# How to compute $\sum_{n=1}^\infty\frac{H_n^2}{n^32^n}$?

Can we evaluate $$\displaystyle\sum_{n=1}^\infty\frac{H_n^2}{n^32^n}$$ ?

where $$H_n=\sum_{k=1}^n\frac1n$$ is the harmonic number.

A related integral is $$\displaystyle\int_0^1\frac{\ln^2(1-x)\operatorname{Li}_2\left(\frac x2\right)}{x}dx$$.

where $$\operatorname{Li}_2(x)=\sum_{n=1}^\infty\frac{x^n}{n^2}$$ is the dilogarithmic function.

Here is how the integral and the sum are related:

From here we have

$$\int_0^1x^{n-1}\ln^2(1-x)\ dx=\frac{H_n^2+H_n^{(2)}}{n}$$

Divide both sides by $$n^22^n$$ then sum up we get

$$\sum_{n=1}^\infty \frac{H_n^2+H_n^{(2)}}{n^32^n}=\int_0^1\frac{\ln^2(1-x)}{x}\sum_{n=1}^\infty \frac{x^n}{n^22^n}dx=\int_0^1\frac{\ln^2(1-x)\operatorname{Li}_2(x/2)}{x}dx$$

• We can certainly evaluate it numerically. It's unlikely to have a closed form solution in terms of "nice" functions. Are you looking for a closed form solution? – parsiad Aug 28 at 9:07
• Well, it is at least doable for $n^2$ instead of $n^3$ in the denominator (of course, only by considering $\operatorname{Li}_4\left(\frac12\right)$ as given as usual). – mrtaurho Aug 28 at 11:13
• @parsiad yes solution should be in closed form. – Ali Shather Aug 28 at 15:03
• @mrtaurho the $n^2$ version is already calculated and I posted the problem/ solution before. – Ali Shather Aug 28 at 15:06
• @AliShather Ah, okay. I wasn't aware of this. However, I can't adapt my solution in order to evaluate this sum. – mrtaurho Aug 28 at 15:07

Using the generating function of $$\displaystyle\{H_k^2\}_{k=1}^\infty$$: $$\frac{\text{Li}_2(x) +\ln^2(1-x)}{1-x} = \sum_{k=1}^\infty H_k^2 x^k$$ we can observe that \begin{align*} S =& \sum_{k=1}^\infty {H_k^2 \over 2^k} \frac 1{k^2}\\ =& \sum_{k=1}^\infty {H_k^2 \over 2^k} \frac 1 2\int_0^1 x^{k-1}\ln^2 x\ dx\\ =&\frac 1 2 \int_0^1 \left(\sum_{k=1}^\infty H_k^2\left(\frac x 2\right)^k\right)\frac{\ln^2 x}{x} dx\\ =& \frac 1 2 \int_0^{\frac 1 2}\left(\sum_{k=1}^\infty H_k^2 x ^k\right) \frac{\ln^2 (2x)}{x} dx \\ =& \frac 1 2 \int_0^{\frac 1 2} \frac{\text{Li}_2(x) +\ln^2(1-x)}{1-x} \frac{\ln^2 x}{x} dx \\ &+ \ln 2\int_0^{\frac 1 2} \frac{\text{Li}_2(x) +\ln^2(1-x)}{1-x} \frac{\ln x}{x} dx \\ &+\frac{\ln^2 2}{2}\int_0^{\frac 1 2} \frac{\text{Li}_2(x) +\ln^2(1-x)}{1-x} \frac{dx}{x}\\ =&: I_1 + I_2 + I_3. \end{align*}

For $$I_1$$, we have \begin{align*} I_1=&\frac 1 2 \int_0^{\frac 12 } \frac{\big[\text{Li}_2(x)+\ln^2(1-x)\big]\ln^2 x}{x(1-x)}dx \\ =& \frac 1 2 \int_0^{\frac 12 } \frac{\text{Li}_2(x)\ln^2 x}{x}dx+\frac 1 2 \int_0^{\frac 12 } \frac{\text{Li}_2(x)\ln^2 x}{1-x}dx +\frac 1 2 \int_0^{\frac 1 2} \frac{\ln^2 (1-x)\ln^2 x}{x(1-x)} dx\\ =&:I_1'+I_1''+I_1'''. \end{align*}

For $$I_1'$$, we integrate by parts twice to get \begin{align*} I_1' \underset{\text{IBP}}{=}& \frac 1 2\left[ \text{Li}_3(x)\ln^2 x\right]^{1/2}_0 -\int_0^{\frac 1 2}\frac{\text{Li}_3(x) \ln x}{x} dx\\ \underset{\text{IBP}}{=}&\frac{\ln^2 2\ \text{Li}_3(1/2)}2 - \left[\text{Li}_4(x)\ln x\right]^{1/2}_0 +\int_0^{\frac 1 2 }\frac{\text{Li}_4(x)}x dx\\ =&\frac{\ln^2 2\ \text{Li}_3(1/2)}2 + \ln 2\ \text{Li}_4(1/2)+\text{Li}_5(1/2)\\ =&\boxed{\text{Li}_5(1/2)+\ln 2\ \text{Li}_4(1/2)+\frac{7\ln^2 2}{16}\zeta(3)-\frac {\pi^2\ln^3 2}{24}+\frac{\ln^5 2}{12}} \end{align*} where the well-known value of $$\text{Li}_3(1/2) = \frac 78 \zeta(3) -\frac{\pi^2\ln 2}{12}+\frac{\ln^3 2}{6}$$ is used to simplify.

For $$I_1''$$, by integrating by parts, \begin{align*} I_1'' \underset{\text{IBP}}{=}& \frac 1 2 \int_0^{\frac 1 2} \ln(1-x)\left[\frac{2\ln x\text{Li}_2(x)}{x} - \frac{\ln(1-x)\ln^2 x}{x}\right]dx +\frac{\ln^3 2}{2}\text{Li}_2(1/2)\\ =& {\int_0^{\frac 1 2} \ln x\frac{\ln(1-x)\text{Li}_2(x)}{x} dx}-\underbrace{\frac 1 2 \int_0^{\frac 1 2} \frac{\ln^2(1-x)\ln^2 x}{x}dx}_{=:J} +\frac{\ln^3 2}{2}\text{Li}_2(1/2)\\ \underset{\text{IBP}}{=}&\frac{\ln 2}2\text{Li}_2^2(1/2) +{\frac 1 2\int_0^{\frac 1 2}\frac{\text{Li}_2^2(x)}{x} dx}-J +\frac{\ln^3 2}{2}\text{Li}_2(1/2)\\ =&\frac{\pi^4\ln 2}{288} -\frac{\ln^5 2}{8}+{\frac 1 2\int_0^{\frac 1 2}\frac{\text{Li}_2^2(x)}{x} dx}-J. \end{align*} The well-known value of $$\text{Li}_2(1/2) = \frac{\pi^2}{12} - \frac{\ln^2 2}{2}$$ is used to simplify. In fact, the integral $${\int_0^{1/2}\frac{\text{Li}_2^2(x)}{x} dx}$$ was already evaluated in my previous answer here: \begin{align*} {\int_0^{\frac 1 2}\frac{\text{Li}_2^2(x)}{x} dx} = &-2\text{Li}_5(1/2) -2\ln 2\ \text{Li}_4(1/2)+\frac{27}{32}\zeta(5) +\frac{7\pi^2}{48}\zeta(3)-\frac{7\ln^2 2}{8}\zeta(3) \\ &-\frac{\pi^4\ln 2}{144} +\frac{\pi^2\ln^3 2}{12} - \frac{7\ln^5 2}{60}. \end{align*}

