How to tackle
$$I=\int_0^1\frac{\text{Li}_2(x^2)\arcsin^2(x)}{x}dx\ ?$$
This integral came up while I was working on finding $\sum_{n=1}^\infty\frac{4^nH_n}{n^4{2n\choose n}}$.
First attempt: By writing $\text{Li}_2(x^2)=-\int_0^1\frac{x^2\ln(y)}{1-x^2y}dy$ we have
$$I=-\int_0^1\ln(y)\left(\int_0^1\frac{x\arcsin^2(x)}{1-x^2y}dx\right)dy$$
and Mathematica gave a complicated expression for the inner integral and that made me stop.
Second attempt: $x=\sin\theta$
$$I=\int_0^{\pi/2}\theta^2\cot\theta\ \text{Li}_2(\sin^2\theta)d\theta$$
$$=\sum_{n=1}^\infty\frac{1}{n^2}\int_0^{\pi/2}\theta^2\cot\theta \sin^{2n}(\theta) d\theta$$
and I have no idea how to continue. Any suggestion?
Thanks
How $I$ appeared in my calculations:
$$\frac{\arcsin x}{\sqrt{1-x^2}}=\sum_{n=1}^\infty\frac{(2x)^{2n-1}}{n{2n\choose n}}$$
we can write
$$\frac{2\sqrt{x}\arcsin \sqrt{x}}{\sqrt{1-x}}=\sum_{n=1}^\infty\frac{2^{2n}x^{n}}{n{2n\choose n}}$$
Divide both sides by $x$ then $\int_0^y$ we have
$$\sum_{n=1}^\infty\frac{2^{2n}y^n}{n^2{2n\choose n}}=2\int_0^y \frac{\arcsin \sqrt{x}}{\sqrt{x}\sqrt{1-x}}dx$$
Next multiply both sides by $\frac{\text{Li}_2(y)}{y}$ then $\sum_{n=1}^\infty$ and use that $\int_0^1 y^{n-1}\text{Li}_2(y)dy=\frac{\zeta(2)}{n^2}-\frac{H_n}{n^2}$ we get
$$\sum_{n=1}^\infty\frac{\zeta(2)2^{2n}}{n^3{2n\choose n}}-\sum_{n=1}^\infty\frac{2^{2n}H_n}{n^4{2n\choose n}}=2\int_0^1\int_0^y \frac{\arcsin \sqrt{x}\text{Li}_2(y)}{y\sqrt{x}\sqrt{1-x}}dxdy$$
$$=2\int_0^1 \frac{\arcsin \sqrt{x}}{\sqrt{x}\sqrt{1-x}}\left(\int_x^1\frac{\text{Li}_2(y)}{y}dy\right)dx$$
$$=2\int_0^1 \frac{\arcsin \sqrt{x}}{\sqrt{x}\sqrt{1-x}}\left(\zeta(3)-\text{Li}_3(x)\right)dx$$
$$\overset{\sqrt{x}\to x}{=}4\int_0^1\frac{\arcsin x}{\sqrt{1-x^2}}(\zeta(3)-\text{Li}_3(x^2))dx$$
$$\overset{\text{IBP}}{=}4\int_0^1\frac{\text{Li}_2(x^2)\arcsin^2(x)}{x}dx$$
Substitute $\sum_{n=1}^\infty\frac{\zeta(2)2^{2n}}{n^3{2n\choose n}}=15\ln(2)\zeta(4)-\frac72\zeta(2)\zeta(3)$ we get
$$\sum_{n=1}^\infty\frac{2^{2n}H_n}{n^4{2n\choose n}}=15\ln(2)\zeta(4)-\frac72\zeta(2)\zeta(3)-4\int_0^1\frac{\text{Li}_2(x^2)\arcsin^2(x)}{x}dx$$
Best Answer
Here is a solution for:
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.