Closed form for ∫∞0(∫101√1−y2√1+x2y2dy)3dx.\int_0^\infty\left(\int_0^1\frac1{\sqrt{1-y^2}\sqrt{1+x^2\,y^2}}\mathrm dy\right)^3\mathrm dx.

Question 1

I need to find a closed form for these nested definite integrals:
$I=\int_0^\infty\left(\int_0^1\frac1{\sqrt{1-y^2}\sqrt{1+x^2\,y^2}}\mathrm dy\right)^3\mathrm dx.$
The inner integral can be represented using the hypergeometric function $_2F_1$ or the complete elliptic integral of the 1st kind $K$ with an imaginary argument:
$\int_0^1\frac1{\sqrt{1-y^2}\sqrt{1+x^2\,y^2}}\mathrm dy=\frac\pi2 {_2F_1}\left(\frac12,\frac12;1;-x^2\right)=K(x\,\sqrt{-1}).$
My conjecture is the integral $I$ has a closed-form representation: $I\stackrel{?}{=}\frac{3\,\Gamma (\frac14)^8}{1280\,\pi^2}=7.09022700484626946098980237...,$
but I was neither able to find a proof of it, nor disprove the equality using numerical integration. Could you please help me with resolving this question?

Question 2

Using
$K(ik) = \frac{1}{\sqrt{1+k^2}}K\left(\sqrt{\frac{k^2}{k^2+1}}\right)$
and a substitution $t^2 = \frac{k^2}{1+k^2}$ , rewrite the integral as
$\int_0^\infty K(i k)^3\,dk = \int_0^1 K(t)^3\,dt.$

There is a paper “Moments of elliptic integrals and critical L-values”
by Rogers, Wan and Zucker (http://arxiv.org/abs/1303.2259; also one of
the authors’ earlier papers: http://arxiv.org/abs/1101.1132), and the
authors, by relating this integral to an L-series of a modular form (their theorems 1 and 2),
show that
$\int_0^1 K(k)^3\,dk = \frac{3}{5}K(1/\sqrt{2})^4 = \frac{3\Gamma(\frac14)^8}{1280\pi^2},$
using $K(1/\sqrt{2}) = \frac14 \pi^{-1/2}\Gamma(\frac14)^2$ .

Closed form for ∫∞0(∫101√1−y2√1+x2y2dy)3dx.\int_0^\infty\left(\int_0^1\frac1{\sqrt{1-y^2}\sqrt{1+x^2\,y^2}}\mathrm dy\right)^3\mathrm dx.

Answer

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