Following the same technique in this question , one can find
$$\sum_{n=1}^\infty\frac{4^nH_{2n}}{n^3{2n\choose n}}=-4\int_0^1\frac{\arcsin^2(x)\ln(1-x)}{x}dx=-4\int_0^{\pi/2}x^2\cot x\ln(1-\sin x)dx$$
First attempt:
$$\int_0^1\frac{\arcsin^2(x)\ln(1-x^2)}{x}dx=\int_0^1\frac{\arcsin^2(x)\ln(1-x)}{x}dx+\int_0^1\frac{\arcsin^2(x)\ln(1+x)}{x}dx$$
The LHS integral is already calculated in the link above but I have no idea how to tackle the RHS integral which seems as hard as the main integral.
Second attempt: Using $1-\sin x=2\sin^2\left(\frac{x}{2}-\frac{\pi}{4}\right)$ then setting $\frac{x}{2}-\frac{\pi}{4}=y$ will complicate the problem.
Third attempt: Expanding $\ln(1-x)$ in series
$$\int_0^1\frac{\arcsin^2(x)\ln(1-x)}{x}dx=-\sum_{n=1}^\infty\frac1n\int_0^1 \arcsin^2(x)x^{n-1}dx$$
$$=-\sum_{n=1}^\infty\frac1n \int_0^{\pi/2} x^2\cot x\sin^{n-1}(x)dx$$
which is hard to crack as well. Any thought?
Thanks in advance
Note: I am not sure if this integral has a closed form and I am just giving it a try.
The following problem is just for fun:
Prove without breaking the summand that
$$\sum_{n=1}^\infty\frac{4^nH_{2n}-4^nH_n}{n^3{2n\choose n}}=4\int_0^1\frac{\arcsin^2(x)\ln(1+x)}{x}dx$$
Best Answer
I am going to evaluate the main sum $\displaystyle\sum_{n=1}^\infty\frac{4^nH_{2n}}{n^3{2n\choose n}}$ from which the integral in the question appeared:
From the Beta function presented in the book, (Almost) Impossible Integrals, Sums, and Series, $\displaystyle \int_0^1 \frac{x^{a-1}+x^{b-1}}{(1+x)^{a+b}} dx = \operatorname{B}(a,b)$, (see pages $72$-$73$).
where if we set $a=b=n$ we have
$$\int_0^1\frac{2x^{n-1}}{(1+x)^{2n}}dx=\frac{\Gamma^2(n)}{\Gamma(2n)}=\frac{2}{n{2n\choose n}}$$
Or $$\frac{1}{n{2n\choose n}}=\int_0^1\frac{x^{n-1}}{(1+x)^{2n}}dx=\int_0^1\frac1x\left(\frac{x}{(1+x)^2}\right)^ndx$$
Differentiate both sides with respect to $n$ we get
$$\frac{H_n}{n{2n\choose n}}-\frac{H_{2n}}{n{2n\choose n}}-\frac{1}{2n^2{2n\choose n}}=\frac12\int_0^1\frac{1}x\ln\left(\frac{x}{(1+x)^2}\right)\left(\frac{x}{(1+x)^2}\right)^ndx$$
Next, multiply both sides by $\frac{4^n}{n^2}$ then $\sum_{n=1}^\infty$ we get
$$\sum_{n=1}^\infty\frac{4^nH_{n}}{n^3{2n\choose n}}-\sum_{n=1}^\infty\frac{4^nH_{2n}}{n^3{2n\choose n}}-\frac12\sum_{n=1}^\infty\frac{4^n}{n^4{2n\choose n}}$$ $$=\frac12\int_0^1\frac{1}x\ln\left(\frac{x}{(1+x)^2}\right)\left[\sum_{n=1}^\infty\frac{\left(\frac{4x}{(1+x)^2}\right)^n}{n^2}\right]dx$$
$$=\frac12\int_0^1\frac{1}x\ln\left(\frac{x}{(1+x)^2}\right)\left[\text{Li}_2\left(\frac{4x}{(1+x)^2}\right)\right]dx$$
$$\overset{IBP}{=}-\frac54\zeta(4)-\frac12\int_0^1\left(\frac12\ln^2x+2\text{Li}_2(-x)\right)\left[\frac{2(x-1)}{x(1+x)}\ln\left(\frac{1-x}{1+x}\right)\right]dx$$
$$=-\frac54\zeta(4)+\frac12\underbrace{\int_0^1\frac{\ln^2x\ln(1-x)}{x}dx}_{\mathcal{\Large{I}_1}}-\frac12\underbrace{\int_0^1\frac{\ln^2x\ln(1+x)}{x}dx}_{\mathcal{\Large{I}_2}}$$ $$+2\underbrace{\int_0^1\frac{\ln(1-x)\text{Li}_2(-x)}{x}dx}_{\mathcal{\Large{I}_3}}-2\underbrace{\int_0^1\frac{\ln(1+x)\text{Li}_2(-x)}{x}dx}_{\mathcal{\Large{I}_4}}$$
$$-\underbrace{\int_0^1\frac{\ln^2x\ln(1-x)}{1+x}dx}_{\mathcal{\Large{I}_5}}+\underbrace{\int_0^1\frac{\ln^2x\ln(1+x)}{1+x}dx}_{\mathcal{\Large{I}_6}}$$ $$-4\underbrace{\int_0^1\frac{\ln(1-x)\text{Li}_2(-x)}{1+x}dx}_{\mathcal{\Large{I}_7}}+4\underbrace{\int_0^1\frac{\ln(1+x)\text{Li}_2(-x)}{1+x}dx}_{\mathcal{\Large{I}_8}}$$
