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Prove that

$$S=\sum_{n=1}^\infty\frac{H_{2n}H_n^{(2)}}{(2n)^2}=\frac{101}{64}\zeta(5)-\frac5{16}\zeta(2)\zeta(3)$$ where $H_n^{(m)}=\sum_{k=1}^n\frac1{k^m}$ is the n$th$ generalized harmonic number of order $m$ and $\zeta$ is the Riemann zeta function.

This problem is proposed by Cornel Valean and can be found here.

Here is how I managed to find the integral representation:

We have $\int_0^1 x^{2n-1}\ln(1-x)\ dx=-\frac{H_{2n}}{2n}$ , then we can write

$$\sum_{n=1}^\infty\frac{H_{2n}H_n^{(2)}}{(2n)^2}=-\frac12\int_0^1\frac{\ln(1-x)}{x}\sum_{n=1}^\infty\frac{H_n^{(2)}}{n}(x^2)^n\ dx\\=\small{-\frac12\int_0^1\frac{\ln(1-x)}{x}\left(\operatorname{Li}_3(x^2)+2\operatorname{Li}_3(1-x^2)-\ln(1-x^2)\operatorname{Li}_2(1-x^2)-\zeta(2)\ln(1-x^2)-2\zeta(3)\right)\ dx}$$

So any idea how to crack this integral or different approach?

Thanks.


UPDATE:

This result was mentioned by @nospoon here in equation $(3)$. He didn't post the solution but he provided the idea.