For $$J$$, we make substitution $$y= \frac{x}{1-x}$$ to get \begin{align*} J=&\frac 1 2\int_0^1 \frac{\ln^2(1+y)\ln^2\left(\frac{y}{1+y}\right)}{y(1+y)}dy\\ =&\frac 1 2\int_0^1 \frac{\ln^2(1+y)\ln^2\left(\frac{y}{1+y}\right)}{y}dy-\frac 1 2\int_0^1 \frac{\ln^2(1+y)\ln^2\left(\frac{y}{1+y}\right)}{1+y}dy\\ =&:K-L. \end{align*}

For $$K$$, expanding $$\ln^2\left(\frac y {1+y}\right)=\big[\ln y -\ln(1+y)\big]^2$$ and integrating by parts we get \begin{align*} K =&\frac 1 2\int_0^1 \frac{\ln^2(1+y)\ln^2\left(\frac{y}{1+y}\right)}{y}dy\\ =& \frac 12{ \int_0^1\frac{\ln^2 y\ln^2(1+y)}{y} dy}-{\int_0^1 \frac{\ln y\ln^3(1+y)}{y} dy}+\frac 12 \int_0^1 \frac{\ln^4(1+y)}{y}dy\\ \underset{\text{IBP}}{=}&-\frac 1 3 \int_0^1 \frac{\ln^3 y \ln(1+y)}{1+y}dy + \frac 3 2\int_0^1 \frac{\ln^2 y\ln^2(1+y)}{1+y}dy +\frac 12 \int_0^1 \frac{\ln^4(1+y)}{y}dy \end{align*} Doing the same thing for $$L$$, \begin{align*} L = & \frac 1 2\int_0^1 \frac{\ln^2(1+y)\ln^2\left(\frac{y}{1+y}\right)}{1+y}dy\\ =& \frac 12 \int_0^1 \frac{\ln^2 y\ln^2(1+y)}{1+y}dy- {\int_0^1 \frac{\ln y\ln^3(1+y)}{1+y} dy} +\frac 12 \int_0^1 \frac{\ln^4(1+y)}{1+y} dy\\ \underset{\text{IBP}}{=}& \frac 12 \int_0^1 \frac{\ln^2 y\ln^2(1+y)}{1+y}dy+\frac 1 4 \int_0^1 \frac{\ln^4(1+y)}y dy +\frac{\ln^5 2}{10}. \end{align*} This gives that \begin{align*} J=&K-L\\ =&-\frac 1 3 \int_0^1 \frac{\ln^3 y \ln(1+y)}{1+y}dy + \int_0^1 \frac{\ln^2 y\ln^2(1+y)}{1+y}dy +\frac 14 \int_0^1 \frac{\ln^4(1+y)}{y}dy -\frac{\ln^5 2}{10}\\ =&:-V_1+V_2+V_3 -\frac{\ln^5 2}{10}. \end{align*}

For $$V_1$$, we can use the Maclaurin series of $$\frac{\ln (1+y)}{1+y} = \sum_{k=0}^\infty (-1)^{k-1} H_k y^k$$ to get \begin{align*} V_1=&\frac 1 3\int_0^1 \frac{\ln^3 y\ln(1+y)}{1+y} dy \\ =& \frac 1 3\sum_{k=0}^\infty (-1)^{k-1}H_k {\int_0^1 y^k\ln^3 y\ dy} \\ =& \frac{-6}{3}\sum_{k=0}^\infty \frac{(-1)^{k-1}H_k}{(k+1)^4}\\ =&2\sum_{k=0}^\infty \frac{(-1)^k \left(H_{k+1}-\frac 1{k+1}\right)}{(k+1)^4} \\ =&2 \sum_{k=1}^\infty \frac{(-1)^{k-1}H_k}{k^4} -2 \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k^5}\tag{k+1\mapsto k}\\ =&2\left(\frac{59}{32}\zeta(5) - \frac{\pi^2\zeta(3)}{12}\right)-2\cdot \frac{15}{16}\zeta(5)\\ =&\frac{29}{16}\zeta(5) - \frac{\pi^2}{6}\zeta(3) \end{align*} where the known value of alternating Euler sum $$\sum_{k=1}^\infty \tfrac{(-1)^{k-1}H_k}{k^4}$$ is used.

For $$V_2$$, we consider the algebraic identity $$6a^2b^2 = (a-b)^4 - a^4 +4a^3b +4ab^3 -b^4$$ with $$a=\ln y$$ and $$b = \ln(1+y)$$ to get \begin{align*} V_2 =&\frac 1 6{\int_0^1 \frac{\ln^4\left(\frac y{1+y}\right)}{1+y} dy}-\frac 1 6\int_0^1 \frac{\ln^4 y}{1+y} dy +\underbrace{\frac 2 3\int_0^1 \frac{\ln^3 y\ln(1+y)}{1+y} dy}_{=2V_1}\\ &+\frac 2 3\underbrace{\int_0^1 \frac{\ln y\ln^3(1+y)}{1+y} dy}_{=- V_3\text{ by IBP}} -\frac 1 6\int_0^1 \frac{\ln^4(1+y)}{1+y} dy\\ =&\frac 1 6 \int_0^{\frac 1 2} \frac{\ln^4 x}{1-x} dx -\frac 1 6\int_0^1 \frac{\ln^4 y}{1+y} dy +2V_1-\frac 2 3 V_3 -\frac{\ln^5 2}{30}.\tag{\tfrac y{1+y}= x} \end{align*} For the first integral, we have \begin{align*} W:=&\frac 1 6{\int_0^{\frac 1 2 }\frac{\ln^4 x}{1-x} dx}\tag{2x\mapsto x} \\ =& \frac 1 6\int_0^1 \frac{\ln^4(\tfrac x 2)}{2-x}dx\\ =&\frac 1 6\sum_{k=1}^\infty \frac{1}{2^k}\int_0^1 x^{k-1}\Big[\ln^4 x -4\ln 2 \ln^3 x + 6\ln^2 2\ln^2 x - 4\ln^3 2 \ln x + \ln^4 2\Big]dx\\ =&\frac 1 6\sum_{k=1}^\infty \frac{1}{2^k}\left[\frac{24}{k^5} + \frac{24\ln 2}{k^4} +\frac {12\ln^2 2}{k^3} +\frac{4\ln^3 2}{k^2} +\frac{\ln^4 2}{k}\right]\\ =&4\text{Li}_5(1/2) +4\ln 2\text{Li}_4(1/2) + 2\ln^2 2\text{Li}_3(1/2) + \frac{2\ln^3 2}{3}\text{Li}_2(1/2) + \frac{\ln^5 2}6\\ =&4\text{Li}_5(1/2) +4\ln 2\text{Li}_4(1/2) + \frac{7\ln^2 2}{4}\zeta(3)-\frac{\pi^2\ln^3 2}{9}+ \frac{\ln^5 2}6. \end{align*}

