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How to prove

$$\sum_{n=1}^\infty\frac{H_{2n}H_n^{(2)}}{(2n+1)^2}= \\ \small{\frac43\ln^32\zeta(2)-\frac72\ln^22\zeta(3)-\frac{21}{16}\zeta(2)\zeta(3)+\frac{713}{64}\zeta(5)-\frac4{15}\ln^52-8\ln2\operatorname{Li}_4\left(\frac12\right)-8\operatorname{Li}_5\left(\frac12\right)}$$

where $H_n^{(q)}=\sum_{k=1}^n\frac{1}{n^q}$ is the harmonic number, $\operatorname{Li}_r(x)=\sum_{n=1}^\infty\frac{x^n}{n^r}$ is the polylogarithm function and $\zeta$ is the Riemann zeta function.

This problem is proposed by Cornel with no solution submitted.


My trial

By applying integration by parts we have

$$\int_0^1 x^{2n}(\operatorname{Li}_2(x)-\zeta(2))\ dx=-\frac{H_{2n}}{(2n+1)^2}-\frac{1}{(2n+1)^3}$$

now multiply both sides by $H_n^{(2)}$ then sum both sides from $n=1$ to $\infty$ we get

$$\int_0^1(\operatorname{Li}_2(x)-\zeta(2))\sum_{n=1}^\infty H_n^{(2)}x^{2n}\ dx=-\sum_{n=1}^\infty\frac{H_{2n}H_n^{(2)}}{(2n+1)^2}-\sum_{n=1}^\infty\frac{H_n^{(2)}}{(2n+1)^3}$$

$$\int_0^1\frac{(\operatorname{Li}_2(x)-\zeta(2))\operatorname{Li}_2(x^2)}{1-x^2}\ dx=-\sum_{n=1}^\infty\frac{H_{2n}H_n^{(2)}}{(2n+1)^2}-\color{blue}{\sum_{n=1}^\infty\frac{H_n^{(2)}}{(2n+1)^3}}$$

I managed to find the blue sum using Abel's summation. As for the integral, I tried integration by parts but still resistant.

QUESTION

Any idea how to crack the integral or a different approach to find the target sum?

Thanks.

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I managed to evaluate the integral based on many integrals/ sums results and solution turned out really long but I did my best to make it shorter. If any step is not clear please let me know.


In the body, we reached

$$\int_0^1\frac{(\operatorname{Li}_2(x)-\zeta(2))\operatorname{Li}_2(x^2)}{1-x^2}\ dx=-\sum_{n=1}^\infty\frac{H_{2n}H_n^{(2)}}{(2n+1)^2}-\color{blue}{\sum_{n=1}^\infty\frac{H_n^{(2)}}{(2n+1)^3}}$$

Or $$\sum_{n=1}^\infty\frac{H_{2n}H_n^{(2)}}{(2n+1)^2}=-\int_0^1\frac{(\operatorname{Li}_2(x)-\zeta(2))\operatorname{Li}_2(x^2)}{1-x^2}\ dx-\color{blue}{\sum_{n=1}^\infty\frac{H_n^{(2)}}{(2n+1)^3}}$$

.


Evaluation of the integral:

By applying integration by parts we have

$$I=\int_0^1\frac{(\operatorname{Li}_2(x)-\zeta(2))\operatorname{Li}_2(x^2)}{1-x^2}\ dx\\=\int_0^1\frac{\tanh^{-1}x\ln(1-x)\operatorname{Li}_2(x^2)}{x}\ dx+2\int_0^1\frac{\tanh^{-1}x\ln(1-x^2)}{x}(\operatorname{Li}_2(x)-\zeta(2))\ dx\\ =A+2B$$


The first integral $A$: Using $\tanh^{-1}x=\frac12\ln\left(\frac{1+x}{1-x}\right)$

$$A=\frac12\int_0^1\frac{\ln(1+x)\ln(1-x)\operatorname{Li}_2(x^2)}{x}\ dx-\frac12\int_0^1\frac{\ln^2(1-x)\operatorname{Li}_2(x^2)}{x}\ dx\\ =\frac12(A_1-A_2)$$

where $A_1$ is nicely calculated by Cornel here:

$$A_1=\frac{275}{32}\zeta (5)-\frac{5 }{8}\zeta (2) \zeta (3)+\frac{4}{3} \ln ^32\zeta (2)-\frac{7}{2} \ln ^22\zeta (3)-\frac{4}{15}\ln ^52\\-8 \ln 2\operatorname{Li}_4\left(\frac{1}{2}\right)-8 \operatorname{Li}_5\left(\frac{1}{2}\right).$$