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  • $\begingroup$ I would try summation by parts: since $$ \sum_{n=1}^{N}\frac{H_n^{(2)}}{n^2} = \frac{\left(H_N^{(2)}\right)^2+H_N^{(4)}}{2} $$ the computation of the original series boils down to the computation of $\lim_{N\to +\infty}\sum_{n=1}^{N}a_n$, with a_n being one of the following objects: $$ \frac{\left(H_n^{(2)}\right)^2}{n(2n+1)},\qquad \frac{H_n^{(4)}}{n(2n+1)} .$$ $\endgroup$ – Jack D'Aurizio Sep 7 at 18:26
  • $\begingroup$ The latter series can be tackled by integration, the former is probably best dealt with Fourier-* expansions. $\endgroup$ – Jack D'Aurizio Sep 7 at 18:27
  • $\begingroup$ @Jack D'Aurizio so $a_k=H_{2k}$ and $b_k=\frac{H_k^{(2)}}{k^2}$ ? $$\sum_{k=1}^n a_k b_k=A_nb_{n+1}+\sum_{k=1}^{n}A_k\left(b_k-b_{k+1}\right)$$ $\endgroup$ – Ali Shather Sep 7 at 19:26
  • $\begingroup$ The other way around, $a_k=\frac{H_k^{(2)}}{k^2}$ and $b_k=H_{2k}$. $\endgroup$ – Jack D'Aurizio Sep 7 at 19:35
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    $\begingroup$ @Jack D'Aurizio the sum $\sum_{n=1}^\infty \frac{H_n H_n^{(2)}}{n^2}$ you referring to, i think i did it before. I'll check. $\endgroup$ – Ali Shather Sep 7 at 20:08
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We will look into the integral \begin{align*} I = &\int_{0}^{1}\arcsin^4 x\frac{ \ln x}{\sqrt{1-x^2}}\ \mathrm dx \end{align*} taking the @nospoon's novel approach presented here. Using the MacLaurin series of $\arcsin^4 x$ $$ \arcsin^4 x =\frac 3 2 \sum_{n=1}^\infty \frac{4^{n}H_{n-1}^{(2)}}{n^2{2n \choose n}}x^{2n} $$ and the fact that $$ \small\operatorname{B}(n+\tfrac 1 2,\tfrac 1 2) = \int_0^1 x^{n-1/2}(1-x)^{-1/2}\ \mathrm dx = 2\int_0^{\frac\pi 2} \sin^{2n}\theta\ \mathrm d\theta = \frac{\pi}{4^n}{2n \choose n},\tag{$\small x\mapsto \sin^2\theta$} $$ \begin{align*}\small \psi(n+\tfrac 12 ) -\psi(n+1) =&\small \sum_{k=1}^\infty \frac 1{\scriptsize k+n} - \frac 1{\scriptsize k+n-\tfrac 1 2} \\ =&\small\sum_{k=1}^\infty \left(\frac 1{\scriptsize k} - \frac 1{\scriptsize k-\tfrac 1 2}\right)-\sum_{k=1}^n\frac 1 {\scriptsize k} + \sum_{k=1}^n\frac 1{\scriptsize k-\tfrac 1 2}\\ =&\small-2\ln 2 -H_n +2(H_{2n}-\tfrac 1 2H_n)\\ =&\small 2(H_{2n}-H_n-\ln 2), \end{align*} \begin{align*} \Longrightarrow \ {\int_{ 0}^{1 }x^{2n}\frac{ \ln x}{\sqrt{1-x^2}}\ \mathrm dx} = & \frac 1 4\int_{0 }^{1 }x^{n-1/2} { \ln x \over \sqrt{1-x}}\ \mathrm dx\tag{$\small x^2\mapsto x$}\\ =& \frac 1 4 \left[\frac{\partial }{\partial x}\operatorname{B}(x,y) \right]_{x=n+1/2,y=1/2}\\ =&\frac 1 4\Big[ \operatorname{B}(x,y)\big[\psi(x) -\psi(x+y) \big]\Big]_{x=n+1/2,y=1/2}\\ =& \frac 1 4 \operatorname{B}(n+\tfrac 1 2,\tfrac 1 2)\big[\psi(n+\tfrac 12 ) -\psi(n+1) \big]\\ =& \frac{\pi}2\frac{{2n \choose n}}{4^{n}} \left(H_{2n} - H_n -\ln 2\right), \end{align*} where $\operatorname{B}(x,y)$ and $\psi(x)$ are the Beta and digamma function, respectively, we have \begin{align*} I = &\frac 3 2\sum_{n=1}^\infty \frac{4^{n}H_{n-1}^{(2)}}{n^2{2n \choose n}}\int_{0}^{1}x^{2n}\frac{ \ln x}{\sqrt{1-x^2}}\ \mathrm dx \\ =&\frac {3\pi}4 \sum_{n=1}^\infty \frac{H^{(2)}_{n-1}}{n^2}\left(H_{2n} - H_n -\ln 2\right) \\ =&\frac {3\pi}4\sum_{n=1}^\infty \frac{H^{(2)}_{n-1}H_{2n}}{n^2}-\frac {3\pi}4\underbrace{\sum_{n=1}^\infty \frac{H^{(2)}_{n-1}H_{n}}{n^2}}_{=-2\zeta(5) +2\zeta(2)\zeta(3)}-\frac {3\pi\ln 2}4\underbrace{\sum_{n=1}^\infty \frac{H^{(2)}_{n-1}}{n^2}}_{=\frac{3}4 \zeta(4)}\\ =&\frac{3\pi}{4} \sum_{n=1}^\infty \frac{H^{(2)}_{n}H_{2n}}{n^2} -\frac{3\pi}4\underbrace{\sum_{n=1}^\infty \frac{H_{2n}}{n^4}}_{=\frac{37}{4}\zeta(5)-4\zeta(2)\zeta(3)} +\frac{3\pi}2 \zeta(5) -\frac{\pi^3}4\zeta(3) -\frac{\pi^5\ln 2}{160}\\ =&\boxed{3\pi S -\frac{87\pi}{16} \zeta(5) +\frac{\pi^3}{4}\zeta(3) -\frac{\pi^5\ln 2}{160}} \end{align*} where $S = \sum_{n=1}^\infty \frac{H_{2n}H^{(2)}_{n}}{4n^2}$ is the sum in question, and the known values of several Euler sums $$ \sum_{n=1}^\infty \frac{H^{(2)}_{n-1}H_{n}}{n^2}=-2\zeta(5) +2\zeta(2)\zeta(3),\tag{1} $$ $$\sum_{n=1}^\infty \frac{H^{(2)}_{n}}{n^2}=\frac{7}4 \zeta(4),\tag{2} $$ \begin{align*}\sum_{n=1}^\infty \frac{H_{2n}}{n^4} =& 8\sum_{n=1}^\infty \frac{H_{n}}{n^4}-8\sum_{n=1}^\infty \frac{(-1)^{n-1} H_{n}}{n^4}\\ =&8\big(3\zeta(5)-\zeta(2)\zeta(3)\big)-8\left(\frac{59}{32}\zeta(5)-\frac 1 2\zeta(2)\zeta(3)\right)\\ =&\frac{37}4\zeta(5) - 4\zeta(2)\zeta(3)\tag{3} \end{align*} are used.

