Computing $\sum_{n=1}^\infty\frac{4^nH_n^{(3)}}{n^2{2n\choose n}}$

Question

I managed to find

$$\sum_{n=1}^\infty\frac{4^nH_n^{(3)}}{n^2{2n\choose n}}=3\zeta(2)\zeta(3)-8\underbrace{\int_0^{\pi/2}x^2\tan x\ln^2(\sin x)dx}_{I}\tag1$$

but I could not finish $I$:

My attempt:

In the book, Almost Impossible Integrals, Sums and series, page $243$, Eq$(3.281)$ we have

$$\tan x\ln(\sin x)=-\sum_{n=1}^\infty\left(\psi\left(\frac{n+1}{2}\right)-\psi\left(\frac{n}{2}\right)-\frac1n\right)\sin(2nx)$$

$$=-\sum_{n=1}^\infty\left(\int_0^1\frac{1-t}{1+t}t^{n-1}dt\right)\sin(2nx),\quad 0<x<\frac{\pi}{2}$$

So $$I=-\sum_{n=1}^\infty\left(\int_0^1\frac{1-t}{1+t}t^{n-1}dt\right)\underbrace{\left(\int_0^{\pi/2}x^2\sin(2nx)\ln(\sin x)dx\right)}_{J}$$

For $J$, we use the Fourier series of $\ln(\sin x)=-\ln(2)-\sum_{k=1}^\infty\frac{\cos(2kx)}{k}$

$$J=-\ln(2)\underbrace{\int_0^{\pi/2}x^2\sin(2nx)dx}_{J_1}-\sum_{k=1}^\infty \frac1k\underbrace{\int_0^{\pi/2}x^2\sin(2nx)\cos(2kx)dx}_{J_2}$$

$$J_1=\frac{\cos(n\pi)}{4n^3}-\frac{1}{4n^3}-\frac{3\zeta(2)\cos(n\pi)}{4n}+\frac{\pi\sin(n\pi)}{4n^2}$$

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

The last result follows from $\cos(n\pi)=(-1)^n$ and $\sin(n\pi)=0$ for $n=1,2,3,..$

$$J_2=\frac18\left(\frac{1}{(k-n)^3}-\frac{\pi\sin(\pi(k-n))}{(k-n)^3}-\frac{(1-3(k-n)^2\zeta(2))\cos(\pi(k-n))}{(k-n)^3}\right)$$

$$-\frac18\left(\frac{1}{(k+n)^3}-\frac{\pi\sin(\pi(k+n))}{(k+n)^3}-\frac{(1-3(k+n)^2\zeta(2))\cos(\pi(k+n))}{(k+n)^3}\right)$$

I stopped here as I can not simplify $J_2$. But I think it can be simplified considering $n,k\in\mathbb{Z}>0$. Any idea? Also do you have a different path?

Thank you.

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Addendum

I also found that

$$\sum_{n=1}^\infty\frac{4^nH_n^{(3)}}{n^2{2n\choose n}}=\int_0^1\frac{\text{Li}_4(x)-\ln(1-x)\text{Li}_3(x)-\frac12\text{Li}_2^2(x)}{x\sqrt{1-x}}dx$$

$$=\int_0^{\pi/2}\frac{2\text{Li}_4(\sin^2x)-4\ln(\cos x)\text{Li}_3(\sin^2x)-\text{Li}_2^2(\sin^2x)}{\sin x}dx$$

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Proof of $(1)$:

Differentiating both sides of $\int_0^1 x^{n-1}\ln(1-x)dx=-\frac{H_n}{n}$ with respect to $n$ gives

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

multiply both sides by $\frac{4^n}{n^2{2n\choose n}}$ then $\sum_{n=1}^\infty$ we get

$$\sum_{n=1}^\infty\frac{4^nH_n}{n^4{2n\choose n}}+\sum_{n=1}^\infty\frac{4^nH_n^{(2)}}{n^3{2n\choose n}}-\sum_{n=1}^\infty\frac{\zeta(2)4^n}{n^3{2n\choose n}}=\int_0^1\frac{\ln x\ln(1-x)}{x}\left(\sum_{n=1}^\infty\frac{(4x)^n}{n^2{2n\choose n}}\right)dx$$

$$=\int_0^1\frac{\ln x\ln(1-x)}{x}\left(2\arcsin^2(\sqrt{x})\right)dx$$ $$\overset{\sqrt{x}=\sin\theta}{=}16\int_0^{\pi/2}x^2\cot x\ln(\sin x)\ln(\cos x)dx\tag1$$

Next, differentiate both sides of $\int_0^1 x^{n-1}\ln(1-x)dx=-\frac{H_n}{n}$ with respect to $n$ twice we get

$$-\frac12\int_0^1 x^{n-1}\ln^2x\ln(1-x)dx=\frac{H_n}{n^3}+\frac{H_n^{(2)}}{n^2}+\frac{H_n^{(3)}}{n}-\frac{\zeta(2)}{n^2}-\frac{\zeta(3)}{n}$$

multiply both sides by $\frac{4^n}{n{2n\choose n}}$ then $\sum_{n=1}^\infty$ we have

$$\sum_{n=1}^\infty\frac{4^nH_n}{n^4{2n\choose n}}+\sum_{n=1}^\infty\frac{4^nH_n^{(2)}}{n^3{2n\choose n}}+\sum_{n=1}^\infty\frac{4^nH_n^{(3)}}{n^2{2n\choose n}}-\sum_{n=1}^\infty\frac{\zeta(2)4^n}{n^3{2n\choose n}}-\sum_{n=1}^\infty\frac{\zeta(3)4^n}{n^2{2n\choose n}}$$ $$=-\frac12\int_0^1\frac{\ln x\ln(1-x)}{x}\left(\sum_{n=1}^\infty\frac{(4x)^n}{n{2n\choose n}}\right)dx$$ $$=-\frac12\int_0^1\frac{\ln x\ln(1-x)}{x}\left(\frac{2\sqrt{x}\arcsin(\sqrt{x})}{\sqrt{1-x}}\right)dx$$

$$\overset{\sqrt{x}=\sin\theta}{=}-16\int_0^{\pi/2}x\ln(\cos x)\ln^2(\sin x)dx$$

$$\overset{IBP}{=}16\int_0^{\pi/2}x^2\cot x\ln(\sin x)\ln(\cos x)dx-8\int_0^{\pi/2}x^2\tan x\ln^2(\sin x)dx\tag2$$

