How to compute $\int_0^1\frac{\text{Li}_2(x^2)\arcsin^2(x)}{x}dx$ or $\sum_{n=1}^\infty\frac{4^nH_n}{n^4{2n\choose n}}$

Here is a solution for:

$$\sum _{k=1}^{\infty }\frac{4^kH_k}{k^4\binom{2k}{k}}=-\frac{31}{2}\zeta \left(5\right)+3\zeta \left(2\right)\zeta \left(3\right)+16\operatorname{Li}_5\left(\frac{1}{2}\right)$$ $$+2\ln \left(2\right)\zeta \left(4\right)+\frac{16}{3}\ln ^3\left(2\right)\zeta \left(2\right)-\frac{2}{15}\ln ^5\left(2\right),$$ $$\sum _{k=1}^{\infty }\frac{4^kH_k^{\left(2\right)}}{k^3\binom{2k}{k}}=\frac{31}{4}\zeta \left(5\right)+\frac{3}{2}\zeta \left(2\right)\zeta \left(3\right)-16\operatorname{Li}_5\left(\frac{1}{2}\right)$$ $$-2\ln \left(2\right)\zeta \left(4\right)+\frac{8}{3}\ln ^3\left(2\right)\zeta \left(2\right)+\frac{2}{15}\ln ^5\left(2\right).$$

Note that: $$\arcsin ^4\left(x\right)=\frac{3}{2}\sum _{k=1}^{\infty }\frac{4^kH_k^{\left(2\right)}}{k^2\binom{2k}{k}}x^{2k}-\frac{3}{2}\sum _{k=1}^{\infty }\frac{4^k}{k^4\binom{2k}{k}}x^{2k}$$ $$2\int _0^1\frac{\arcsin ^4\left(x\right)}{x}\:dx=\frac{3}{2}\sum _{k=1}^{\infty }\frac{4^kH_k^{\left(2\right)}}{k^3\binom{2k}{k}}-\frac{3}{2}\sum _{k=1}^{\infty }\frac{4^k}{k^5\binom{2k}{k}}$$ $$\sum _{k=1}^{\infty }\frac{4^kH_k^{\left(2\right)}}{k^3\binom{2k}{k}}=-\frac{16}{3}\int _0^{\frac{\pi }{2}}x^3\ln \left(\sin \left(x\right)\right)\:dx-\frac{16}{3}\int _0^{\frac{\pi }{2}}x\ln ^3\left(\sin \left(x\right)\right)\:dx,$$ where the first integral is easy to calculate due to the fourier series expansion of $\ln \left(\sin \left(x\right)\right)$ and the other integral can be found evaluated here, using their closed-forms the previous announced result follows.

Now let's find the other series, consider the result also found in the previous link: $$\int _0^{\frac{\pi }{2}}x\ln ^2\left(\sin \left(x\right)\right)\ln \left(\cos \left(x\right)\right)\:dx=-\frac{155}{128}\zeta \left(5\right)-\frac{1}{32}\zeta \left(2\right)\zeta \left(3\right)+\operatorname{Li}_5\left(\frac{1}{2}\right)$$ $$+\frac{49}{32}\ln \left(2\right)\zeta \left(4\right)-\frac{2}{3}\ln ^3\left(2\right)\zeta \left(2\right)-\frac{1}{120}\ln ^5\left(2\right),$$ and note that: $$\int _0^{\frac{\pi }{2}}x\ln ^2\left(\sin \left(x\right)\right)\ln \left(\cos \left(x\right)\right)\:dx=\frac{1}{2}\int _0^1\frac{\ln ^2\left(x\right)\ln \left(1-x^2\right)\arcsin \left(x\right)}{\sqrt{1-x^2}}\:dx$$ $$=\frac{1}{32}\sum _{k=1}^{\infty }\frac{4^k}{k\binom{2k}{k}}\int _0^1x^{k-1}\ln ^2\left(x\right)\ln \left(1-x\right)\:dx$$ $$\sum _{k=1}^{\infty }\frac{4^kH_k}{k^4\binom{2k}{k}}=-16\int _0^{\frac{\pi }{2}}x\ln ^2\left(\sin \left(x\right)\right)\ln \left(\cos \left(x\right)\right)\:dx-\sum _{k=1}^{\infty }\frac{4^kH_k^{\left(2\right)}}{k^3\binom{2k}{k}}-\sum _{k=1}^{\infty }\frac{4^kH_k^{\left(3\right)}}{k^2\binom{2k}{k}}$$ $$+\zeta \left(2\right)\sum _{k=1}^{\infty }\frac{4^k}{k^3\binom{2k}{k}}+\zeta \left(3\right)\sum _{k=1}^{\infty }\frac{4^k}{k^2\binom{2k}{k}}.$$ The last $2$ series are well-known while the other harmonic series can be found evaluated in large steps here: $$\sum _{k=1}^{\infty }\frac{4^kH_k^{\left(3\right)}}{k^2\binom{2k}{k}}=\frac{217}{8}\zeta \left(5\right)-\frac{9}{2}\zeta \left(2\right)\zeta \left(3\right)-16\operatorname{Li}_5\left(\frac{1}{2}\right)$$ $$-\frac{19}{2}\ln \left(2\right)\zeta \left(4\right)+\frac{8}{3}\ln ^3\left(2\right)\zeta \left(2\right)+\frac{2}{15}\ln ^5\left(2\right),$$ Using these results the closed-form mentioned in the beginning also follows.