For the second integral, we have \begin{align*} \frac 1 6\int_0^1 \frac{\ln^4 y}{1+y} dy =& \frac 1 6\sum_{k=1}^\infty (-1)^{k-1} \int_0^1 y^{k-1}\ln^4 y \ dy \\ =& \frac 1 6 \sum_{k=1}^\infty(-1)^{k-1} \frac{24}{k^5}\\ =&\frac{15}{4}\zeta(5). \end{align*}

This gives $$V_2 = W +2V_1-\frac 2 3 V_3-\frac{15}{4}\zeta(5)-\frac{\ln^5 2}{30}.$$

For $$V_3$$ we have \begin{align*} V_3=&\frac 14 \int_0^1 \frac{\ln^4(1+y)}{y}dy \tag{y\mapsto y+1}\\ =& \frac 1 {4}{ \int_1^2 \frac{\ln^4 y }{y-1} dy} \tag{\tfrac 1 y\mapsto y}\\ =&\frac 1 {4} \int_{\frac 1 2}^1 \frac{\ln^4 y}{y(1-y)} dy\\ =&\frac 1 {4} \int_{\frac 1 2}^1 \frac{\ln^4 y}{y} dy+\frac 1 {4} \int_{\frac 1 2}^1 \frac{\ln^4 y}{1-y} dy\\ =&\frac{\ln^5 2}{20} + \frac 1 {4} \int_0^1 \frac{\ln^4 y}{1-y} dy -\frac 1 {4} \underbrace{\int_0^{\frac 1 2}\frac{\ln^4 y}{1-y}dy}_{=6 W}\\ =&\frac{\ln^5 2}{20} +\frac 1 {4} \sum_{k=1}^\infty \int_0^1 y^{k-1}\ln^4 y\ dy- \frac 3 2 W\\ =&\frac{\ln^5 2}{20} +6\zeta(5)- \frac 3 2 W. \end{align*}

Combining $$V_1$$, $$V_2$$ and $$V_3$$, we get \begin{align*} J = & V_2 -V_1+V_3 -\frac{\ln^5 2}{10}\\ =& \left[W+2V_1 -\frac 2 3 V_3-\frac{15}{4}\zeta(5)-\frac{\ln^5 2}{30}\right]-V_1+V_3 -\frac{\ln^5 2}{10}\\ =& W+V_1+\frac 1 3 V_3-\frac{15}{4}\zeta(5)-\frac{2\ln^5 2}{15}\\ =&\frac 1 2 W+V_1 -\frac 7 4\zeta(5) -\frac{7\ln^5 2}{60}\\ =&2\text{Li}_5(1/2) +2\ln 2\ \text{Li}_4(1/2) +\frac 1 {16}\zeta(5) -\frac{\pi^2}6 \zeta(3) +\frac {7\ln^2 2}{8} \zeta(3) -\frac{\ln^2 2\pi^3}{18}-\frac{\ln^5 2}{30}. \end{align*} This gives \begin{align*} I_1'' =&\frac{\pi^4\ln 2}{288} -\frac{\ln^5 2}{8}+\frac 1 2\int_0^{\frac 1 2}\frac{\text{Li}_2^2(x)}{x} dx-J\\ =&\boxed{\small -3\text{Li}_5(1/2) -3\ln 2\text{Li}_4(1/2) +\frac{23}{64}\zeta(5) +\frac {23\pi^2}{96}\zeta(3) -\frac {21\ln^2 2}{16}\zeta(3) +\frac{7\pi^2\ln^3 2}{72} - \frac{3\ln^5 2}{20}.} \end{align*}

For $$I_1'''$$, we exploit the symmetric nature of the integrand to write \begin{align*} I_1''' :=& \frac 1 2\int_0^{\frac 1 2} \frac{\ln^2 x \ln^2 (1-x)}{x(1-x)} dx\\ =& \frac 1 4\int_0^1 \frac{\ln^2 x \ln^2 (1-x)}{x(1-x)} dx \\ =& \frac 1 4\underbrace{\int_0^1 \frac{\ln^2 x \ln^2 (1-x)}{x} dx}_{1-x\mapsto x}+\frac 1 4{\int_0^1 \frac{\ln^2 x \ln^2 (1-x)}{1-x} dx}\\ =&\frac 1 2\int_0^1 \frac{\ln^2 x \ln^2 (1-x)}{1-x} dx\\ =&\frac 1 2 \left[\frac{\partial^4}{\partial x^2 \partial y^2 } \text{B}(x,y)\right]_{x=1,y=0^+} \end{align*} where $$\text{B}(x,y)=\tfrac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$ is Euler's Beta function. Now we can use the fact that \begin{align*} \lim_{y\to 0^+}\frac{\partial^3 }{\partial x\partial y^2 }\text{B}(x,y) =&-\frac 1 3\psi'''(x)+\psi''(x)\Big[\psi(x) +\gamma\Big] + \psi'(x)\Big[\psi'(x)-\zeta(2) - \big[\psi(x) + \gamma\big]^2\Big] \end{align*} to obtain \begin{align*} I_1'''=& \frac 1 2\frac d{dx}\left[\frac{\partial^3 }{\partial x\partial y^2 }\text{B}(x,0^+)\right]_{x=1} \\ =& -\frac 1 6 \psi''''(1) +\psi'(1)\psi''(1) \\ =&\boxed{4\zeta(5) -\frac{\pi^2}3 \zeta(3)} \end{align*} where the values of $$\psi(1) +\gamma = 0$$, $$\psi'(1) =\zeta(2)$$, $$\psi''(1) =-2\zeta(3)$$ and $$\psi''''(1) = -24\zeta(5)$$ are used.

