}_2(1-x)\left(\ln^2(1-x)-\ln(1-x)\ln x\right)+\operatorname{Li}_2(x)\ln^2 x\\ &\quad\quad-\operatorname{Li}_2\left(1-\frac{1}{x}\right)\left(\ln^2(1-x)-2\ln(1-x)\ln x+\ln^2 x\right)\\ &\quad\quad+\frac{1}{2}\operatorname{Li}_2^2(1-x)\\ &\quad\quad+2\left(\operatorname{Li}_3\left(1-\frac{1}{x}\right)\left(\ln(1-x)-\ln x\right)\right.\\ &\quad\quad\quad+\left.\operatorname{Li}_3(1-x)\ln(1-x)-\operatorname{Li}_3(x)\ln x\right)\\ &\quad\quad\left.-2\left(\operatorname{Li}_4(1-x)+\operatorname{Li}_4\left(1-\frac{1}{x}\right)-\operatorname{Li}_4(x)\right)\right)+C\\ &=\zeta(3)\ln x-\frac{1}{4}\ln^2(1-x)\ln^2x+\frac{1}{2}\ln(1-x)\ln^3x-\frac{1}{8}\ln^4x\\ &\quad-\frac{1}{2}\operatorname{Li}_2(1-x)\left(\ln^2(1-x)-\ln(1-x)\ln x\right)+\frac{1}{2}\operatorname{Li}_2(x)\left(\ln^2 x-\frac{\pi^2}{6}\right)\\ &\quad-\frac{1}{2}\operatorname{Li}_2\left(1-\frac{1}{x}\right)\left(\ln^2(1-x)-2\ln(1-x)\ln x+\ln^2 x\right)\\ &\quad+\frac{1}{4}\operatorname{Li}^2_2(x)-\frac{1}{4}\operatorname{Li}^2_2(1-x)-\operatorname{Li}_3(x)\ln x\\ &\quad+\operatorname{Li}_3\left(1-\frac{1}{x}\right)\left(\ln(1-x)-\ln x\right)+\operatorname{Li}_3(1-x)\left(\ln(1-x)-\ln(x)\right)\\ &\quad-\operatorname{Li}_4(1-x)-\operatorname{Li}_4\left(1-\frac{1}{x}\right)+2\operatorname{Li}_4(x)+C\tag{4} \end{align*}

From (4) we can now determine the integration constant $C$. In order to do so we calculate $C$ by taking the limit as $x\rightarrow 1$. Most of the terms vanish and noting that according to this answer \begin{align*} \sum_{n=1}^\infty \frac{H_n}{n^3}=\frac{\pi^4}{72} \end{align*} we obtain respecting that $\operatorname{Li}_2(1)=\frac{\pi^2}{6}$ and $\operatorname{Li}_4(1)=\frac{\pi^4}{90}$

\begin{align*} \frac{\pi^4}{72}&=\frac{1}{2}\operatorname{Li}_2(1)\left(-\frac{\pi^2}{6}\right)+\frac{1}{4}\operatorname{Li}^2_2(1)+2\operatorname{Li}_4(1)+C\\ &=-\frac{\pi^4}{72}+\frac{\pi^4}{144}+\frac{2\pi^4}{90}+C\\ \text{it follows}\qquad C&=-\frac{\pi^4}{720} \end{align*}

Setting $x=\frac{1}{2}$ in (4) we finally obtain with $C=-\frac{\pi^4}{720}$ and noting that \begin{align*} \operatorname{Li}_2\left(\frac{1}{2}\right)&=\frac{\pi^{2}}{12}-\frac{1}{2}\ln^2(2)\\ \operatorname{Li}_3\left(\frac{1}{2}\right)&=\frac{7}{8}\zeta(3)+\frac{1}{6}\ln^3(2)-\frac{\pi^{2}}{12}\ln 2\\ \operatorname{Li}_4(-1)&=-\frac{7\pi^4}{720} \end{align*}

