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

Question

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

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How $I$ appeared in my calculations:

Since

$$\frac{\arcsin x}{\sqrt{1-x^2}}=\sum_{n=1}^\infty\frac{(2x)^{2n-1}}{n{2n\choose n}}$$

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 $\int_0^1 y^{n-1}\text{Li}_2(y)dy=\frac{\zeta(2)}{n}-\frac{H_n}{n^2}$ 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

$$\sum_{n=1}^\infty\frac{2^{2n}H_n}{n^4{2n\choose n}}=15\ln(2)\zeta(4)-\frac72\zeta(2)\zeta(3)-4\int_0^1\frac{\text{Li}_2(x^2)\arcsin^2(x)}{x}dx$$

Answer 1

A different path.

Since $$\int _0^1\frac{\arcsin ^2\left(x\right)\operatorname{Li}_2\left(x^2\right)}{x}\:dx=\int _0^{\frac{\pi }{2}}x^2\cot \left(x\right)\operatorname{Li}_2\left(\sin ^2\left(x\right)\right)\:dx,$$ we may use that $$\operatorname{Li}_2\left(\sin ^2\left(x\right)\right)$$ $$=\zeta \left(2\right)-2\ln ^2\left(2\right)+4\sum _{k=1}^{\infty }\left(-1\right)^k\left(\int _0^1t^{k-1}\ln \left(\frac{1+t}{2\sqrt{t}}\right)\:dt\right)\cos \left(2kx\right),\quad x\in \mathbb{R},$$ which is derived in More (Almost) Impossible Integrals, Sums, and Series: A New Collection of Fiendish Problems and Surprising Solutions (2023), pages $742$–$748$. This means that if we employ this to our integral, we obtain $$\int _0^{\frac{\pi }{2}}x^2\cot \left(x\right)\operatorname{Li}_2\left(\sin ^2\left(x\right)\right)\:dx=\left(\zeta \left(2\right)-2\ln ^2\left(2\right)\right)\underbrace{\int _0^{\frac{\pi }{2}}x^2\cot \left(x\right)\:dx}_{-\frac{7}{8}\zeta \left(3\right)+\frac{3}{2}\ln \left(2\right)\zeta \left(2\right)}$$ $$+4\sum _{k=1}^{\infty }\left(-1\right)^k\left(\int _0^1t^{k-1}\ln \left(\frac{1+t}{2\sqrt{t}}\right)\:dt\right)\underbrace{\int _0^{\frac{\pi }{2}}x^2\cot \left(x\right)\cos \left(2kx\right)\:dx}_{I_k}\tag{1}.$$ Moreover, by considering the difference $I_{n+1}-I_n$, we can prove in a straightforward but laborious manner that $$\int _0^{\frac{\pi }{2}}x^2\cot \left(x\right)\cos \left(2kx\right)\:dx=\frac{1}{2}H_k^{\left(3\right)}+\frac{1}{2}\overline{H}_k^{\left(3\right)}-\frac{3}{2}\zeta \left(2\right)\overline{H}_k-\frac{1}{4}\frac{1}{k^3}-\frac{1}{4}\frac{\left(-1\right)^{k+1}}{k^3}$$ $$+\frac{3}{4}\zeta \left(2\right)\frac{\left(-1\right)^{k+1}}{k}-\frac{7}{8}\zeta \left(3\right)+\frac{3}{2}\ln \left(2\right)\zeta \left(2\right),\quad k\in \mathbb{N}.\tag{2}$$ Thus, if we replace $\left(2\right)$ in $\left(1\right)$ and reduce the resulting expressions, we arrive at $$\int _0^1\frac{\arcsin ^2\left(x\right)\operatorname{Li}_2\left(x^2\right)}{x}\:dx=\int _0^{\frac{\pi }{2}}x^2\cot \left(x\right)\operatorname{Li}_2\left(\sin ^2\left(x\right)\right)\:dx$$ $$=\int _0^1\frac{\ln \left(\frac{1+t}{2\sqrt{t}}\right)\operatorname{Li}_3\left(-t\right)}{t}\:dt-\int _0^1\frac{\ln \left(\frac{1+t}{2\sqrt{t}}\right)\operatorname{Li}_3\left(t\right)}{t}\:dt-2\int _0^1\frac{\ln \left(\frac{1+t}{2\sqrt{t}}\right)\operatorname{Li}_3\left(-t\right)}{1+t}\:dt$$ $$+2\int _0^1\frac{\ln \left(\frac{1+t}{2\sqrt{t}}\right)\operatorname{Li}_3\left(t\right)}{1+t}\:dt-3\zeta \left(2\right)\int _0^1\frac{\ln \left(1-t\right)\ln \left(\frac{1+t}{2\sqrt{t}}\right)}{t}\:dt$$ $$+6\zeta \left(2\right)\int _0^1\frac{\ln \left(1-t\right)\ln \left(\frac{1+t}{2\sqrt{t}}\right)}{1+t}\:dt.$$ Now, we can observe that many known integrals arose, which may allow us to complete the calculations.

As an additional note, it is better to keep the integral inside the Fourier series from the beginning as this allows us for better manipulations.

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A second straightforward path.