$$\mathcal{I}_1=\int_0^1\frac{\ln^2x\ln(1-x)}{x}dx=-\sum_{n=1}^\infty\frac1{n}\int_0^1 x^{n-1}\ln^2xdx=-2\sum_{n=1}^\infty\frac{1}{n^4}=\boxed{-2\zeta(4)}$$
$$\mathcal{I}_2=\int_0^1\frac{\ln^2x\ln(1+x)}{x}dx=-\sum_{n=1}^\infty\frac{-1)^n}{n}\int_0^1 x^{n-1}\ln^2xdx=-2\sum_{n=1}^\infty\frac{(-1)^n}{n^4}=\boxed{\frac74\zeta(4)}$$
$$\mathcal{I}_3=\int_0^1\frac{\ln(1-x)\text{Li}_2(-x)}{x}dx=\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\int_0^1 x^{n-1}\ln(1-x)dx=\boxed{-\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}}$$
$$\mathcal{I}_4=\int_0^1\frac{\ln(1+x)\text{Li}_2(-x)}{x}dx=-\frac12\text{Li}_2^2(-1)=\boxed{-\frac{5}{16}\zeta(4)}$$
$\mathcal{I}_5$ is calculated here (see the integral $Q$):
$$\mathcal{I}_5=\int_0^1\frac{\ln^2x\ln(1-x)}{1+x}dx=\boxed{\zeta(4)+\ln^22\zeta(2)-\frac16\ln^42-4\operatorname{Li}_4\left(\frac12\right)}$$
$$\mathcal{I}_6=\int_0^1\frac{\ln^2x\ln(1+x)}{1+x}dx=\sum_{n=1}^\infty (-1)^nH_{n-1}\int_0^1 x^{n-1}\ln^2xdx=2\sum_{n=1}^\infty \frac{(-1)^nH_{n-1}}{n^3}$$ $$=\boxed{2\sum_{n=1}^\infty \frac{(-1)^nH_n}{n^3}+\frac74\zeta(4)}$$
$$\mathcal{I}_7=\int_0^1\frac{\ln(1-x)\text{Li}_2(-x)}{1+x}dx=-\sum_{n=1}^\infty (-1)^n H_{n-1}^{(2)}\int_0^1 x^{n-1}\ln(1-x)dx$$ $$=\sum_{n=1}^\infty\frac{(-1)^nH_{n-1}^{(2)}H_n}{n}=\boxed{\sum_{n=1}^\infty\frac{(-1)^nH_n^{(2)}H_n}{n}-\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}}$$
$$\mathcal{I}_8=\int_0^1\frac{\ln(1+x)\text{Li}_2(-x)}{1+x}dx\overset{IBP}{=}-\frac14\ln^2(2)\zeta(2)+\frac12\int_0^1\frac{\ln^3(1+x)}{x}dx$$
$$=\boxed{3\zeta(4)-\frac{21}{8}\ln(2)\zeta(3)+\frac12\ln^2(2)\zeta(2)-\frac18\ln^4(2)-3\text{Li}_4\left(\frac12\right)}$$
where the last result follows from the generalization here.
Collect all integrals we get
$$\sum_{n=1}^\infty\frac{4^nH_{2n}}{n^3{2n\choose n}}=8\text{Li}_4\left(\frac12\right)-\frac{41}{4}\zeta(4)+\frac{21}{2}\ln(2)\zeta(3)-\ln^2(2)\zeta(2)+\frac13\ln^4(2)$$
$$+4\sum_{n=1}^\infty\frac{(-1)^nH_n^{(2)}H_n}{n}-4\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}+\sum_{n=1}^\infty\frac{4^nH_{n}}{n^3{2n\choose n}}-\frac12\sum_{n=1}^\infty\frac{4^n}{n^4{2n\choose n}}$$
For the last sum, use the identity
$$\sum_{n=1}^\infty\frac{4^nx^{2n}}{n^2{2n\choose n}}=\arcsin^2(x)$$
Multiply both sides by $-\frac{\ln x}{4x}$ then $\int_0^1$ we get
$$\sum_{n=1}^\infty\frac{4^n}{n^4{2n\choose n}}=-\frac14\int_0^1\frac{\arcsin^2(x)\ln x}{x}dx$$
$$\overset{IBP}{=}4\int_0^1\frac{\arcsin(x)\ln^2x}{\sqrt{1-x^2}}dx=4\int_0^{\pi/2}x\ln^2(\sin x)dx$$
$$=4\operatorname{Li}_4\left(\frac{1}{2}\right)-\frac{19}{8}\zeta(4)+2\ln^2(2)\zeta(2)+\frac{1}{6}\ln^4(2)$$
Where the last integral is calculated here.
Substitute this result along with the following results:
$$\sum_{n=1}^\infty\frac{(-1)^nH_nH_n^{(2)}}{n}=-2\text{Li}_4\left(\frac12\right)+\zeta(4)-\frac{7}{8}\ln(2)\zeta(3)+\frac14\ln^2(2)\zeta(2)-\frac1{12}\ln^4(2)$$
$$\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}=2\operatorname{Li_4}\left(\frac12\right)-\frac{11}4\zeta(4)+\frac74\ln(2)\zeta(3)-\frac12\ln^2(2)\zeta(2)+\frac{1}{12}\ln^4(2)$$
$$\sum_{n=1}^\infty\frac{4^nH_n}{n^3{2n\choose n}}=-8\text{Li}_4\left(\frac12\right)+\zeta(4)+8\ln^2(2)\zeta(2)-\frac{1}{3}\ln^4(2)$$
we finally obtain
$$\sum_{n=1}^\infty\frac{4^nH_{2n}}{n^3{2n\choose n}}=-20\text{Li}_4\left(\frac12\right)+\frac{65}{8}\zeta(4)+8\ln^2(2)\zeta(2)-\frac56\ln^4(2)$$