And that's all there is to it.

I wasn't able to find a closed-form for this, but I was able to simplify it to

$$\frac{\pi^2}{48} \left( 2\pi^2 \ln(2) - 7\zeta(3) \right) - \sum_{n=1}^{\infty} \frac{2^{2n-2} H_n}{n^4 \binom{2n}{n}}$$

Evaluate $$I = \int_0^1\frac{\text{Li}_2(x^2)\arcsin^2(x)}{x}dx$$

Expanding $\arcsin^2(x)$ using the power series yields: $$\int_0^1 \text{Li}_2(x^2) \sum_{n=1}^{\infty} \frac{2^{2n-1}}{n^2 \binom{2n}{n}} x^{2n-1} dx$$

Swapping integration and sum:

$$\sum_{n=1}^{\infty} \frac{2^{2n-1}}{n^2 \binom{2n}{n}}\int_0^1 \text{Li}_2(x^2) x^{2n-1} dx$$

Making the substitution $u = x^2$:

$$\sum_{n=1}^{\infty} \frac{2^{2n-2}}{n^2 \binom{2n}{n}}\int_0^1 \text{Li}_2(u) u^{n-1}du$$

The inner integral would be $$\int_0^1 \sum_{k=1}^{\infty} \frac{u^k}{k^2} u^{n-1} du = \sum_{k=1}^{\infty} \frac{1}{k^2} \frac{1}{k+n} = \frac{\pi^2}{6n} - \frac{H_n}{n^2}$$

Which makes the overall integral into $$\sum_{n=1}^{\infty} \frac{2^{2n-2}}{n^2 \binom{2n}{n}}\left(\frac{\pi^2}{6n} - \frac{H_n}{n^2}\right)$$

Or splitting the sums up: $$\frac{\pi^2}{24}\sum_{n=1}^{\infty} \frac{2^{2n}}{n^3 \binom{2n}{n}} - \sum_{n=1}^{\infty} \frac{2^{2n-2} H_n}{n^4 \binom{2n}{n}}$$

Let $f(x) = \sum_{n=1}^{\infty} \frac{x^{2n}}{n^3 \binom{2n}{n}}$. Then $f'(x) = 2\sum_{n=1}^{\infty} \frac{x^{2n-1}}{n^2 \binom{2n}{n}} = \frac{4\arcsin^2\left( \frac{x}{2} \right)}{x}$

Then the integral to solve for the first sum is $$\int_{0}^{2}\frac{4\arcsin^{2}\left(\frac{x}{2}\right)}{x}dx = 4\int_{0}^{1}\frac{\arcsin^{2}\left(x\right)}{x}dx$$

Making the substitution $x \to \arcsin(x)$ yields $$4\int_0^{\pi/2} x^2 \cot(x) dx$$

This can be done by complex methods (substituting $u = e^{2ix}-1$ and then doing partial fractions) to get the indefinite integral in closed form. Then the integral would be $$\pi^2 \ln(2) - \frac{7}{2}\zeta(3)$$

This then makes the original integral to $$\frac{\pi^2}{48} \left( 2\pi^2 \ln(2) - 7\zeta(3) \right) - \sum_{n=1}^{\infty} \frac{2^{2n-2} H_n}{n^4 \binom{2n}{n}}$$

I will start from your second attempt: $$I=\sum_{n=1}^\infty\frac{1}{n^2}\underbrace{\int_0^{\pi/2}x^2\cot x \sin^{2n}(x) dx}_{I_n}$$