Note: $(1)$ is in @nospoon's answer here, $(2)$ can be found here, and for $(3)$ you can see Euler's formula and here.


Evaluation of $I$: By making substitution $x = \sin \theta$ and using the Fourier series of $$ \ln (\sin\theta) = -\ln 2 -\sum_{k=1}^\infty \frac{ \cos(2k \theta)}{k}, $$ we get \begin{align*} I =& \int_{0}^{\frac\pi 2} \theta^4 \ln(\sin\theta)\ \mathrm d\theta\\ =&\int_{0}^{\frac\pi 2} \theta^4\left(-\ln 2 -\sum_{k=1}^\infty \frac{ \cos(2k \theta)}{k}\right)\ \mathrm d\theta\\ =& -\ln 2\int_0^{\frac \pi 2}\theta^4\ \mathrm d\theta-\sum_{k=1}^\infty \frac{1}{k}\underbrace{\int_{0}^{\frac\pi 2}\theta^4 \cos(2k \theta) \ \mathrm d\theta}_{\text{IBP}\times 4}\\ =& -\frac{\pi^5\ln 2}{160}-\sum_{k=1}^\infty \frac{1}{k}\cdot\left(-\frac{\pi^3}{8}\frac{(-1)^{k-1}}{k^2} +\frac{3\pi}{4}\frac{(-1)^{k-1}}{k^4}\right)\\ =&-\frac{\pi^5\ln 2}{160}+\frac{\pi^3}8\underbrace{\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k^3}}_{=\frac 3 4 \zeta(3)} - \frac{3\pi}4\underbrace{\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k^5}}_{=\frac{15}{16}\zeta(5)}\\ =&\boxed{-\frac{\pi^5\ln 2}{160}+\frac{3\pi^3}{32}\zeta(3) -\frac{45\pi}{64}\zeta(5).} \end{align*}


Combining these, we get the equation $$ 3\pi S-\frac{87\pi}{16} \zeta(5) +\frac{\pi^3}{4}\zeta(3) -\frac{\pi^5\ln 2}{160}=-\frac{\pi^5\ln 2}{160} +\frac{3\pi^3}{32}\zeta(3)-\frac{45\pi}{64}\zeta(5), $$hence it follows $$ \boxed{S = \frac{101}{64}\zeta(5) -\frac{5\pi^2}{96}\zeta(3).} $$


Addendum: By considering MacLaurin series of \begin{align*} \ln(1-x)\ln(1+x) =&-\sum_{k=1}^\infty \left(\frac{ H_{2k}}k-\frac{H_k}{k} + \frac1{2k^2}\right)x^{2k} \end{align*} and \begin{align*} \frac{H_k}{k^2} + \frac{H_k^{(2)}}{k} -\frac{\zeta(2)}{k} =& \frac{\partial }{\partial k}\left[-\frac{H_k}{k}\right]\\ =& \int_0^1 x^{k-1}\ln x\ln(1-x)\ \mathrm dx\\ =&4\int_0^1 x^{2k-1}\ln x \ln(1-x^2)\ \mathrm dx \end{align*} we have that \begin{align*} &\int_{0}^{1}\ln(1-x)\ln(1+x) \frac{\ln x\ln(1-x^2)}x \ \mathrm dx \\&=-\sum_{k=1}^\infty \left(\frac{ H_{2k}}k-\frac{H_k}{k} + \frac1{2k^2}\right)\int_{0}^{1}x^{2k-1} \ln x \ln(1-x^2)\ \mathrm dx \\ &=-\frac 1 4\sum_{k=1}^\infty \left(\frac{ H_{2k}}k-\frac{H_k}{k} + \frac1{2k^2}\right)\left(\frac{H_k}{k^2} + \frac{H_k^{(2)}}{k} -\frac{\zeta(2)}{k}\right). \end{align*} The integral can be attacked by considering algebraic identity $$ ab(a+b) = \frac 1 3 (a+b)^3 - \frac {a^3}3 -\frac{b^3}3 $$ with $a=\ln(1-x)$ and $b=\ln(1+x)$, and extant results.

