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

Question

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

Answer 1

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.

Answer 2

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...

Answer 3

An elementary approach \begin{align} I &= \int_0^1 \frac{\arcsin x\arcsin\frac x{\sqrt2}}{\sqrt{2-x^2}} \, dx \\ &= \frac12 \int_0^1 \arcsin x \ d\left(\arcsin^2\frac x{\sqrt2} -\frac{\pi^2}{16}\right)\\ &\overset{ibp}= \frac12 \int_0^1 \left(\frac{\pi^2}{16}-\arcsin^2\frac x{\sqrt2} \right)\frac1{\sqrt{1-x^2}}dx \end{align} Substitute $x=\sqrt2\sin\left(\frac\pi4-\theta\right)$ and then $\theta\to\frac\pi2-\theta$ \begin{align} I &= \frac1{4\sqrt2}\int_0^{\pi/2}\theta\left(\frac\pi2-\theta\right)\left(\sqrt{\tan\theta}+\sqrt{\cot\theta}\right) \overset{\tan \theta =t^2}{d\theta} \\ &= \frac1{2\sqrt2} \int_0^\infty \frac{(t^2+1)\tan^{-1} t^2 \cot^{-1} t^2}{t^4+1}dt\\ &= \frac1{2\sqrt2}\left( \int_0^\infty \frac{\tan^{-1} t^2 \cot^{-1} t^2}{t^2+\sqrt2t+1}dt + \sqrt2\int_0^\infty\frac{t\tan^{-1} t^2 \cot^{-1} t^2}{t^4+1} dt\right)\\ &= \frac1{2\sqrt2}\left(\frac\pi{4\sqrt2}\ln^22+\sqrt2\cdot \frac{\pi^3}{48} \right)=\frac\pi{16}\left( \ln^22+\frac{\pi^2}{12}\right) \end{align}

Answer 4

A slight detour from Ron Gordon's answer, using the square of the $\arcsin$ series:

$$\begin{align*} I &= \int_0^1 \frac{\arcsin(x) \arcsin\left(\frac x{\sqrt2}\right)}{\sqrt{2-x^2}} \, dx \\[1ex] &= \frac{\pi^3}{64} - \frac12 \int_0^1 \arcsin^2\left(\frac x{\sqrt2}\right) \, \frac{dx}{\sqrt{1-x^2}} \tag{1} \\[1ex] &= \frac{\pi^3}{64} - \frac14 \sum_{n=1}^\infty \frac{2^n}{n^{2} \binom{2n}n} \int_0^1 \frac{x^{2n}}{\sqrt{1-x^2}} \, dx \tag{2} \\[1ex] &= \frac{\pi^3}{64} - \frac18 \sum_{n=1}^\infty \frac{2^n}{n^2 \binom{2n}n} \int_0^1 x^{n-\frac12} (1-x)^{-\frac12} \, dx \tag{3} \\[1ex] &= \frac{\pi^3}{64} - \frac18 \sum_{n=1}^\infty \frac{2^n}{n^2 \binom{2n}n} \frac{\Gamma\left(n+\frac12\right)\Gamma\left(\frac12\right)}{\Gamma(n+1)} \tag{4} \\[1ex] &= \frac{\pi^3}{64} - \frac\pi8 \sum_{n=1}^\infty \frac1{2^nn^2} \tag{5} \\[1ex] &= \frac{\pi^3}{64} - \frac\pi8 \operatorname{Li}_2\left(\frac12\right) \tag{6} \\[1ex] &= \frac{\pi^3}{64} - \frac\pi8 \left(\frac{\pi^2}{12} - \frac{\log^2(2)}2\right) \\[1ex] &= \boxed{\frac{\pi^3}{192} + \frac{\pi\log^2(2)}{16}} \end{align*}$$

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• $(1)$ : integrate by parts

• $(2)$ : power series of $\arcsin^2(x)$

• $(3)$ : substitute $x\mapsto\sqrt x$

• $(4)$ : beta function

• $(5)$ : simplify gammas and central binomial coefficient; in particular, we have the identity $\dfrac{\Gamma\left(n+\frac12\right)}{\Gamma\left(\frac12\right)\Gamma(n+1)} = \dfrac1{4^n} \dbinom{2n}n$

• $(6)$ : dilogarithm

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I suspect one could adapt Jack D'Aurizio's general solution for a related integral... an exercise for the reader.