For $A_2$,

\begin{align} A_2&=\sum_{n=1}^\infty\frac1{n^2}\int_0^1 x^{2n-1} \ln^2(1-x)\ dx\\ &=\sum_{n=1}^\infty\frac1{n^2}\left(\frac{H_{2n}^2+H_{2n}^{(2)}}{2n}\right)\\ &=2\sum_{n=1}^\infty\frac{H_n^2+H_n^{(2)}}{n^3}(1+(-1)^n) \end{align}

collecting these results we get

$$A_2=-\frac{1}{8}\zeta (5)+\frac{11 }{2}\zeta (2) \zeta (3)+\frac{4}{3} \ln ^32\zeta (2)-\frac{7}{2}\ln^22\zeta (3)-\frac{4}{15}\ln ^52\\-8 \ln 2\operatorname{Li}_4\left(\frac{1}{2}\right)-8 \operatorname{Li}_5\left(\frac{1}{2}\right).$$

Combining $A_1$ and $A_2$ we get

$$\boxed{A=\frac{279}{64}\zeta(5)-\frac{49}{16}\zeta(2)\zeta(3)}$$


The second integral $B$:

By using the identity

$$\tanh^{-1}x\ln(1-x^2)=-2\sum_{n=1}^\infty\frac{H_{2n}}{2n+1}x^{2n+1}$$

we can write

\begin{align} B&=-2\sum_{n=1}^\infty\frac{H_{2n}}{2n+1}\int_0^1 x^{2n}(\operatorname{Li}_2(x)-\zeta(2))\ dx\\ &=-2\sum_{n=1}^\infty\frac{H_{2n}}{2n+1}\left(-\frac{H_{2n}}{(2n+1)^2}-\frac{1}{(2n+1)^3}\right)\\ &=2\sum_{n=1}^\infty\frac{H_{2n}^2}{(2n+1)^3}+2\sum_{n=1}^\infty\frac{H_{2n}}{(2n+1)^4}\\ &=\sum_{n=1}^\infty\frac{H_{n}^2}{(n+1)^3}+\sum_{n=1}^\infty\frac{(-1)^nH_{n}^2}{(n+1)^3}+\sum_{n=1}^\infty\frac{H_{n}}{(n+1)^4}+\sum_{n=1}^\infty\frac{(-1)^nH_{n}}{(n+1)^4}\\ &=\sum_{n=1}^\infty\frac{H_{n-1}^2}{n^3}-\sum_{n=1}^\infty\frac{(-1)^nH_{n-1}^2}{n^3}+\sum_{n=1}^\infty\frac{H_{n-1}}{n^4}-\sum_{n=1}^\infty\frac{(-1)^nH_{n-1}}{n^4}\\ &=\sum_{n=1}^\infty\frac{H_n^2}{n^3}-\sum_{n=1}^\infty\frac{(-1)^nH_n^2}{n^3}+\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}-\sum_{n=1}^\infty\frac{H_n}{n^4} \end{align}

collecting these results we get

$$\boxed{\small{B=-\frac{31}{16}\zeta (5)-\frac{7 }{8}\zeta (2) \zeta (3)-\frac{2}{3} \ln ^32\zeta (2)+\frac{7}{4}\ln^22\zeta (3)+\frac{2}{15}\ln ^52+4 \ln 2\operatorname{Li}_4\left(\frac{1}{2}\right)+4 \operatorname{Li}_5\left(\frac{1}{2}\right)}}$$


Finally, combine the boxed results of $A$ and $B$ we get

$$I=\frac{31}{64}\zeta (5)-\frac{77 }{16}\zeta (2) \zeta (3)-\frac{4}{3} \ln ^32\zeta (2)+\frac{7}{2}\ln^22\zeta (3)+\frac{4}{15}\ln ^52\\+8 \ln 2\operatorname{Li}_4\left(\frac{1}{2}\right)+8 \operatorname{Li}_5\left(\frac{1}{2}\right).$$


Evaluation of the blue sum:

By Abel's summation we have

$$\sum_{n=1}^\infty\frac{H_n^{(2)}-\zeta(2)}{(2n-1)^3}=\frac18\sum_{n=1}^\infty\frac{H_n^{(3)}}{(n+1)^2}-\sum_{n=1}^\infty\frac{H_{2n}^{(3)}}{(n+1)^2}$$

where

$$\sum_{n=1}^\infty\frac{H_n^{(3)}}{(n+1)^2}=\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^2}-\zeta(5)=\frac92\zeta(5)-2\zeta(2)\zeta(3)$$

and

$$\sum_{n=1}^\infty\frac{H_{2n}^{(3)}}{(n+1)^2}=\sum_{n=1}^\infty\frac{H_{2n-2}^{(3)}}{n^2}=\sum_{n=1}^\infty\frac{H_{2n}^{(3)}-\frac1{(2n)^3}-\frac1{(2n-1)^3}}{n^2}\\ =\sum_{n=1}^\infty\frac{H_{2n}^{(3)}}{n^2}-\frac18\zeta(5)-\sum_{n=1}^\infty\frac1{n^2(2n-1)^3}\\$$

combining the two sums we get

$$\sum_{n=1}^\infty\frac{H_n^{(2)}-\zeta(2)}{(2n-1)^3}=\frac{11}{16}\zeta(5)-\frac14\zeta(2)\zeta(3)-\sum_{n=1}^\infty\frac{H_{2n}^{(3)}}{n^2}+\sum_{n=1}^\infty\frac1{n^2(2n-1)^3}\tag1$$


On the other hand:

\begin{align} \sum_{n=1}^\infty\frac{H_n^{(2)}-\zeta(2)}{(2n-1)^3}&=\sum_{n=1}^\infty\frac{H_n^{(2)}}{(2n-1)^3}-\sum_{n=1}^\infty\frac{\zeta(2)}{(2n-1)^3}\\ &=\sum_{n=0}^\infty\frac{H_{n+1}^{(2)}}{(2n+1)^3}-\frac78\zeta(2)\zeta(3)\\ &=\sum_{n=0}^\infty\frac{H_{n}^{(2)}}{(2n+1)^3}+\sum_{n=0}^\infty\frac{1}{(n+1)^2(2n+1)^3}-\frac78\zeta(2)\zeta(3)\\ &=\sum_{n=0}^\infty\frac{H_{n}^{(2)}}{(2n+1)^3}+\sum_{n=1}^\infty\frac{1}{n^2(2n-1)^3}-\frac78\zeta(2)\zeta(3)\tag{2} \end{align}


From (1) and (2) we get

$$\sum_{n=0}^\infty\frac{H_{n}^{(2)}}{(2n+1)^3}=\frac{11}{16}\zeta(5)+\frac58\zeta(2)\zeta(3)-\sum_{n=1}^\infty\frac{H_{2n}^{(3)}}{n^2}\\ =\frac{11}{16}\zeta(5)+\frac58\zeta(2)\zeta(3)-2\sum_{n=1}^\infty\frac{H_{n}^{(3)}}{n^2}-2\sum_{n=1}^\infty\frac{(-1)^nH_{n}^{(3)}}{n^2}$$

Thus

$$\sum_{n=1}^\infty\frac{H_{n}^{(2)}}{(2n+1)^3}=\frac{49}{8}\zeta(2)\zeta(3)-\frac{93}{8}\zeta(5)$$


Finally, by collecting the results of $I$ and the blue sum we get our closed form.


References

$\sum_{n=1}^\infty\frac{(-1)^{n}H_n^{(2)}}{n^3}=\frac{11}{32}\zeta(5)-\frac58\zeta(2)\zeta(3)\\ \small{\sum_{n=1}^\infty\frac{(-1)^{n}H_n^2}{n^3}=-4\operatorname{Li}_5\left(\frac12\right)-4\ln2\operatorname{Li}_4\left(\frac12\right)+\frac{19}{32}\zeta(5)+\frac{11}8\zeta(2)\zeta(3)-\frac74\ln^22\zeta(3)+\frac23\ln^32\zeta(2)-\frac2{15}\ln^52} $ $\sum_{n=1}^\infty\frac{(-1)^nH_n^{(3)}}{n^2}=\frac{21}{32}\zeta(5)-\frac34\zeta(2)\zeta(3)$

$\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}=3\zeta(2)\zeta(3)-\frac92\zeta(5)\\\sum_{n=1}^\infty\frac{H_n^2}{n^3}=\frac72\zeta(5)-\zeta(2)\zeta(3)$

$\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^2}=\frac{11}2\zeta(5)-2\zeta(2)\zeta(3)$

$\sum^\infty_{n=1}\frac{(-1)^nH_n}{n^4}=-\frac{59}{32}\zeta(5)+\frac12\zeta(2)\zeta(3)$

$\sum^\infty_{n=1}\frac{H_n}{n^4}=3\zeta(5)-\zeta(2)\zeta(3)$ ( Euler Identity)

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