\begin{align*} \sum_{n=1}^\infty \frac{H_n}{n^32^n}&=-\zeta(3)\ln(2)+\frac{1}{8}\ln^4(2) +\frac{1}{2}\operatorname{Li}_2\left(\frac{1}{2}\right)\left(\ln^2(2)-\frac{\pi^2}{6}\right)\\ &\qquad+\operatorname{Li}_3\left(\frac{1}{2}\right)\ln 2-\operatorname{Li}_4(-1)+\operatorname{Li}_4\left(\frac{1}{2}\right)-\frac{\pi^4}{720}\\ &=-\frac{1}{8}\ln 2\zeta(3)+\frac{1}{24}\ln^4(2)+\frac{\pi^4}{720}+ \operatorname{Li}_4\left(\frac{1}{2}\right)\\ &\stackrel{.}{=}0.55824 \end{align*} and the claim follows.

Note: Two aspects remain open. The important one is a derivation of \begin{align*} \int&\frac{\ln(1-x)}{x}\operatorname{Li}_2{(1-x)}dx \end{align*} without support from WA. It would also be nice to find some further simplifications of the final expression (4).

Answer 4

Start with the series $$\sum_{n=1}^\infty H_n z^n = - \dfrac{\ln(1-z)}{1-z} = f_0(z) $$

Then (according to Maple 18) $$ \sum_{n=1}^\infty \dfrac{H_n}{n} z^n = \int_0^z \dfrac{f_0(t)}{t}\; dt = \operatorname{Li}_{2}(1-z) + \dfrac{\ln(1-z)^2}{2} = f_1(z)$$

$$\displaystyle \sum_{n=1}^\infty \dfrac{H_n}{n^2} z^n = \int_0^z \dfrac{f_1(t)}{t} dt$$

$$= \zeta \left( 3 \right) +\dfrac{1}{2}\, \ln^2 (1-z) \ln \left( z \right) +\ln (1-z) \operatorname{Li}_{2} (z) -\operatorname{Li}_{3}(1-z) + \operatorname{Li}_{3}(z) $$

But for the next integration it fails to find a closed form. $$\sum_{n=1}^\infty \dfrac{H_n}{n^3} z^n = \int_0^z f_2(t)\; dt$$

Answer 5

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

Since values of $\ds{\,\mrm{Li}_{2}\pars{1/2}}$ and $\ds{\,\mrm{Li}_{3}\pars{1/2}}$ are well known and $\ds{\,\mrm{Li}_{3}\pars{1} = \zeta\pars{3}}$:

\begin{align} \sum_{n = 1}^{\infty}{H_{n} \over n^{3}\,2^{n}} & = -\,{1 \over 12}\,\ln^{4}\pars{2} - {1 \over 8}\,\ln\pars{2}\zeta\pars{3} + \,\mrm{Li}_{4}\pars{1 \over 2} - {1 \over 2}\ \underbrace{\int_{1/2}^{1}{\ln^{2}\pars{1 - x}\ln\pars{x} \over x}\,\dd x} _{\ds{\equiv\ \mc{I}}} \label{1}\tag{1} \end{align}