Finally, from $$I_1 = I_1'+I_1''+I_1'''$$ we get

\begin{align*} I_1 =& -2\text{Li}_5(1/2) - 2\ln 2\text{Li}_4(1/2) + \frac {279}{64}\zeta(5) -\frac {3\pi^2}{32}\zeta(3)-\frac {7\ln^2 2}{8} \zeta(3)+\frac {\pi^2\ln^3 2}{18}-\frac{\ln^5 2}{15}. \end{align*}

For $$I_2$$, we observe that \begin{align*}\require{cancel} I_2 =& \ln 2 \int_0^{\frac 12} \frac{\big[\text{Li}_2(x) + \ln^2(1-x)\big]\ln x}{x(1-x)} dx \\ =& \ln 2 {\int_0^{\frac 12} \frac{\text{Li}_2(x) \ln x}{1-x} dx}+\ln 2\int_0^{\frac 12} \frac{\ln^2(1-x)\ln x}{1-x} dx+ \ln 2 \int_0^{\frac 12} \frac{\big[\text{Li}_2(x) + \ln^2(1-x)\big]\ln x}{x} dx\\ \underset{\text{IBP}}{=}&-\ln^3 2\ \text{Li}_2(1/2)+\ln 2 \int_0^{\frac 12} \ln(1-x)\frac{-\cancel{\ln(1-x)\ln x}+\text{Li}_2(x)}{x} dx \\ &+\ln2 \int_0^{\frac 1 2} \frac{\ln^2(1-x)\ln x}{1-x} dx + \ln 2{ \int_0^{\frac 12} \frac{\big[\text{Li}_2(x)+\cancel{ \ln^2(1-x)}\big] \ln x}{x} dx}\\ \underset{\text{IBP}}{=}&\small-\ln^3 2 \text{Li}_2(1/2)-\tfrac{\ln 2}{2} \left[\text{Li}^2_2(x)\right]^{1/2}_0-\frac{\ln^5 2}3+{\frac{\ln 2}3{\int_0^{\frac 1 2} \frac{\ln^3(1-x)}{x} dx}} -\ln^2 2\ \text{Li}_3(1/2)-\ln 2\int_0^{\frac 1 2} \frac{\text{Li}_3(x)}{x} dx\normalsize\\ =&-\ln^3 2\ \text{Li}_2(1/2) -\tfrac{\ln 2}{2} \text{Li}^2_2(1/2)-\frac{\ln^5 2}3+ \small\underbrace{\frac{\ln 2}3{\int_{\frac 1 2}^1 \frac{\ln^3 x}{1-x} dx}}_{1-x\mapsto x, \ =:I_2'}\normalsize-\ln^2 2\ \text{Li}_3(1/2)-\ln 2\ \text{Li}_4(1/2)\\ =&-\ln 2\ \text{Li}_4(1/2) -\frac{7\ln^2 2}8 \zeta(3) -\frac{\pi^4 \ln 2}{288}+\frac {\pi^2\ln^3 2}{24} -\frac{\ln^5 2}{8} + I_2'. \end{align*}

For $$I_2'$$, by integrating by parts, we have \begin{align*} I_2' =& \frac{\ln 2}3\int_{\frac 1 2}^1 \frac{\ln^3 x}{1-x} dx\\ =&\frac{\ln 2}3\int_{0}^1 \frac{\ln^3 x}{1-x} dx -\frac{\ln 2}3{\int_{0}^{\frac 12} \frac{\ln^3 x}{1-x} dx}\tag{x=\tfrac y 2}\\ =&\frac{\ln 2}3\sum_{k=1}^\infty {\int_0^1 x^{k-1}\ln^3 x\ dx}-\underbrace{\frac{\ln 2}3\int_{0}^{1} \frac{\ln^3 (\tfrac y 2)}{2-y} dy}_{=:A}\\ =&-\frac{\pi^4\ln 2}{45}-A. \end{align*} \begin{align*} A=&\frac{\ln 2}3\sum_{k=1}^\infty \frac 1 {2^k}\int_{0}^{1} y^{k-1}\ln^3 (\tfrac y 2) dy\\ =&\frac{\ln 2}3\sum_{k=1}^\infty \frac 1 {2^k}\int_{0}^{1} y^{k-1}\left[\ln^3 y - 3\ln 2\ln^2 y +3\ln^2 2\ln y -\ln^3 2\right]dy\\ =&-\frac{\ln 2}3\sum_{k=1}^\infty \frac 1 {2^k}\left[\frac 6 {k^4}+\frac{6\ln 2}{k^3} +\frac{3\ln^2 2}{k^2} +\frac{\ln^3 2}{k}\right]\\ =& -2\ln 2\ \text{Li}_4(1/2) - 2\ln^2 2\ \text{Li}_3(1/2)-\ln^3 2\ \text{Li}_2(1/2)-\frac{\ln^5 2}3\\ =&-2\ln 2\ \text{Li}_4(1/2)-\frac{7\ln^2 2}{4}\zeta(3)+\frac{\pi^2\ln^3 2}{12}-\frac{\ln^5 2}{6}. \end{align*}

This gives $$I_2'= 2\ln 2\ \text{Li}_4(1/2)+\frac{7\ln^2 2}{4}\zeta(3)-\frac{\pi^4\ln 2}{45}-\frac{\pi^2\ln^3 2}{12}-\frac{\ln^5 2}{6}$$ and

\begin{align*} I_2=\ln 2\ \text{Li}_4(1/2)+\frac{7\ln^2 2}{8}\zeta(3)-\frac{37\pi^4\ln 2}{1440}-\frac{\pi^2\ln^3 2}{24}+\frac{\ln^5 2}{24}. \end{align*}

For $$I_3$$, we have \begin{align*} I_3=&\frac{\ln^2 2}{2}\int_0^{\frac 1 2} \frac{\text{Li}_2(x)+\ln^2(1-x)}{x(1-x)}dx\\ =&\frac{\ln^2 2}{2}\int_0^{\frac 1 2} \frac{\text{Li}_2(x)}{x}dx+\frac{\ln^2 2}{2}\int_0^{\frac 1 2} {\left[\frac{\ln^2(1-x)}{x}+\frac{\text{Li}_2(x)}{1-x}\right]}dx+\frac{\ln^2 2}{2}\int_0^{\frac 1 2} \frac{\ln^2(1-x)}{1-x}dx\\ =&\frac{\ln^2 2}{2}\text{Li}_3(1/2) +\frac{\ln^2 2}{2}\big[-\ln(1-x)\text{Li}_2(x)\big]^{1/2}_0+\frac{\ln^5 2}6. \end{align*} Using the well-known values of $$\text{Li}_3(1/2)$$ and $$\text{Li}_2(1/2)$$, this simplifies to

$$I_3 =\frac {7\ln^2 2}{16}\zeta(3).$$

From $$S = I_1+I_2 + I_3$$, we finally get

\begin{align*} \sum_{k=1}^\infty \frac{H_k^2}{k^32^k} =& -2\text{Li}_5(1/2) -\ln 2\ \text{Li}_4(1/2) + \frac{279}{64}\zeta(5) - \frac{3\pi^2}{32}\zeta(3) + \frac{7\ln^2 2}{16}\zeta(3) - \frac{37\pi^4 \ln 2}{1440}\\ & + \frac{\pi^2 \ln^3 2}{72} - \frac{\ln^5 2}{40}. \end{align*}