For the sum, after expanding the summand, the only tricky part is $$ \sum_{k=1}^\infty\frac{H_{2k}H_k}{k^3}, $$ which can be found here. Then, the sum $\sum_{k=1}^\infty \frac{H_{2k}H_k^{(2)}}{4k^2}$ can be evaluated by solving the equation obtained.

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  • $\begingroup$ Very nice.. good job for doing all these tedious calculations. I'll present my solution soon. (+1) $\endgroup$ – Ali Shather Sep 10 at 5:18
  • $\begingroup$ @AliShather Thank you :) I'm looking forward to your solution. $\endgroup$ – Song Sep 10 at 5:30
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I am going to establish two relations as Cornel suggested and solve them by elimination.


From here we have

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

Multiply both sides by $\frac{H_{2n}}{4n}$ then sum them from $n=1$ to $\infty$ we get

\begin{align} R_1&=\sum_{n=1}^\infty\frac{H_{2n}}{(2n)^2}\left(H_n^2+H_n^{(2)}\right)=\frac12\int_0^1\frac{\ln^2(1-x)}{x}\sum_{n=1}^\infty\frac{x^nH_{2n}}{2n}\ dx\quad \color{red}{x\mapsto x^2}\\ &=\int_0^1\frac{\ln^2(1-x^2)}{x}\sum_{n=1}^\infty\frac{x^{2n}H_{2n}}{2n}\ dx, \quad \color{red}{2\sum_{n=1}^\infty a_{2n}=\sum_{n=1}^\infty a_{n}+\sum_{n=1}^\infty (-1)^na_{n}}\\ &=\frac12\int_0^1\frac{\ln^2(1-x^2)}{x}\left(\sum_{n=1}^\infty\frac{x^{n}H_{n}}{n}+\sum_{n=1}^\infty\frac{(-x)^{n}H_{n}}{n}\right)\ dx\\ &=\frac12\int_0^1\frac{\ln^2(1-x^2)}{x}\left(\underbrace{\operatorname{Li}_2(x)+\operatorname{Li}_2(-x)}_{\frac12\operatorname{Li}_2(x^2)}+\frac12\ln^2(1-x)+\frac12\ln^2(1+x)\right)\ dx\\ &=\small{\frac14\underbrace{\int_0^1\frac{\ln^2(1-x^2)\operatorname{Li}_2(x^2)}{x}\ dx}_{x^2\mapsto x}+\frac14\int_0^1\frac{\ln^2(1-x^2)}{x}\left(\ln^2(1-x)+\ln^2(1+x)\right)\ dx}\\ &=\frac18\int_0^1\frac{\ln^2(1-x)\operatorname{Li}_2(x)}{x}\ dx+\frac14\int_0^1\frac{\ln^2(1-x^2)}{x}\left(\ln^2(1-x)+\ln^2(1+x)\right)\ dx\\ &=\frac18A+\frac14B\tag{1} \end{align}