Subtracting $(2)$ from $(1)$ and using $\sum_{n=1}^\infty\frac{\zeta(3)4^n}{n^2{2n\choose n}}=3\zeta(2)\zeta(3)$ yields

$$\sum_{n=1}^\infty\frac{4^nH_n^{(3)}}{n^2{2n\choose n}}=3\zeta(2)\zeta(3)-8\int_0^{\pi/2}x^2\tan x\ln^2(\sin x)dx$$

Answer 1

A solution in large steps for $$\sum _{k=1}^{\infty }\frac{4^kH_k^{\left(3\right)}}{k^2\binom{2k}{k}}$$

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Note that $$\sum _{k=1}^{\infty }\frac{4^kH_k^{\left(3\right)}}{k^2\binom{2k}{k}}=\zeta \left(3\right)\sum _{k=1}^{\infty }\frac{4^k}{k^2\binom{2k}{k}}+\frac{1}{2}\sum _{k=1}^{\infty }\frac{4^k}{k^2\binom{2k}{k}}\psi ^{\left(2\right)}\left(k+1\right)$$ $$=2\zeta \left(3\right)\arcsin ^2\left(1\right)-4\int _0^1\left(\sum _{k=1}^{\infty }\frac{4^kx^{2k}}{k^2\binom{2k}{k}}\right)\frac{x\ln ^2\left(x\right)}{1-x^2}\:dx$$ $$=3\zeta \left(2\right)\zeta \left(3\right)-8\int _0^{\frac{\pi }{2}}x^2\tan \left(x\right)\ln ^2\left(\sin \left(x\right)\right)\:dx$$ $$=3\zeta \left(2\right)\zeta \left(3\right)-2\pi ^2\int _0^{\frac{\pi }{2}}\cot \left(x\right)\ln ^2\left(\cos \left(x\right)\right)\:dx+8\pi \underbrace{\int _0^{\frac{\pi }{2}}x\cot \left(x\right)\ln ^2\left(\cos \left(x\right)\right)\:dx}_{I}$$ $$-8\underbrace{\int _0^{\frac{\pi }{2}}x^2\cot \left(x\right)\ln ^2\left(\cos \left(x\right)\right)\:dx}_{J}$$ You can find $I$ here.

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A solution for $J$ also in large steps. $$\int _0^{\frac{\pi }{2}}x^2\cot \left(x\right)\ln ^2\left(\cos \left(x\right)\right)\:dx=\frac{1}{4}\int _0^{\infty }\frac{\arctan ^2\left(x\right)\ln ^2\left(1+x^2\right)}{x\left(1+x^2\right)}\:dx$$ $$=-\frac{1}{6}\operatorname{\mathfrak{R}} \left\{\int _0^{\infty }\frac{\ln ^4\left(\frac{i}{i+x}\right)}{x\left(1+x^2\right)}\:dx\right\}+\frac{1}{6}\int _0^{\infty }\frac{\arctan ^4\left(x\right)}{x\left(1+x^2\right)}\:dx+\frac{1}{96}\int _0^{\infty }\frac{\ln ^4\left(1+x^2\right)}{x\left(1+x^2\right)}\:dx$$ $$=-\frac{1}{6}\operatorname{\mathfrak{R}} \left\{24\zeta \left(5\right)-12\operatorname{Li}_5\left(2\right)-i\pi \ln ^4\left(2\right)\right\}-\frac{2}{3}\int _0^{\frac{\pi }{2}}x^3\ln \left(\sin \left(x\right)\right)\:dx+\frac{1}{192}\int _0^1\frac{\ln ^4\left(x\right)}{1-x}\:dx$$

Therefore $$\int _0^{\frac{\pi }{2}}x^2\cot \left(x\right)\ln ^2\left(\cos \left(x\right)\right)\:dx=-\frac{217}{64}\zeta \left(5\right)-\frac{9}{16}\zeta \left(2\right)\zeta \left(3\right)+2\operatorname{Li}_5\left(\frac{1}{2}\right)$$ $$+\frac{79}{16}\ln \left(2\right)\zeta \left(4\right)+\frac{2}{3}\ln ^3\left(2\right)\zeta \left(2\right)-\frac{1}{60}\ln ^5\left(2\right)$$

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And finally we can say that $$\sum _{k=1}^{\infty }\frac{4^kH_k^{\left(3\right)}}{k^2\binom{2k}{k}}=\frac{217}{8}\zeta \left(5\right)-\frac{9}{2}\zeta \left(2\right)\zeta \left(3\right)-16\operatorname{Li}_5\left(\frac{1}{2}\right)$$ $$-\frac{19}{2}\ln \left(2\right)\zeta \left(4\right)+\frac{8}{3}\ln ^3\left(2\right)\zeta \left(2\right)+\frac{2}{15}\ln ^5\left(2\right)$$