---

$\ds{\large\mc{I}:\ ?}$. \begin{align} \mc{I} & \equiv \int_{1/2}^{1}{\ln^{2}\pars{1 - x}\ln\pars{x} \over x}\,\dd x \\[5mm] & = {1 \over 3}\int_{1/2}^{1}\!{\ln^{3}\pars{1 - x} \over x}\dd x - {1 \over 3}\int_{1/2}^{1}\!{\ln^{3}\pars{x} \over x}\dd x - {1 \over 3}\int_{1/2}^{1}\!\ln^{3}\pars{1 - x \over x}{\dd x \over x} + \int_{1/2}^{1}\!{\ln\pars{1 - x}\ln^{2}\pars{x} \over x}\,\dd x \\[5mm] & = {1 \over 3}\int_{0}^{1/2}{\ln^{3}\pars{x} \over 1 - x}\dd x + {1 \over 12}\,\ln^{4}\pars{2} + {1 \over 3}\int_{0}^{-1}{\ln^{3}\pars{-x} \over 1 - x}\,\dd x - \int_{1/2}^{1}\mrm{Li}_{2}'\pars{x}\ln^{2}\pars{x}\,\dd x \\[1cm] & = {1 \over 3}\bracks{-\ln^{4}\pars{2} - 3\int_{0}^{1/2}\mrm{Li}_{2}'\pars{x}\ln^{2}\pars{x}\dd x} + {1 \over 12}\,\ln^{4}\pars{2} - \int_{0}^{-1}\mrm{Li}_{2}'\pars{x}\ln^{2}\pars{-x}\,\dd x \\[5mm] & -\int_{1/2}^{1}\mrm{Li}_{2}'\pars{x}\ln^{2}\pars{x}\,\dd x \\[1cm] & = -\,{1 \over 4}\,\ln^{4}\pars{2} -\int_{0}^{1}\mrm{Li}_{2}'\pars{x}\ln^{2}\pars{x}\,\dd x - \int_{0}^{-1}\mrm{Li}_{2}'\pars{x}\ln^{2}\pars{-x}\,\dd x \end{align}

The remaining integrals can be straightforward evaluated by successive integration by parts and by using the $\ds{\,\mrm{Li}_{s}}$ recursive property. Namely,

\begin{align} &\int\mrm{Li}_{2}'\pars{x}\ln^{2}\pars{\pm x}\,\dd x = \mrm{Li}_{2}\pars{x}\ln^{2}\pars{\pm x} - 2\int\mrm{Li}_{3}'\pars{x}\ln\pars{\pm x}\,\dd x \\[5mm] & = \mrm{Li}_{2}\pars{x}\ln^{2}\pars{\pm x} - 2\,\mrm{Li}_{3}\pars{x}\ln\pars{\pm x} + 2\int\mrm{Li}_{4}'\pars{x}\,\dd x \\[5mm] & =\ \bbox[15px,#ffe,border:1px dotted navy]{\ds{% \mrm{Li}_{2}\pars{x}\ln^{2}\pars{\pm x} - 2\,\mrm{Li}_{3}\pars{x}\ln\pars{\pm x} + 2\,\mrm{Li}_{4}\pars{x}}} \end{align} such that \begin{equation} \mc{I} \equiv \int_{1/2}^{1}{\ln^{2}\pars{1 - x}\ln\pars{x} \over x}\,\dd x =\ \bbox[15px,#ffe,border:1px dotted navy]{\ds{% -\,{1 \over 4}\,\ln^{4}\pars{2} - {\pi^{4} \over 360}}}\label{2}\tag{2} \end{equation}

---

With \eqref{1} and \eqref{2}: \begin{align} \sum_{n = 1}^{\infty}{H_{n} \over n^{3}\,2^{n}} & = -\,{1 \over 12}\,\ln^{4}\pars{2} - {1 \over 8}\,\ln\pars{2}\zeta\pars{3} + \,\mrm{Li}_{4}\pars{1 \over 2} - {1 \over 2} \bracks{-\,{1 \over 4}\,\ln^{4}\pars{2} - {\pi^{4} \over 360}} \\[5mm] & =\ \bbox[25px,#ffe,border:1px dotted navy]{\ds{% {1 \over 720}\,\pi^{4} + {1 \over 24}\,\ln^{4}\pars{2} - {1 \over 8}\,\ln\pars{2}\zeta\pars{3} + \,\mrm{Li}_{4}\pars{1 \over 2}}}\ \approx 0.5582 \end{align}