Let's start with $A$ and by seting $1-x\mapsto x$ we can write

\begin{align} A&=\int_0^1\frac{\ln^2x\color{blue}{\operatorname{Li}_2(1-x)}}{1-x}\ dx\\ &=\int_0^1\frac{\ln^2x}{1-x}\left[\color{blue}{\zeta(2)-\ln x\ln(1-x)-\operatorname{Li}_2(x)}\right]\ dx\\ &=\zeta(2)\int_0^1\frac{\ln^2x}{1-x}\ dx-\int_0^1\frac{\ln^3x\ln(1-x)}{1-x}\ dx-\int_0^1\frac{\ln^2x\operatorname{Li}_2(x)}{1-x}\ dx\\ &=2\zeta(2)\zeta(3)+\sum_{n=1}^\infty H_n\int_0^1 x^n\ln^3x\ dx-\sum_{n=1}^\infty H_n^{(2)}\int_0^1 x^n \ln^2x\ dx\\ &=2\zeta(2)\zeta(3)-6\sum_{n=1}^\infty\frac{H_n}{(n+1)^4}-2\sum_{n=1}^\infty\frac{H_n^{(2)}}{(n+1)^3}\\ &=2\zeta(2)\zeta(3)-6\sum_{n=1}^\infty\frac{H_n}{n^4}+6\zeta(5)-2\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}+2\zeta(5)\\ &\boxed{A=2\zeta(2)\zeta(3)-\zeta(5)} \end{align}

where we used $\sum_{n=1}^\infty\frac{H_n}{n^4}=3\zeta(5)-\zeta(2)\zeta(3)$ and $\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}=3\zeta(2)\zeta(3)-\frac92\zeta(5)$


To evaluate $B$, we are going to use the key identity

$$(a+b)^2(a^2+b^2)=\frac23a^4+\frac23b^4+\frac5{12}(a+b)^4-\frac1{12}(a-b)^4$$

and by taking $a=\ln(1-x)$ and $b=\ln(1+x)$ , we get

$$B=\int_0^1\frac{\ln^2(1-x^2)}{x}\left(\ln^2(1-x)+\ln^2(1+x)\right)\ dx\\ =\small{\frac23\int_0^1\frac{\ln^4(1-x)}{x}\ dx+\frac23\int_0^1\frac{\ln^4(1+x)}{x}\ dx+\frac5{12}\underbrace{\int_0^1\frac{\ln^4(1-x^2)}{x}\ dx}_{x^2\mapsto x}-\frac1{12}\underbrace{\int_0^1\frac{\ln^4\left(\frac{1-x}{1+x}\right)}{x}\ dx}_{\frac{1-x}{1+x}\mapsto x}}\\ =\frac78\int_0^1\frac{\ln^4(1-x)}{x}\ dx+\frac23\int_0^1\frac{\ln^4(1+x)}{x}\ dx-\frac16\int_0^1\frac{\ln^4x}{1-x^2}\ dx\\ =\frac78(24\zeta(5))+\frac23\int_0^1\frac{\ln^4(1+x)}{x}\ dx-\frac16\left(\frac{93}{4}\zeta(5)\right)\\ =\frac{137}{8}\zeta(5)+\frac23\int_0^1\frac{\ln^4(1+x)}{x}\ dx$$

Since \begin{align} \int_0^1\frac{\ln^4(1+x)}{x}&=\int_{1/2}^1\frac{\ln^4x}{x}\ dx+\int_{1/2}^1\frac{\ln^4x}{1-x}\ dx\\ &=\frac15\ln^52+\sum_{n=1}^\infty\int_{1/2}^1 x^{n-1}\ln^4x\ dx\\ &=\frac15\ln^52+\sum_{n=1}^\infty\left(\frac{24}{n^5}-\frac{24}{n^52^n}-\frac{24\ln2}{n^42^n}-\frac{12\ln^22}{n^32^n}-\frac{4\ln^32}{n^22^n}-\frac{\ln^42}{n2^n}\right)\\ &=\small{4\ln^32\zeta(2)-\frac{21}2\ln^22\zeta(3)+24\zeta(5)-\frac45\ln^52-24\ln2\operatorname{Li}_4\left(\frac12\right)-24\operatorname{Li}_5\left(\frac12\right)} \end{align}

Then

$$\boxed{B=\frac83\ln^32\zeta(2)-7\ln^22\zeta(3)+\frac{265}{8}\zeta(5)-\frac8{15}\ln^52-16\ln2\operatorname{Li}_4\left(\frac12\right)-16\operatorname{Li}_5\left(\frac12\right)}$$

Plugging the boxed results of $A$ and $B$ in (1) we get our first relation:

$$R_1=\sum_{n=1}^\infty\frac{H_{2n}}{(2n)^2}\left(H_n^2+H_n^{(2)}\right)\\ =\small{\frac23\ln^32\zeta(2)-\frac74\ln^22\zeta(3)+\frac14\zeta(2)\zeta(3)+\frac{261}{32}\zeta(5)-\frac2{15}\ln^52-4\ln2\operatorname{Li}_4\left(\frac12\right)-4\operatorname{Li}_5\left(\frac12\right)}$$


