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$$I=\int_0^1\frac{\arcsin^2(x^2)}{\sqrt{1-x^2}}dx\stackrel?=\frac{5}{24}\pi^3-\frac{\pi}2\log^2 2-2\pi\chi_2\left(\frac1{\sqrt 2}\right)$$ This result seems to me digitally correct? Can we prove that the equality is exact?

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We have:

$$ \arcsin^2(z^2)=\sum_{n\geq 0}\frac{2^{2n+1} n!^2}{(2n+2)!}z^{4n+4} \tag{1}$$ hence:

$$ I = \frac{\pi}{4}\sum_{n\geq 0}\frac{n!^2 (4n+3)!}{2^{2n+1}(2n+2)!^2 (2n+1)!}=\frac{3\pi}{16}\cdot\phantom{}_5 F_4\left(1,1,1,\frac{5}{4},\frac{7}{4};\frac{3}{2},\frac{3}{2},2,2;1\right).\tag{2} $$

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    $\begingroup$ How does this confirm the OP's conjecture? I don't see why this answer was accepted. $\endgroup$ – Antonio DJC Apr 17 '17 at 4:51
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\begin{align}\int_0^{1} \frac{\arcsin^2 x^2}{\sqrt{1-x^2}}\,dx &= \frac{1}{2}\int_0^{1} \frac{\arcsin^2 x}{\sqrt{x}\sqrt{1-x}}\,dx \tag{1}\\&= \frac{1}{2}\int_0^{\pi/2} \frac{\theta^2\cos \theta}{\sqrt{\sin \theta - \sin^2 \theta}}\,d\theta \tag{2}\\&= \frac{1}{\sqrt{2}}\int_0^{\pi/2} \frac{\left(\frac{\pi}{2} - \theta\right)^2\cos \frac{\theta}{2}}{\sqrt{1-2\sin^2 \frac{\theta}{2}}}\,d\theta \tag{3}\\&= \int_0^{\pi/2} \left(\frac{\pi}{2} - 2\arcsin \left(\frac{\sin \alpha}{\sqrt{2}}\right)\right)^2\,d\alpha \tag{4}\\&= \frac{\pi^3}{8} - 2\pi \int_0^{\pi/2} \arcsin \left(\frac{\sin \alpha}{\sqrt{2}}\right)\,d\alpha + 4\int_0^{\pi/2} \left(\arcsin \left(\frac{\sin \alpha}{\sqrt{2}}\right)\right)^2\,d\alpha \tag{5}\end{align}

where, we made $x \mapsto \sqrt{x}$ in step $(1)$. In step $(2)$ we made $\theta = \arcsin x$ and finally in $(3)$ we made the change of variable $\sin \dfrac{\theta}{2} = \dfrac{\sin \alpha}{\sqrt{2}}$.

Now we recall the famous series expansion: $\displaystyle \arcsin^2 x = \frac{1}{2}\sum\limits_{n=1}^{\infty} \dfrac{(2x)^{2n}}{n^2\binom{2n}{n}}$,

Hence, \begin{align*}\int_0^{\pi/2} \left(\arcsin \left(\frac{\sin \alpha}{\sqrt{2}}\right)\right)^2\,d\alpha &= \frac{1}{2}\sum\limits_{n=1}^{\infty} \dfrac{2^n}{n^2\binom{2n}{n}}\int_0^{\pi/2} \sin^{2n} \alpha \,d\alpha\\&= \frac{\pi}{4}\sum\limits_{n=1}^{\infty} \dfrac{1}{n^22^{n}} = \frac{\pi}{4}\operatorname{Li}_2 \left(\frac{1}{2}\right) = \frac{\pi}{8}\left(\zeta(2) - \log^2 2\right)\end{align*}

also, the infinite series expansion for $\displaystyle \arcsin x = \sum\limits_{n=0}^{\infty} \dfrac{\binom{2n}{n}x^{2n+1}}{(2n+1)4^n}$ give us,

\begin{align*}\int_0^{\pi/2} \arcsin \left(\frac{\sin \alpha}{\sqrt{2}}\right)\,d\alpha &= \frac{1}{\sqrt{2}}\sum\limits_{n=0}^{\infty} \dfrac{\binom{2n}{n}}{(2n+1)8^n}\int_0^{\pi/2} \sin^{2n+1} \alpha \,d\alpha\\&= \frac{1}{\sqrt{2}}\sum\limits_{n=0}^{\infty} \dfrac{1}{(2n+1)^2 2^n} = \chi_2 \left(\frac{1}{\sqrt{2}}\right)\end{align*}

Combining the results, $$\int_0^{1} \frac{\arcsin^2 x^2}{\sqrt{1-x^2}}\,dx = \frac{5\pi^3}{24} - \frac{\pi}{2}\log^2 2 - 2\pi \chi_2 \left(\frac{1}{\sqrt{2}}\right)$$

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    $\begingroup$ Well,beautifull(+1) $\endgroup$ – user178256 Feb 7 '16 at 14:30
  • $\begingroup$ @user178256 Thanks! Glad you like it :) $\endgroup$ – r9m Feb 7 '16 at 15:04

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