We can observe that the values of $$I_2$$ and $$I_3$$ can be used to evaluate sums of lower order in a similar way:

$$\begin{eqnarray*} &\sum_{k=1}^\infty \frac{H_k^2}{k^2 2^k} = -\frac 1{\ln 2} I_2 - \frac{2}{\ln 2} I_3=-\text{Li}_4(1/2) -\frac{7\ln 2}4\zeta(3) +\frac{37 \pi^4}{1440} +\frac{\pi^2\ln^2 2}{24} -\frac{\ln^4 2}{24},\\ &\sum_{k=1}^\infty \frac{H_k^2}{k 2^k} = \frac{2}{\ln^2 2} I_2 = \frac{7}{8}\zeta(3). \end{eqnarray*}$$

• nice work (+1). my approach is somewhat shorter. – Ali Shather Aug 29 at 0:54
• @AliShather Thank you :) It took too long to type all these equations... – Song Aug 29 at 0:59
• its good to see different approaches though and that is the point of posting most of my problems. – Ali Shather Aug 29 at 1:07
• the nice thing about your solution is that you didnt use results of other integrals/ sums unlike me where i used couple results. – Ali Shather Aug 29 at 1:10
• @AliShather I wanted to see if this Euler sum involving a factor $2^{-k}$ can be transformed into the alternating one $\sum_{k=1}^\infty \frac{(-1)^{k-1}H_k}{k^4}$ ($+$ some manageable integrals.) – Song Aug 29 at 1:21

We proved here

$$\frac{\ln^2(1-x)}{1-x}=\sum_{n=1}^\infty x^n\left(H_n^2-H_n^{(2)}\right)\tag{1}$$

multiply both sides by $$\frac{\ln^2x}{x}$$ then integrate from $$x=0$$ to $$1/2$$ we have

\begin{align} I&=\int_0^{1/2}\frac{\ln^2(1-x)\ln^2x}{x(1-x)}\ dx=\sum_{n=1}^\infty\left(H_n^2-H_n^{(2)}\right)\int_0^{1/2}x^{n-1}\ln^2x\ dx\\ &=\sum_{n=1}^\infty\left(H_n^2-H_n^{(2)}\right)\left(\frac{\ln^22}{n2^n}+\frac{2\ln2}{n^22^n}+\frac{2}{n^32^n}\right)\\ &=\ln^22\sum_{n=1}^\infty\frac{H_n^2-H_n^{(2)}}{n2^n}+2\ln2\sum_{n=1}^\infty\frac{H_n^2-H_n^{(2)}}{n^22^n}+2\sum_{n=1}^\infty\frac{H_n^2}{n^32^n}-2\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^32^n}\\ &=\ln^22S_1+2\ln2S_2+2\sum_{n=1}^\infty\frac{H_n^2}{n^32^n}-2S_3 \end{align}

Rearranging the terms we have

$$\sum_{n=1}^\infty\frac{H_n^2}{n^32^n}=\frac12I-\frac12\ln^22S_1-\ln2S_2+S_3\tag{2}$$

Evaluation of $$I$$:

\begin{align} I&=\int_0^{1/2}\frac{\ln^2(1-x)\ln^2x}{x(1-x)}\ dx\overset{1-x\mapsto x}{=}\int_{1/2}^1\frac{\ln^2(1-x)\ln^2x}{x(1-x)}\ dx\\ 2I&=\int_0^{1}\frac{\ln^2(1-x)\ln^2x}{x(1-x)}\ dx=\int_0^{1}\frac{\ln^2(1-x)\ln^2x}{x}\ dx+\underbrace{\int_0^{1}\frac{\ln^2(1-x)\ln^2x}{1-x}\ dx}_{1-x\mapsto x}\\ I&=\int_0^{1}\frac{\ln^2(1-x)\ln^2x}{x}\ dx=2\sum_{n=1}^\infty\frac{H_n}{n+1}\int_0^1x^n\ln^2x\ dx\\ &=4\sum_{n=1}^\infty\frac{H_n}{(n+1)^4}=4\sum_{n=1}^\infty\frac{H_n}{n^4}-4\zeta(5)=\boxed{8\zeta(5)-4\zeta(2)\zeta(3)} \end{align}

where we used $$\sum_{n=1}^\infty\frac{H_n}{n^4}=3\zeta(5)-\zeta(2)\zeta(3)$$

Evaluation of $$S_1$$:

Divide both sides of (1) by $$x$$ then integrate from $$x=0$$ to $$1/2$$ and use the fact that $$\int_0^{1/2}x^{n-1}=\frac1{n2^n}$$

\begin{align} S_1&=\sum_{n=1}^\infty \frac{H_n^2-H_n^{(2)}}{n2^n}=\int_0^{1/2}\frac{\ln^2(1-x)}{x(1-x)}\ dx\\ &=\int_{1/2}^{1}\frac{\ln^2x}{x(1-x)}\ dx=\sum_{n=0}^\infty\int_{1/2}^1x^{n-1}\ln^2x\ dx\\ &=\frac13\ln^32+\sum_{n=1}^\infty\int_{1/2}^1x^{n-1}\ln^2x\ dx\\ &=\frac13\ln^32+\sum_{n=1}^\infty\left(\frac2{n^3}-\frac{\ln^22}{n2^n}-\frac{2\ln2}{n^22^n}-\frac{2}{n^32^n}\right)\\ &=\frac13\ln^32+2\zeta(3)-\ln^32-2\ln2\operatorname{Li}_2\left(\frac12\right)-2\operatorname{Li}_3\left(\frac12\right)=\boxed{\frac14\zeta(3)} \end{align}

where we used $$\operatorname{Li}_2\left(\frac12\right)=\frac12\zeta(2)-\frac12\ln^22$$ and $$\operatorname{Li}_3\left(\frac12\right)=\frac78\zeta(3)-\frac12\ln2\zeta(2)+\frac16\ln^32$$

Evaluation of $$S_2$$:

integrate both sides of (1) from $$x=0$$ to $$x$$ to have

$$-\frac13\ln^3(1-x)=\sum_{n=1}^\infty\frac{x^{n+1}}{n+1}\left(H_n^2-H_n^{(2)}\right)=\sum_{n=1}^\infty\frac{x^{n}}{n}\left(H_n^2-H_n^{(2)}-\frac{2H_n}{n}+\frac{2}{n^2}\right)\tag{3}$$