We have

$$\frac{\ln^2(1-y)}{1-y}=\sum_{n=1}^\infty y^n(H_n^2-H_n^{(2)})$$

integrate both sides from $y=0$ to $y=x$ to get

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

Now replace $x$ with $x^2$ then multiply both sides by $-\frac{\ln(1-x)}{x}$ and integrate from $x=0$ to $x=1$, also note that $\int_0^1 -x^{2n-1}\ln(1-x)\ dx=\frac{H_{2n}}{2n}$ we get

$$\frac13\underbrace{\int_0^1\frac{\ln^3(1-x^2)\ln(1-x)}{x}\ dx}_{\large C}=\sum_{n=1}^\infty\frac{H_{2n}}{2n^2}\left(H_n^2-H_n^{(2)}-\frac{2H_n}{n}+\frac{2}{n^2}\right)$$

Rearranging the terms to get

$$R_2=\sum_{n=1}^\infty\frac{H_{2n}}{(2n)^2}(H_n^2-H_n^{(2)})=4\sum_{n=1}^\infty\frac{H_{2n}H_n}{(2n)^3}-8\sum_{n=1}^\infty\frac{H_{2n}}{(2n)^4}+\frac16C\tag{2}$$

Cornel elegantly calculated the first sum here

$$\boxed{\small{\sum _{n=1}^{\infty } \frac{H_{2 n}H_n }{(2 n)^3}=\frac{307}{128}\zeta(5)-\frac{1}{16}\zeta (2) \zeta (3)+\frac{1}{3}\ln ^3(2)\zeta (2) -\frac{7}{8} \ln ^2(2)\zeta (3)-\frac{1}{15} \ln ^5(2) -2 \ln (2) \operatorname{Li}_4\left(\frac{1}{2}\right) -2 \operatorname{Li}_5\left(\frac{1}{2}\right)}}$$

For the second sum: $$\sum_{n=1}^\infty\frac{H_{2n}}{(2n)^4}=\frac12\sum_{n=1}^\infty\frac{H_{n}}{n^4}+\frac12\sum_{n=1}^\infty(-1)^n\frac{H_{n}}{n^4}$$

plugging the common results:

$$\sum_{n=1}^\infty\frac{H_{n}}{n^4}=3\zeta(5)-\zeta(2)\zeta(3)$$

$$\sum_{n=1}^\infty(-1)^n\frac{H_{n}}{n^4}=\frac12\zeta(2)\zeta(3)-\frac{59}{32}\zeta(5)$$

we get

$$\boxed{\sum_{n=1}^\infty\frac{H_{2n}}{(2n)^4}=\frac{37}{64}\zeta(5)-\frac14\zeta(2)\zeta(3)}$$

For the remaining integral $C$, we use the magical identity

$$(a+b)^3a=a^4-b^4+\frac12(a+b)^4-\frac12(a-b)^4-(a-b)^3b$$

with $a=\ln(1-x)$ and $b=\ln(1+x)$ we can write

$$C=\int_0^1\frac{\ln^4(1-x)}{x}\ dx-\int_0^1\frac{\ln^4(1+x)}{x}\ dx+\frac12\underbrace{\int_0^1\frac{\ln^4(1-x^2)}{x}\ dx}_{x^2\mapsto x}\\-\underbrace{\frac12\int_0^1\frac{\ln^4\left(\frac{1-x}{1+x}\right)}{x}\ dx}_{\frac{1-x}{1+x}\mapsto x}-\underbrace{\int_0^1\frac{\ln^3\left(\frac{1-x}{1+x}\right)\ln(1+x)}{x}\ dx}_{\frac{1-x}{1+x}\mapsto x}$$