Using integration by parts, $I_n$ is equal to $$I_n = x^2 \frac{\sin^{2n}(x)}{2n} \Big|^{\pi/2}_0 - \int_0^{\pi/2} x \frac{\sin^{2n}(x)}{n} dx$$

Which simplifies to $$\frac{\pi^2}{8n} - \frac{1}{n} \int_0^{\pi/2} x\sin^{2n}(x) dx$$

Splitting the $\sin^{2n}(x)$ as $\sin^{2n-1}(x)\sin(x)$ so that I can integrate by parts:

$$J_n = \int_0^{\pi/2} x\sin^{2n}(x) dx = \int_0^{\pi/2} \sin^{2n-1}(x) x \sin(x)dx$$

Integrating by parts:

$$1-\int_{0}^{\frac{\pi}{2}}\left(-x\cos\left(x\right)+\sin\left(x\right)\right)\left(2n-1\right)\cos\left(x\right)\sin\left(x\right)^{\left(2n-2\right)}dx$$

Separating and evaluating gives the relation $$J_n = \frac{1}{2n} - (2n-1) J_n + (2n-1)J_{n-1}$$

which has the solution $$J_n = \frac{1}{4n^2} + \frac{2n-1}{2n} J_{n-1}$$ with $J_0 = \frac{\pi^2}{8}$

The explicit solution to this is $$\frac{\binom{2n}{n}}{4^n}\left(\frac{\pi^2}{8} + \sum_{m=1}^{n} \frac{4^{m-1}}{\binom{2m}{m} m^2}\right)$$

Which then makes $I_n$ $$\frac{\pi^2}{8n} - \frac{1}{n} \frac{\binom{2n}{n}}{4^n}\left(\frac{\pi^2}{8} + \sum_{m=1}^{n} \frac{4^{m-1}}{\binom{2m}{m} m^2} \right)$$

The original integral/sum is then $$\sum_{n=1}^{\infty} \frac{1}{n^2} \left( \frac{\pi^2}{8n} - \frac{1}{n} \frac{\binom{2n}{n}}{4^n}\left(\frac{\pi^2}{8} + \sum_{m=1}^{n} \frac{4^{m-1}}{\binom{2m}{m} m^2} \right) \right)$$

This can be simplified to $$\frac{\pi^2}{8} \zeta(3) - \frac{\pi^2}{8}\underbrace{\sum_{n=1}^{\infty} \frac{\binom{2n}{n}}{4^n n^3}}_{S_1} - \underbrace{\sum_{n=1}^{\infty}\frac{\binom{2n}{n}}{4^n n^3} \sum_{m=1}^{n} \frac{4^{m-1}}{\binom{2m}{m} m^2}}_{S_2} \tag 1$$

Focusing on $S_2$, $\sum_{n=1}^{\infty}\frac{\binom{2n}{n}}{4^n n^3} \sum_{m=1}^{n} \frac{4^{m-1}}{\binom{2m}{m} m^2}$: This can be rewritten as $$\sum_{m=1}^{\infty} \frac{4^{m-1}}{\binom{2m}{m} m^2}\left(\sum_{n=1}^{\infty}\frac{\binom{2n}{n}}{4^n n^3} - \sum_{n=1}^{m-1} \frac{\binom{2n}{n}}{4^n n^3} \right) = S_1\underbrace{\sum_{m=1}^{\infty} \frac{4^{m-1}}{\binom{2m}{m} m^2}}_{S_3} - \sum_{m=1}^{\infty} \frac{4^{m-1}}{\binom{2m}{m} m^2}\sum_{n=1}^{m-1} \frac{\binom{2n}{n}}{4^n n^3} $$

$S_3$ can be simplified using the series expansion of $\arcsin^2(x)$ to get $S_3 = \frac{\pi^2}{8}$

This then simplifies the overall integral/sums to $$\frac{\pi^2}{8} \zeta(3) - \frac{\pi^2}{4}\sum_{n=1}^{\infty} \frac{\binom{2n}{n}}{4^n n^3} + \sum_{m=1}^{\infty} \frac{4^{m-1}}{\binom{2m}{m} m^2}\sum_{n=1}^{m-1} \frac{\binom{2n}{n}}{4^n n^3} \tag 2$$

Using Mathematica, I found $S_1 = \frac{-\pi^2 \ln(4) + \ln^3(4) + 12\zeta(3)}{6}$, but don't have a proof for this. I feel like there might be a proof of this somewhere on MSE, but unfortunately Approach0 is down right now (so I can't search as effectively).