Now divide both sides of (3) by $$x$$ then integrate from $$x=0$$ to $$1/2$$ and use the fact that $$\int_0^{1/2}x^{n-1}=\frac1{n2^n}$$

$$-\frac13\int_0^{1/2}\frac{\ln^3(1-x)}{x}\ dx=\sum_{n=1}^\infty\frac{1}{n^22^n}\left(H_n^2-H_n^{(2)}-\frac{2H_n}{n}+\frac{2}{n^2}\right)$$

Rearranging the terms

$$S_2=\sum_{n=1}^\infty\frac{H_n^2-H_n^{(2)}}{n^22^n}=\boxed{2\sum_{n=1}^\infty\frac{H_n}{n^32^n}-\frac13\int_0^{1/2}\frac{\ln^3(1-x)}{x}\ dx-2\operatorname{Li}_4\left(\frac12\right)}$$

Evaluation of $$S_3$$:

By Cauchy product we have

$$\operatorname{Li}_2^2(x)=\sum_{n=1}^\infty x^n\left(\frac{4H_n}{n^3}+\frac{2H_n^{(2)}}{n^2}-\frac{6}{n^4}\right)$$

divide both sides by $$x$$ then integrate from $$x=0$$ to $$1/2$$ and use the fact that $$\int_0^{1/2}x^{n-1}=\frac1{n2^n}$$ we have

$$\int_0^{1/2}\frac{\operatorname{Li}_2^2(x)}{x}\ dx=\sum_{n=1}^\infty \frac{1}{n2^n}\left(\frac{4H_n}{n^3}+\frac{2H_n^{(2)}}{n^2}-\frac{6}{n^4}\right)$$

rearrange to get

$$S_3=\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^32^n}=\boxed{3\operatorname{Li}_5\left(\frac12\right)-2\sum_{n=1}^\infty\frac{H_n}{n^42^n}+\frac12\int_0^{1/2}\frac{\operatorname{Li}_2^2(x)}{x}\ dx}$$

Substituting the results of $$I$$, $$S_1$$, $$S_2$$ and $$S_3$$ in (2) we have

$$\sum_{n=1}^\infty\frac{H_n^2}{n^32^n}=3\operatorname{Li}_5\left(\frac12\right)+2\ln2\operatorname{Li}_4\left(\frac12\right)+4\zeta(5)-2\zeta(2)\zeta(3)-\frac18\ln^22\zeta(3)-2\left(\color{blue}{\ln2\sum_{n=1}^\infty\frac{H_n}{n^32^n}+\sum_{n=1}^\infty\frac{H_n}{n^42^n}}\right)+\frac13\ln2\int_0^{1/2}\frac{\ln^3(1-x)}{x}\ dx+\frac12\int_0^{1/2}\frac{\operatorname{Li}_2^2(x)}{x}\ dx$$

I managed here to prove

$$\color{blue}{\ln2\sum_{n=1}^{\infty}\frac{H_n}{2^n n^3}+\sum_{n=1}^{\infty}\frac{H_n}{2^nn^4} }=-\frac12\ln^22\sum_{n=1}^{\infty}\frac{H_n}{2^n n^2}-\frac16\ln^32\sum_{n=1}^{\infty}\frac{H_n}{2^n n}+\frac12\sum_{n=1}^{\infty}\frac{H_n}{n^4}-\frac{47}{32}\zeta(5)+\frac{1}{15}\ln^52+\frac{1}{3}\ln^32\operatorname{Li_2}\left( \frac12\right)+\ln^22\operatorname{Li_3}\left( \frac12\right)+2\ln2\operatorname{Li_4}\left( \frac12\right) +2\operatorname{Li_5}\left( \frac12\right)$$

plugging the trivial sums $$\sum_{n=1}^{\infty}\frac{H_n}{ n^22^n}=\zeta(3)-\frac{1}{2}\ln(2)\zeta(2)$$ and $$\sum_{n=1}^\infty\frac{H_n}{n2^n}=\frac12\zeta(2)$$ we get

$$\color{blue}{\ln2\sum_{n=1}^{\infty}\frac{H_n}{2^n n^3}+\sum_{n=1}^{\infty}\frac{H_n}{2^nn^4} }=2\operatorname{Li}_5\left( \frac12\right)+2\ln2\operatorname{Li}_4\left( \frac12\right)+\frac1{32}\zeta(5)-\frac12\zeta(2)\zeta(3)+\frac38\ln^22\zeta(3)\\-\frac16\ln^32\zeta(2)+\frac1{15}\ln^52$$

Also @Song nicely proved here

$$\int_0^{1/2}\frac{\operatorname{Li}_2^2(x)}{x}\ dx=\frac12\ln^32\zeta(2)-\frac78\ln^22\zeta(3)-\frac58\ln2\zeta(4)+\frac{27}{32}\zeta(5)+\frac78\zeta(2)\zeta(3)\\-\frac{7}{60}\ln^52-2\ln2\operatorname{Li}_4\left(\frac12\right)-2\operatorname{Li}_5\left(\frac12\right)$$

for the integral: \begin{align} \int_0^{1/2}\frac{\ln^3(1-x)}{x}\ dx&=\int_{1/2}^{1}\frac{\ln^3x}{1-x}\ dx\\ &=\sum_{n=1}^\infty\int_{1/2}^1 x^{n-1}\ln^3x\ dx\\ &=\sum_{n=1}^\infty\left(\frac{\ln^32}{n2^n}+\frac{3\ln^22}{n^22^n}+\frac{6\ln2}{n^32^n}+\frac{6}{n^42^n}-\frac{6}{n^4}\right)\\ &=\ln^42+3\ln^32\operatorname{Li}_2\left(\frac12\right)+6\ln2\operatorname{Li}_3\left(\frac12\right)+6\operatorname{Li}_4\left(\frac12\right)-6\zeta(4)\\ &=6\operatorname{Li}_4\left(\frac12\right)-6\zeta(4)+\frac{21}4\ln2\zeta(3)-\frac32\ln^22\zeta(2)+\frac12\ln^42 \end{align}

Combining these results we get

$$\sum_{n=1}^\infty\frac{H_n^2}{n^32^n}=-2\operatorname{Li}_5\left(\frac12\right)-\ln2\operatorname{Li}_4\left(\frac12\right)+\frac{279}{64}\zeta(5)-\frac{37}{16}\ln2\zeta(4)-\frac{9}{16}\zeta(2)\zeta(3)\\+\frac{7}{16}\ln^22\zeta(3)+\frac1{12}\ln^32\zeta(2)-\frac{1}{40}\ln^52$$

BONUS:

In our solution we got

$$\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^32^n}=3\operatorname{Li}_5\left(\frac12\right)-2\sum_{n=1}^\infty\frac{H_n}{n^42^n}+\frac12\int_0^{1/2}\frac{\operatorname{Li}_2^2(x)}{x}\ dx$$