$$C=\frac54\underbrace{\int_0^1\frac{\ln^4(1-x)}{x}\ dx}_{4!\zeta(5)}-\underbrace{\int_0^1\frac{\ln^4(1+x)}{x}\ dx}_{K}-\underbrace{\int_0^1\frac{\ln^4x}{1-x^2}\ dx}_{\frac{93}{4}\zeta(5)}+\underbrace{2\int_0^1\frac{\ln^3x\ln\left(\frac{1+x}{2}\right)}{1-x^2}\ dx}_{J}$$

$$C=\frac{27}{4}\zeta(5)-K+J\tag{3}$$

we have already evaluated $K$ above: $$K=4\ln^32\zeta(2)-\frac{21}2\ln^22\zeta(3)+24\zeta(5)-\frac45\ln^52-24\ln2\operatorname{Li}_4\left(\frac12\right)-24\operatorname{Li}_5\left(\frac12\right)$$

for $J$

$$J=2\int_0^1\frac{\ln^3x\ln\left(\frac{1+x}{2}\right)}{1-x^2}\ dx=\int_0^1\frac{\ln^3x\ln\left(\frac{1+x}{2}\right)}{1-x}\ dx+\int_0^1\frac{\ln^3x\ln\left(\frac{1+x}{2}\right)}{1+x}\ dx$$

using the rule

$$\int_0^1\frac{\ln^ax\ln\left(\frac{1+x}{2}\right)}{1-x}\ dx=(-1)^aa!\sum_{n=1}^\infty\frac{(-1)^nH_n^{a+1}}{n}$$

allows us to write

\begin{align} J&=-6\sum_{n=1}^\infty\frac{(-1)^nH_n^{(4)}}{n}+\int_0^1\frac{\ln^3x\ln(1+x)}{1+x}\ dx-\ln2\int_0^1\frac{\ln^3x}{1+x}\ dx\\ &=-6\sum_{n=1}^\infty\frac{(-1)^nH_n^{(4)}}{n}-\sum_{n=1}^\infty(-1)^n H_n\int_0^1x^n\ln^3x\ dx-\ln2\left(-\frac{21}4\zeta(4)\right)\\ &=-6\sum_{n=1}^\infty\frac{(-1)^nH_n^{(4)}}{n}+6\sum_{n=1}^\infty\frac{(-1)^n H_n}{(n+1)^4}+\frac{21}{4}\ln2 \zeta(4)\\ &=-6\sum_{n=1}^\infty\frac{(-1)^nH_n^{(4)}}{n}-6\sum_{n=1}^\infty\frac{(-1)^n H_n}{n^4}-\frac{45}{8}\zeta(5)+\frac{21}{4}\ln2 \zeta(4) \end{align}

Plugging

$$\sum_{n=1}^\infty\frac{(-1)^nH_n^{(4)}}{n}=\frac78\ln2\zeta(4)+\frac38\zeta(2)\zeta(3)-2\zeta(5)$$

we get

$$J=\frac{279}{16}\zeta(5)-\frac{21}{4}\zeta(2)\zeta(3)$$

Plugging the results of $K$ and $J$ in (3) we get

$$\boxed{\small{C=24\operatorname{Li}_5\left(\frac12\right)+24\ln2\operatorname{Li}_4\left(\frac12\right)+\frac3{16}\zeta(5)-\frac{21}{4}\zeta(2)\zeta(3)+\frac{21}2\ln^22\zeta(3)-4\ln^32\zeta(2)+\frac45\ln^52}}$$

and by substituting the boxed results in (2) we get our second relation

$$R_2=\sum _{n=1}^{\infty } \frac{H_{2 n} }{(2 n)^2}(H_n^2-H_n^{(2)}) =\frac78\zeta(2)\zeta(3)+5\zeta (5)+\frac{2}{3} \ln ^3(2)\zeta (2) -\frac{7}{4} \ln ^2(2)\zeta (3)\\ -\frac{2}{15} \ln^5(2) -4 \ln2\operatorname{Li}_4\left(\frac{1}{2}\right) -4 \operatorname{Li}_5\left(\frac{1}{2}\right)$$


