definite integralsharmonic-numbersintegrationpolylogarithmsequences-and-series
Is there a closed form for the integral
$$ \int_{0}^{1}\frac{\mathrm{Li}_2(-x^2)}{x^2+1} \mathrm{d}x $$
where $ \displaystyle \mathrm{Li}_2(x) \;\;=\;\; \sum_{n=1}^\infty \frac{x^n}{n^2} $
a related sum is $\displaystyle \sum_{n=0}^{\infty} \frac{(-1)^n H_n^{(2)}}{2n+1} $
By coincidence, I asked virtually the same question. I managed to answer it myself which is,
$$F_a(z) = \sum_{n=1}^{\infty}\frac{H_n}{n^a}z^n= S_{a-1,2}(z) + \rm{Li}_{a+1}(z)$$
with polylogarithm $\rm{Li}_{a+1}(z)$ and Nielsen generalized polylogarithm $S_{n,p}(z)$. The situation then is subsumed by the Nielsen polylogs.
There are formulas for general $z$ when $a=1,2$ (as you mentioned) as well as for $a=3$, though I am not sure if there is for higher.
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*}
Best Answer
\begin{align} I&=\int_0^1\frac{\operatorname{Li}_2(-x^2)}{1+x^2}\ dx\\ &=-\frac{\pi^3}{48}+2\int_0^1\frac{\arctan x\ln(1+x^2)}{x}\ dx\tag1\\ &=-\frac{\pi^3}{48}-4\sum_{n=0}^\infty\frac{(-1)^nH_{2n}}{2n+1}\int_0^1 x^{2n}\ dx\tag2\\ &=-\frac{\pi^3}{48}-4\sum_{n=0}^\infty\frac{(-1)^nH_{2n}}{(2n+1)^2}\tag3\\ &=-\frac{\pi^3}{48}-4\sum_{n=0}^\infty\frac{(-1)^nH_{2n+1}}{(2n+1)^2}+4\underbrace{\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^3}}_{\beta(3)=\frac{\pi^3}{32}}\tag4\\ &=\frac{5\pi^3}{48}-4\Im\sum_{n=1}^\infty\frac{(i)^nH_n}{n^2}\tag5\\ &=\frac{5\pi^3}{48}-4\left(-\frac{\pi}{16}\ln^22-\frac12\ln2\ G-\Im\operatorname{Li}_3(1-i)\right)\tag6\\ &=\frac{5\pi^3}{48}+\frac{\pi}{4}\ln^22+2\ln2\ G+4\Im\operatorname{Li}_3(1-i) \end{align}
The related sum can be obtained from the the generating function
$$\sum_{n=1}^\infty H_n^{(q)}x^n=\frac{\operatorname{Li}_q(x)}{1-x}$$
set $q=2$ and replace $x$ with $-x^2$ we get
$$\sum_{n=1}^\infty (-1)^nH_n^{(2)}x^{2n}=\frac{\operatorname{Li}_2(-x^2)}{1+x^2}$$
now integrate both sides from $x=0$ to $1$
$$\sum_{n=1}^\infty \frac{(-1)^nH_n^{(2)}}{2n+1}=\int_0^1\frac{\operatorname{Li}_2(-x^2)}{1+x^2}\ dx$$
The index $n$ can start from $0$ as $H_{0}^{(2)}=0$
.
Explanation:
$(1)$ Integration by parts.
$(2)$ Using the identity $\arctan x\ln(1+x^2)=-2\sum_{n=0}^\infty\frac{(-1)^nH_{2n}}{2n+1}x^{2n+1}$.
$(3)$ $\int_0^1 x^{a-1}\ dx=\frac1a$
$(4)$ Using $H_n=H_{n+1}-\frac1{n+1}$
$(5)$ Using $\sum_{n=0}^\infty (-1)^n a_{2n+1}=\Im\sum_{n=1}^\infty i ^n a_n$
$(6)$ Using the generating function $\sum_{n=1}^\infty\frac{H_{n}}{n^2}x^{n}=\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x)+\ln(1-x)\operatorname{Li}_2(1-x)+\frac12\ln x\ln^2(1-x)+\zeta(3)$