Substitute

\begin{align} \displaystyle\sum_{n=1}^{\infty}\frac{H_n}{n^42^n}&=2\operatorname{Li_5}\left( \frac12\right)+\ln2\operatorname{Li_4}\left( \frac12\right)-\frac16\ln^32\zeta(2) +\frac12\ln^22\zeta(3)\\ &\quad-\frac18\ln2\zeta(4)- \frac12\zeta(2)\zeta(3)+\frac1{32}\zeta(5)+\frac1{40}\ln^52 \end{align}

along with @Song's result we get

$$\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^32^n}=-2\operatorname{Li}_5\left(\frac12\right)-3\ln2\operatorname{Li}_4\left(\frac12\right)+\frac{23}{64}\zeta(5)-\frac1{16}\ln2\zeta(4)+\frac{23}{16}\zeta(2)\zeta(3)\\-\frac{23}{16}\ln^22\zeta(3)+\frac7{12}\ln^32\zeta(2)-\frac{13}{120}\ln^52$$

Different Solution with a big bonus:

First lets define the following:

$$M=\sum_{n=1}^\infty\frac{H_n^2}{n^32^n}$$

$$N=\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^32^n}$$

I am going to establish two relations of $$M$$ and $$N$$ and solve them as a system of equations.

The first relation:

In the question body, we reached

$$M+N=\int_0^1\frac{\ln^2(1-x)\operatorname{Li}_2(x/2)}{x}dx$$

and by using $$\quad\displaystyle\frac{\operatorname{Li}_2(x)}{x}=-\int_0^1\frac{\ln y}{1-xy}dy\quad$$ we can write

$$M+N=-\frac12\int_0^1\ln y\left[\int_0^1\frac{\ln^2(1-x)}{1-(y/2)x}dx\right]dy$$

$$=2\int_0^1\frac{\ln y}{y}\operatorname{Li}_3\left(\frac{y}{y-2}\right)dy\overset{IBP}{=}-\int_0^1\frac{\operatorname{Li}_2(-y)}{y}\ln^2\left(\frac{2y}{1+y}\right)dy$$

$$\small{=-\int_0^1\frac{\operatorname{Li}_2(-y)}{y}\left[\ln^22+2\ln2\ln y+\ln^2y-2\ln2\ln(1+y)-2\ln y\ln(1+y)+\ln^2(1+y)\right]dy}$$

$$=-\ln^22\underbrace{\int_0^1\frac{\operatorname{Li}_2(-y)}{y}dy}_{I_1}-2\ln2\underbrace{\int_0^1\frac{\operatorname{Li}_2(-y)\ln y}{y}dy}_{I_2}-\underbrace{\int_0^1\frac{\operatorname{Li}_2(-y)\ln^2y}{y}dy}_{I_3}\\+2\ln2\underbrace{\int_0^1\frac{\operatorname{Li}_2(-y)\ln(1+y)}{y}dy}_{I_4}+2\underbrace{\int_0^1\frac{\operatorname{Li}_2(-y)\ln y\ln(1+y)}{y}dy}_{I_5}-\underbrace{\int_0^1\frac{\operatorname{Li}_2(-y)\ln^2(1+y)}{y}dy}_{I_6}$$

$$I_1=\operatorname{Li}_3(-1)=-\frac34\zeta(3)$$

$$I_2\overset{IBP}{=}-\int_0^1\frac{\operatorname{Li}_3(-y)}{y}dy=-\operatorname{Li}_4(-1)=\frac78\zeta(4)$$

$$I_3\overset{IBP}{=}-2\int_0^1\frac{\operatorname{Li}_3(-y)\ln y}{y}dy\overset{IBP}{=}2\int_0^1\frac{\operatorname{Li}_4(-y)}{y}dy=2\operatorname{Li}_5(-1)=-\frac{15}8\zeta(5)$$

$$I_4=-\frac12\operatorname{Li}_2^2(-1)=-\frac5{16}\zeta(4)$$

$$I_5\overset{IBP}{=}\frac12\int_0^1\frac{\operatorname{Li}_2^2(-y)}{y}dy=\frac38\zeta(2)\zeta(3)-\frac{17}{32}\zeta(5)$$

where the last result follows from this solution, check Eq$$(3)$$

$$I_6$$ is elegantly evaluated by Cornel here in page $$5$$ and by me here,

$$\small{I_6=4\operatorname{Li}_5\left(\frac12\right)+4\ln2\operatorname{Li}_4\left(\frac12\right)-\frac{125}{32}\zeta(5)-\frac{1}{8}\zeta(2)\zeta(3)+\frac{7}{4}\ln^22\zeta(3)-\frac2{3}\ln^32\zeta(2)+\frac{2}{15}\ln^52}$$

Combine these results to obtain

$$M+N=-4\operatorname{Li}_5\left(\frac12\right)-4\ln2\operatorname{Li}_4\left(\frac12\right)+\frac{131}{32}\zeta(5)$$ $$-\frac{19}{8}\ln2\zeta(4)+\frac{7}{8}\zeta(2)\zeta(3)-\ln^22\zeta(3)+\frac2{3}\ln^32\zeta(2)-\frac{2}{15}\ln^52\tag1$$

The second relation:

From here

$$\frac{\ln^2(1-x)}{1-x}=\sum_{n=1}^\infty x^n\left(H_n^2-H_n^{(2)}\right)\tag2$$

multiply both sides by $$\frac{\ln^2x}{x}$$ then integrate from $$x=0$$ to $$1/2$$ we have

\begin{align} I&=\int_0^{1/2}\frac{\ln^2(1-x)\ln^2x}{x(1-x)}\ dx=\sum_{n=1}^\infty\left(H_n^2-H_n^{(2)}\right)\int_0^{1/2}x^{n-1}\ln^2x\ dx\\ &=\sum_{n=1}^\infty\left(H_n^2-H_n^{(2)}\right)\left(\frac{\ln^22}{n2^n}+\frac{2\ln2}{n^22^n}+\frac{2}{n^32^n}\right)\\ &=\ln^22\sum_{n=1}^\infty\frac{H_n^2-H_n^{(2)}}{n2^n}+2\ln2\sum_{n=1}^\infty\frac{H_n^2-H_n^{(2)}}{n^22^n}+2\sum_{n=1}^\infty\frac{H_n^2}{n^32^n}-2\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^32^n}\\ &=\ln^22S_1+2\ln2S_2+2M-2N \end{align}