We have $$\int _0^{\frac{\pi }{2}}x^2\cot \left(x\right)\operatorname{Li}_2\left(\sin ^2\left(x\right)\right)\:dx=\frac{3}{4}\zeta \left(2\right)\zeta \left(3\right)-\int _0^{\frac{\pi }{2}}x\operatorname{Li}_3\left(\sin ^2\left(x\right)\right)\:dx,$$ and if we have at our disposal the more advanced Fourier series $$\operatorname{Li}_3\left(\sin ^2\left(x\right)\right)$$ $$\small=2\zeta \left(3\right)-2\ln \left(2\right)\zeta \left(2\right)+\frac{4}{3}\ln ^3\left(2\right)+4\sum _{k=1}^{\infty }\left(-1\right)^k\left(\int _0^1t^{k-1}\ln ^2\left(\frac{1+t}{2\sqrt{t}}\right)\:dt\right)\cos \left(2kx\right),\quad x\in \mathbb{R},$$ derived in the same book, pages $748$–$750$, we would obtain $$\int _0^{\frac{\pi }{2}}x^2\cot \left(x\right)\operatorname{Li}_2\left(\sin ^2\left(x\right)\right)\:dx$$ $$=\frac{3}{4}\zeta \left(2\right)\zeta \left(3\right)-\left(2\zeta \left(3\right)-2\ln \left(2\right)\zeta \left(2\right)+\frac{4}{3}\ln ^3\left(2\right)\right)\int _0^{\frac{\pi }{2}}x\:dx$$ $$-4\sum _{k=1}^{\infty }\left(-1\right)^k\left(\int _0^1t^{k-1}\ln ^2\left(\frac{1+t}{2\sqrt{t}}\right)\:dt\right)\int _0^{\frac{\pi }{2}}x\cos \left(2kx\right)\:dx$$ $$=-\frac{3}{4}\zeta \left(2\right)\zeta \left(3\right)+\frac{15}{4}\ln \left(2\right)\zeta \left(4\right)-\ln ^3\left(2\right)\zeta \left(2\right)-\int _0^1\frac{\ln ^2\left(\frac{1+t}{2\sqrt{t}}\right)\left(\operatorname{Li}_2\left(t\right)-\operatorname{Li}_2\left(-t\right)\right)}{t}\:dt.$$ Here, we encounter known integrals once again.

Answer 2

Update: The approach below can be found as part of this paper.

Here is an approach 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 $$\arcsin ^4\left(x\right)=\frac{3}{2}\sum _{k=1}^{\infty }\frac{4^kH_k^{\left(2\right)}}{k^2\binom{2k}{k}}x^{2k}-\frac{3}{2}\sum _{k=1}^{\infty }\frac{4^k}{k^4\binom{2k}{k}}x^{2k}$$ $$2\int _0^1\frac{\arcsin ^4\left(x\right)}{x}\:dx=\frac{3}{2}\sum _{k=1}^{\infty }\frac{4^kH_k^{\left(2\right)}}{k^3\binom{2k}{k}}-\frac{3}{2}\sum _{k=1}^{\infty }\frac{4^k}{k^5\binom{2k}{k}}.$$ Therefore $$\sum _{k=1}^{\infty }\frac{4^kH_k^{\left(2\right)}}{k^3\binom{2k}{k}}=-\frac{16}{3}\int _0^{\frac{\pi }{2}}x^3\ln \left(\sin \left(x\right)\right)\:dx-\frac{16}{3}\int _0^{\frac{\pi }{2}}x\ln ^3\left(\sin \left(x\right)\right)\:dx.$$ The first integral is easy to calculate due to the fourier series of $\ln \left(\sin \left(x\right)\right)$, while the other integral can be found calculated here. By 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 $$\int _0^{\frac{\pi }{2}}x\ln ^2\left(\sin \left(x\right)\right)\ln \left(\cos \left(x\right)\right)\:dx=\frac{1}{2}\int _0^1\frac{\ln ^2\left(x\right)\ln \left(1-x^2\right)\arcsin \left(x\right)}{\sqrt{1-x^2}}\:dx$$ $$=\frac{1}{32}\sum _{k=1}^{\infty }\frac{4^k}{k\binom{2k}{k}}\int _0^1x^{k-1}\ln ^2\left(x\right)\ln \left(1-x\right)\:dx,$$ which means that $$\sum _{k=1}^{\infty }\frac{4^kH_k}{k^4\binom{2k}{k}}=-16\int _0^{\frac{\pi }{2}}x\ln ^2\left(\sin \left(x\right)\right)\ln \left(\cos \left(x\right)\right)\:dx-\sum _{k=1}^{\infty }\frac{4^kH_k^{\left(2\right)}}{k^3\binom{2k}{k}}-\sum _{k=1}^{\infty }\frac{4^kH_k^{\left(3\right)}}{k^2\binom{2k}{k}}$$ $$+\zeta \left(2\right)\sum _{k=1}^{\infty }\frac{4^k}{k^3\binom{2k}{k}}+\zeta \left(3\right)\sum _{k=1}^{\infty }\frac{4^k}{k^2\binom{2k}{k}}.$$ The last $2$ series are well-known while the other harmonic series can be found evaluated in large steps here: $$\sum _{k=1}^{\infty }\frac{4^kH_k^{\left(3\right)}}{k^2\binom{2k}{k}}=\frac{217}{8}\zeta \left(5\right)-\frac{9}{2}\zeta \left(2\right)\zeta \left(3\right)-16\operatorname{Li}_5\left(\frac{1}{2}\right)$$ $$-\frac{19}{2}\ln \left(2\right)\zeta \left(4\right)+\frac{8}{3}\ln ^3\left(2\right)\zeta \left(2\right)+\frac{2}{15}\ln ^5\left(2\right).$$ By using these results, the closed form mentioned in the beginning also follows.

And that's all there is to it.