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How to prove \begin{align} I &= \int_0^1\frac{\arcsin{(x)}\arcsin{(x\sqrt\frac{1}{2})}}{\sqrt{2-x^2}}dx \\ &= \frac{\pi}{256}\left[ \frac{11\pi^4}{120}+2{\pi^2}\ln^2{2}-2\ln^4{2}-12\zeta{(3)}\ln{2} \right] \end{align} By asking $$x=\sqrt{2}y$$ then using integration by parts, we have $$I=\frac{\pi^5}{2048}-\frac{1}{4}\int_0^1{\arcsin^4\left( \frac{z}{\sqrt{2}}\right) }\frac{dz}{\sqrt{1-x^2}}$$

But how to calculate this integral? I would appreciate your help

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    $\begingroup$ i think you can reduce this to something like $I=\text{const}\times\int_0^{\pi/2}\frac{x^2}{\sqrt{\cos(x)}}dx$ which i think was already discussed on MSE $\endgroup$ – tired Oct 29 '16 at 17:38
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    $\begingroup$ Please improve your question by adding more context: where did the integral come from? Why is it of interest? There are infinitely many integrals that could be posed - what makes this particular one worth posting? How do you know the exact value of the integral? Additional context of this sort will make the question more valuable for others. $\endgroup$ – Carl Mummert Oct 29 '16 at 18:22
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We want to calculate

$$ I=\int_0^1 dx\frac{\arcsin(x)\arcsin\left(\frac{x}{\sqrt{2}}\right)}{\sqrt{2-x^2}} $$

Let us start by a substitution $x=\sqrt{2}\sin(y)$ and integrate by parts so we can write $I$ as

$$ \int_0^{\pi/4}dy y \arcsin(\sqrt{2}\sin(y))\underbrace{= }_{i.b.p}\frac{\pi^3}{64}-\frac{1}{2}\underbrace{\int_0^{\pi/4}dy\frac{y^2\cos(y)}{ \sqrt{1-2\sin(y)^2}}}_{J}$$

but since $1-2\sin(y)^2=\cos(2y)$ and by another substitution $2y =q$

$$ J=\frac{1}{8}\int_0^{\pi/2}dq\frac{ q^2 \cos(q/2) }{\sqrt{\cos(q)}} $$

Now we perform small trick (credit to Mr. Feynman) by rewriting

$$ J=-\frac{1}{8}\left[\frac{d^2}{da^2}\int_0^{\pi/2} dq\cos(q)^{-1/2}\cos(aq)\right]_{a=1/2} $$

which equals according to Tunk-Fey's answer here

$$ J=-\frac{1}{8}\frac{d^2}{da^2}\left[\frac{2\pi}{\sqrt{2}}\frac{1}{B\left(\frac{3/2+a}{2},\frac{3/2-a}{2}\right)}\right]_{a=1/2} $$

Or

$$ I=\frac{\pi^3}{64}+\frac{\pi}{8\sqrt{2}}\frac{d^2}{da^2}\left[\frac{1}{B\left(\frac{3/2+a}{2},\frac{3/2-a}{2}\right)}\right]_{a=1/2} $$

Since i have no CAS at hand today and are way to lazy to do the above derivative by hand that's it for the moment. But everything from here on should be straightforward so i leave that to you...

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This integral can be done by recognizing that

$$\frac{\arcsin{\frac{x}{\sqrt{2}}}}{\sqrt{2-x^2}} = \sum_{n=0}^{\infty} \frac{2^n x^{2 n+1}}{(2 n+1) \binom{2 n}{n}}$$

and that

$$\int_0^1 dx \, x^{2 n+1} \arcsin{x} = \frac{\pi}{4 (n+1)} \left [1- \frac1{2^{2 n+2}} \binom{2 n+2}{n+1}\right ] $$

To see this, integrate by parts and see this answer.

With a bit of algebra, we find that the integral is equal to

$$\frac{\pi}{2} \sum_{n=0}^{\infty} \frac{2^n}{(2 n+1)(2 n+2) \binom{2 n}{n}} - \frac{\pi}{16} \sum_{n=0}^{\infty} \frac1{2^n (n+1)^2}$$

The first sum may be evaluated by recognizing that it is

$$\int_0^1 dx \frac{\arcsin{\frac{x}{\sqrt{2}}}}{\sqrt{2-x^2}} = \frac{\pi^2}{32}$$

The second sum is recognized as $\operatorname{Li_2}{\left ( \frac12 \right )} = \frac{\pi^2}{6}-\log^2{2} $

Putting all of this together, we find that the integral is equal to

$$\int_0^1 dx \, \frac{\arcsin{x} \arcsin{\frac{x}{\sqrt{2}}}}{\sqrt{2-x^2}} = \frac{\pi^3}{192} + \frac{\pi}{16} \log^2{2} $$

Numerical evaluation in Mathematica confirms the result, which differs from that asserted by the OP.

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  • $\begingroup$ Nice! If the formula OP gives is also correct, we have a nice representation of $\zeta(3)$. :) $\endgroup$ – mickep Jan 3 '17 at 9:13

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