Or

$$M-N=\frac12I-\frac12\ln^22S_1-\ln2S_2$$

Evaluation of $$I:$$

\begin{align} I&=\int_0^{1/2}\frac{\ln^2(1-x)\ln^2x}{x(1-x)}\ dx\overset{1-x\mapsto x}{=}\int_{1/2}^1\frac{\ln^2(1-x)\ln^2x}{x(1-x)}\ dx\\ 2I&=\int_0^{1}\frac{\ln^2(1-x)\ln^2x}{x(1-x)}\ dx=\int_0^{1}\frac{\ln^2(1-x)\ln^2x}{x}\ dx+\underbrace{\int_0^{1}\frac{\ln^2(1-x)\ln^2x}{1-x}\ dx}_{1-x\mapsto x}\\ I&=\int_0^{1}\frac{\ln^2(1-x)\ln^2x}{x}\ dx=2\sum_{n=1}^\infty\frac{H_n}{n+1}\int_0^1x^n\ln^2x\ dx\\ &=4\sum_{n=1}^\infty\frac{H_n}{(n+1)^4}=4\sum_{n=1}^\infty\frac{H_n}{n^4}-4\zeta(5)=8\zeta(5)-4\zeta(2)\zeta(3) \end{align}

Evaluation of $$S_1$$:

Divide both sides of (2) by $$x$$ then integrate from $$x=0$$ to $$1/2$$ and use the fact that $$\int_0^{1/2}x^{n-1}=\frac1{n2^n}$$

\begin{align} S_1&=\sum_{n=1}^\infty \frac{H_n^2-H_n^{(2)}}{n2^n}=\int_0^{1/2}\frac{\ln^2(1-x)}{x(1-x)}\ dx\\ &=\int_{1/2}^{1}\frac{\ln^2x}{x(1-x)}\ dx=\sum_{n=0}^\infty\int_{1/2}^1x^{n-1}\ln^2x\ dx\\ &=\frac13\ln^32+\sum_{n=1}^\infty\int_{1/2}^1x^{n-1}\ln^2x\ dx\\ &=\frac13\ln^32+\sum_{n=1}^\infty\left(\frac2{n^3}-\frac{\ln^22}{n2^n}-\frac{2\ln2}{n^22^n}-\frac{2}{n^32^n}\right)\\ &=\frac13\ln^32+2\zeta(3)-\ln^32-2\ln2\operatorname{Li}_2\left(\frac12\right)-2\operatorname{Li}_3\left(\frac12\right)=\frac14\zeta(3) \end{align}

where we used $$\operatorname{Li}_2\left(\frac12\right)=\frac12\zeta(2)-\frac12\ln^22$$ and $$\operatorname{Li}_3\left(\frac12\right)=\frac78\zeta(3)-\frac12\ln2\zeta(2)+\frac16\ln^32$$

Evaluation of $$S_2$$:

integrate both sides of (2) from $$x=0$$ to $$x$$ to have

$$-\frac13\ln^3(1-x)=\sum_{n=1}^\infty\frac{x^{n+1}}{n+1}\left(H_n^2-H_n^{(2)}\right)=\sum_{n=1}^\infty\frac{x^{n}}{n}\left(H_n^2-H_n^{(2)}-\frac{2H_n}{n}+\frac{2}{n^2}\right)\tag{3}$$

Now divide both sides of (3) by $$x$$ then integrate from $$x=0$$ to $$1/2$$ and use the fact that $$\int_0^{1/2}x^{n-1}=\frac1{n2^n}$$

$$-\frac13\int_0^{1/2}\frac{\ln^3(1-x)}{x}\ dx=\sum_{n=1}^\infty\frac{1}{n^22^n}\left(H_n^2-H_n^{(2)}-\frac{2H_n}{n}+\frac{2}{n^2}\right)$$

Rearranging the terms

$$S_2=\sum_{n=1}^\infty\frac{H_n^2-H_n^{(2)}}{n^22^n}=2\sum_{n=1}^\infty\frac{H_n}{n^32^n}-\frac13\int_0^{1/2}\frac{\ln^3(1-x)}{x}\ dx-2\operatorname{Li}_4\left(\frac12\right)$$

$$\sum_{n=1}^\infty \frac{H_n}{2^nn^3}=\operatorname{Li}_4\left(\frac12\right)+\frac18\zeta(4)-\frac18\ln2\zeta(3)+\frac1{24}\ln^42$$

and

\begin{align} \int_0^{1/2}\frac{\ln^3(1-x)}{x}\ dx&=\int_{1/2}^{1}\frac{\ln^3x}{1-x}\ dx\\ &=\sum_{n=1}^\infty\int_{1/2}^1 x^{n-1}\ln^3x\ dx\\ &=\sum_{n=1}^\infty\left(\frac{\ln^32}{n2^n}+\frac{3\ln^22}{n^22^n}+\frac{6\ln2}{n^32^n}+\frac{6}{n^42^n}-\frac{6}{n^4}\right)\\ &=\ln^42+3\ln^32\operatorname{Li}_2\left(\frac12\right)+6\ln2\operatorname{Li}_3\left(\frac12\right)+6\operatorname{Li}_4\left(\frac12\right)-6\zeta(4)\\ &=6\operatorname{Li}_4\left(\frac12\right)-6\zeta(4)+\frac{21}4\ln2\zeta(3)-\frac32\ln^22\zeta(2)+\frac12\ln^42 \end{align}

we get

$$S_2=-2\operatorname{Li}_4\left(\frac12\right)+\frac94\zeta(4)-2\ln2\zeta(3)+\frac12\ln^22\zeta(2)-\frac1{12}\ln^42$$

collect the results of $$I$$, $$S_1$$ and $$S_2$$ we get

$$M-N=2\ln2\operatorname{Li}_4\left(\frac12\right)+4\zeta(5)$$ $$-\frac{9}{4}\ln2\zeta(4)-2\zeta(2)\zeta(3)+\frac{15}8\ln^22\zeta(3)-\frac1{2}\ln^32\zeta(2)+\frac{1}{12}\ln^52\tag4$$

Now we are ready to calculate the two sums:

$$M=\frac{(1)+(4)}{2}=-2\operatorname{Li}_5\left(\frac12\right)-\ln2\operatorname{Li}_4\left(\frac12\right)+\frac{279}{64}\zeta(5)-\frac{37}{16}\ln2\zeta(4)-\frac{9}{16}\zeta(2)\zeta(3)\\+\frac{7}{16}\ln^22\zeta(3)+\frac1{12}\ln^32\zeta(2)-\frac{1}{40}\ln^52$$

$$N=\frac{(1)-(4)}{2}=-2\operatorname{Li}_5\left(\frac12\right)-3\ln2\operatorname{Li}_4\left(\frac12\right)+\frac{23}{64}\zeta(5)-\frac1{16}\ln2\zeta(4)+\frac{23}{16}\zeta(2)\zeta(3)\\-\frac{23}{16}\ln^22\zeta(3)+\frac7{12}\ln^32\zeta(2)-\frac{13}{120}\ln^52$$