Thus

$$\sum_{n=1}^\infty\frac{H_{2n}H_n^{(2)}}{(2n)^2}=\frac{R_1-R_2}{2}=\frac{101}{64}\zeta(5)-\frac5{16}\zeta(2)\zeta(3)$$

and as a bonus

$$\sum_{n=1}^\infty\frac{H_{2n}H_n^2}{(2n)^2}=\frac{R_1+R_2}{2}\\ =\small{\frac{421 }{64}\zeta (5)+\frac{9 }{16}\zeta (2) \zeta (3)+\frac{2}{3} \ln ^32\zeta (2) -\frac{7}{4} \ln ^22\zeta (3) -\frac{2}{15} \ln^52-4 \ln2\operatorname{Li}_4\left(\frac{1}{2}\right) -4 \operatorname{Li}_5\left(\frac{1}{2}\right)}$$


Note:

$\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}$ can be found here and $\sum_{n=1}^\infty\frac{(-1)^nH_n^{(4)}}{n}$ can be found here.

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Update: the details may be found in the preprint The evaluation of a special harmonic series with a weight $5$ structure, involving harmonic numbers of the type $H_{2n}$

The magical way by Cornel Ioan Valean

By the Cauchy product, we have $\operatorname{Li}_2(x^2) \log(1-x^2)= 3\sum _{n=1}^{\infty } \frac{x^{2 n}}{n^3}-2\sum _{n=1}^{\infty } x^{2n}\frac{H_n}{n^2}-\sum _{n=1}^{\infty } x^{2n}\frac{H_n^{(2)}}{n}$, and if we multiply both sides by $\log(1-x)/x$, and integrate from $x=0$ to $x=1$, using that $\int_0^1 x^{n-1}\log(1-x)\textrm{d}x=-H_n/n$, and doing all the reductions, we arrive at

$$2\sum _{n=1}^{\infty } \frac{H_{2 n} H_n^{(2)}}{(2 n)^2}-12\sum _{n=1}^{\infty } \frac{H_n}{n^4}+12\sum _{n=1}^{\infty }(-1)^{n-1} \frac{H_n}{n^4}+\sum _{n=1}^{\infty } \frac{H_n H_{2 n}}{n^3}$$ $$=\int_0^1 \frac{\text{Li}_2\left(x^2\right) \log \left(1-x^2\right) \log (1-x)}{x} \textrm{d}x$$ $$=\int_0^1 \frac{\text{Li}_2\left(x^2\right) \log (1+x) \log (1-x)}{x}\textrm{d}x+2 \int_0^1 \frac{\text{Li}_2(-x) \log ^2(1-x)}{x} \textrm{d}x\\+2 \int_0^1 \frac{\text{Li}_2(x) \log ^2(1-x)}{x} \textrm{d}x$$ $$=\int_0^1 \frac{\text{Li}_2\left(x^2\right) \log (1+x) \log (1-x)}{x} \textrm{d}x+2 \sum _{n=1}^{\infty } \frac{H_n^2}{n^3}-2 \sum _{n=1}^{\infty } \frac{(-1)^{n-1}H_n^2}{n^3}+2 \sum _{n=1}^{\infty } \frac{H_n^{(2)}}{n^3}\\-2 \sum _{n=1}^{\infty }(-1)^{n-1} \frac{ H_n^{(2)}}{n^3},$$ where the last integral is given here Two very advanced harmonic series of weight $5$, and all the last resulting harmonic series are given in the book (Almost) Impossible Integrals, Sums, and Series. The reduction to the last series has been achieved by using the identity, $\displaystyle \int_0^1 x^{n-1}\log^2(1-x)\textrm{d}x=\frac{H_n^2+H_n^{(2)}}{n}$. The series $\sum _{n=1}^{\infty } \frac{H_n H_{2 n}}{n^3}$ maybe found calculated in the paper On the calculation of two essential harmonicseries with a weight 5 structure, involving harmonic numbers of the type H_{2n} by Cornel Ioan Valean. Thus, we have

$$\sum_{n=1}^\infty\frac{H_{2n}H_n^{(2)}}{(2n)^2}=\frac{101}{64}\zeta(5)-\frac5{16}\zeta(2)\zeta(3).$$

All the details will appear in a new paper.

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  • 1
    $\begingroup$ Cornel's solutions are always unique. $\endgroup$ – Ali Shather Oct 3 at 17:54

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