Answer 6

Alternative approach:

first we start with proving the following equality that appeared as Problem 11921 in The American Mathematical Monthly 2016 proposed by Cornel Ioan Valean: \begin{equation*} S=\ln^22\sum_{n=1}^{\infty}\frac{H_n}{(n+1) 2^{n+1}}+\ln2\sum_{n=1}^{\infty}\frac{H_n}{(n+1)^2 2^n}+\sum_{n=1}^{\infty}\frac{H_n}{(n+1)^3 2^n}=\frac14\ln^42+\frac14\zeta(4) \end{equation*} Proof: lets start with the following integral $ I=\displaystyle \int_{1/2}^{1} \frac{\ln(1-x)\ln^2x}{1-x}\,dx $

By using

$$\frac{\ln(1-x)}{1-x}=-\displaystyle \sum_{n=1}^{\infty}H_n x^n$$

we can write

$$I=-\sum_{n=1}^{\infty}H_n\int_{1/2}^{1}x^n \ln^2x\,dx$$

$$=-\sum_{n=1}^{\infty}H_n\left( -\frac{\ln^22}{(n+1)2^{n+1}}-\frac{\ln2}{(n+1)2^{n+1}}-\frac{1}{(n+1)^32^n}+\frac{2}{(n+1)^3}\right)$$

$$=S-2\sum_{n=1}^{\infty}\frac{H_n}{(n+1)^3} \tag{1}$$

On the other hand, upon integrating by parts we obtain \begin{equation*} I=\frac12\ln^42+\int_{1/2}^1 \frac{\ln^2x\ln(1-x)}{x}\,dx\overset{x\mapsto 1-x}{=}\frac12\ln^42+\int_0^{1/2}\frac{\ln^2x\ln(1-x)}{1-x}\,dx \end{equation*} Adding the integral $I=\int_{1/2}^{1}\frac{\ln^2x\ln(1-x)}{1-x}\,dx\ $ to both sides

$$2I=\frac12\ln^42+\int_0^1 \frac{\ln^2x\ln(1-x)}{1-x}\,dx=\frac12\ln^42-\sum_{n=1}^{\infty}H_n\int_{0}^{1}x^n\ln^2x\,dx$$ $$=\frac12\ln^42-2\sum_{n=1}^{\infty}\frac{H_n}{(n+1)^3}\Longrightarrow I=\frac14\ln^42-\sum_{n=1}^{\infty}\frac{H_n}{(n+1)^3} \tag{2}$$

combining $(1)$ and $(2)$ yields

$$S=\frac14\ln^42+\sum_{n=1}^{\infty}\frac{H_n}{(n+1)^3}=\frac14\ln^42-\zeta(4)+\sum_{n=1}^{\infty}\frac{H_n}{n^3}$$

subbing $\sum_{n=1}^{\infty}\frac{H_n}{n^3}=\frac54\zeta(4)$ completes the proof.

---

Using the proved equality: \begin{align*} \frac14\ln^42+\frac14\zeta(4)&=\ln^22\sum_{n=1}^{\infty}\frac{H_n}{(n+1) 2^{n+1}}+\ln2\sum_{n=1}^{\infty}\frac{H_n}{(n+1)^2 2^n}+\sum_{n=1}^{\infty}\frac{H_n}{(n+1)^3 2^n}\\ &=\ln^22\sum_{n=1}^{\infty}\frac{H_{n-1}}{n 2^n}+2\ln2\sum_{n=1}^{\infty}\frac {H_{n-1}}{n^2 2^n}+2\sum_{n=1}^{\infty}\frac{H_{n-1}}{n^3 2^n}\\ &=\ln^22\sum_{n=1}^{\infty}\frac{H_n}{n 2^n}+2\ln2\sum_{n=1}^{\infty}\frac {H_n}{n^2 2^n} +2\sum_{n=1}^{\infty}\frac{H_n}{n^3 2^n}-\ln^22\sum_{n=1}^{\infty}\frac{1}{2^n n^2}\\ &\quad -2\ln2\sum_{n=1}^{\infty}\frac{1}{ n^32^n}-2\sum_{n=1}^{\infty}\frac{1}{n^42^n} \end{align*} rearrange the terms to get

