Closed form for $\int_0^\infty\left(\int_0^1\frac1{\sqrt{1-y^2}\sqrt{1+x^2\,y^2}}\mathrm dy\right)^3\mathrm dx.$

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?

• Mathematica's numerical integration confirmed that the difference between the integral $I$ and your conjectured value differs less than $10^{-100}$. This is a tantalizing result indicating that your conjecture seems true. Anyway, how did you conjecture that result? – Sangchul Lee May 18 '13 at 1:01
• Would you mind editing the title of your question so that it doesn't just consist of LaTeX? If users have their math renderer set to expand on click (as I do) then they won't be able to view this question. – Lee Sleek May 18 '13 at 1:15
• @sos440 I ran this WolframAlpha query from within Mathematica: WolframAlpha["7.09022700484626946098980237", IncludePods -> "PossibleClosedForm", TimeConstraint -> Infinity]. It returned $\frac{3\,L^4}{20}$ (where $L$ is the lemniscate constant) as a possible closed form. – Vladimir Reshetnikov May 18 '13 at 1:21

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 $K(1/\sqrt{2}) = \frac14 \pi^{-1/2}\Gamma(\frac14)^2$.
• The closed-form of $\int_0^1 K(k)\, dk$ and $\int_0^1\big(K(k)\big)^3 dk$ are known. Would you know $\int_0^1\big(K(k)\big)^5 dk$? – Tito Piezas III Feb 8 '19 at 14:56
The first thing I would try is to replace $\int_{0}^{\infty}$ by $\frac12\int_{-\infty}^{\infty}$, then to consider this as a complex integral and pull the integration contour to $i\infty$. Since the only singularity of the hypergeometric function $_2F_1(\ldots,z)$ on the main sheet is the branch point $z=1$, in the end we would have to integrate the jump of $K^3(ix)$ on the cut $B=[i,i\infty)$. The jump of a function on the cut is often simpler than the function itself (like $2\pi$ vs $\ln z$), so I would hope for a simplification of the integral (as the branch point $x=i$ here seems to be logarithmic).