$$\sum_{n=1}^{\infty}\frac{H_n}{n^3 2^n}=-\ln2\sum_{n=1}^{\infty}\frac{H_n}{n^2 2^n}-\frac12\ln^22\sum_{n=1}^{\infty}\frac{H_n}{n 2^n}+\operatorname{Li_4}\left( \frac12\right)+\ln2\operatorname{Li_3}\left( \frac12\right)\\+\frac12\ln^22\operatorname{Li_2}\left( \frac12\right)+\frac18\zeta(4)+\frac18\ln^42$$

plugging the values of the first and second sum proved here and here respectively, along with the values of $\displaystyle\operatorname{Li_3}\left(\frac12\right)$ and $\displaystyle\operatorname{Li_2}\left(\frac12\right)$ we obtain \begin{align} \sum_{n=1}^\infty \frac{H_n}{2^nn^3}=\color{blue}{\operatorname{Li}_4\left(\frac12\right)+\frac18\zeta(4)-\frac18\ln2\zeta(3)+\frac1{24}\ln^42}, \end{align}

Answer 7

You can have instead the equivalent integral representation

$$ I = \int_{0}^{1}\frac{\ln^2(u)\ln(1-u/2)}{u(u-2)}du \sim .5582373010. $$

Try to evaluate the above integral. See my answer. See also here.

Answer 8

By first finding the following integral by using the algebraic identity $a^2b=\frac{1}{6}\left(a+b\right)^3-\frac{1}{6}\left(a-b\right)^3-\frac{1}{3}b^3$ one can easily prove avoiding Euler sums that: $$\int _0^1\frac{\ln ^2\left(1-x\right)\ln \left(1+x\right)}{1+x}\:dx=-\frac{1}{4}\zeta \left(4\right)+2\ln \left(2\right)\zeta \left(3\right)-\ln ^2\left(2\right)\zeta \left(2\right)+\frac{1}{4}\ln ^4\left(2\right)$$ Now: $$\int _0^1\frac{\ln ^2\left(1-x\right)\ln \left(1+x\right)}{1+x}\:dx=\frac{1}{2}\ln \left(2\right)\int _0^1\frac{\ln ^2\left(x\right)}{1-\frac{x}{2}}\:dx+\frac{1}{2}\int _0^1\frac{\ln ^2\left(x\right)\ln \left(1-\frac{x}{2}\right)}{1-\frac{x}{2}}\:dx$$ $$=2\ln \left(2\right)\sum _{k=1}^{\infty }\frac{1}{k^3\:2^k}-2\sum _{k=1}^{\infty }\frac{H_k}{k^3\:2^k}+2\sum _{k=1}^{\infty }\frac{1}{k^4\:2^k}$$ $$=2\ln \left(2\right)\operatorname{Li}_3\left(\frac{1}{2}\right)-2\sum _{k=1}^{\infty }\frac{H_k}{k^3\:2^k}+2\operatorname{Li}_4\left(\frac{1}{2}\right)$$ $$=\frac{7}{4}\ln \left(2\right)\zeta \left(3\right)-\ln ^2\left(2\right)\zeta \left(2\right)+\frac{1}{3}\ln ^4\left(2\right)-2\sum _{k=1}^{\infty }\frac{H_k}{k^3\:2^k}+2\operatorname{Li}_4\left(\frac{1}{2}\right)$$ By making use of the result we find: $$\sum _{k=1}^{\infty }\frac{H_k}{k^3\:2^k}=\frac{1}{8}\zeta \left(4\right)+\operatorname{Li}_4\left(\frac{1}{2}\right)-\frac{1}{8}\ln \left(2\right)\zeta \left(3\right)+\frac{1}{24}\ln ^4\left(2\right)$$