Consider the following integral: $$\mathcal{I}=\int_1^\infty\int_0^1\frac{\mathrm dy\,\mathrm dx}{\sqrt{x^2-1}\sqrt{1-y^2}\sqrt{1-y^2+4\,x^2y^2}}.$$ It can be represented as $$\mathcal{I}=\int_1^\infty\frac{K(\sqrt{1-4\,x^2})}{\sqrt{x^2-1}}\mathrm dx=\Re\int_0^1\frac{K\left(\sqrt\frac{1+3\,y^2}{1-y^2}\right)}{1-y^2}\mathrm dy,$$ where $K(x)$ is the complete elliptic integral of the 1st kind.

I was not able to further simplify any of these integrals, but numerical integration suggests the following conjectural closed form: $$\mathcal{I}\stackrel{?}{=}\frac{3\,\Gamma(1/3)^6}{2^{17/3}\pi^2}.$$ Can you suggest a proof of this conjecture?

Using Mathematica I obtained a closed form for $\mathcal{I}$ in terms of the Meijer G-function: $$\mathcal{I}=\frac{\pi^{3/2}}{4}G_{3,3}^{2,1}\left(\frac{1}{4}\left|\begin{array}{c}1,\ \ 1,\ \ 1\\\frac{1}{2},\frac{1}{2},\frac{1}{2}\\\end{array}\right.\right),$$ but I have no manual proof for that and do not know how to reduce it to $\frac{3\,\Gamma(1/3)^6}{2^{17/3}\pi^2}.$

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    $\begingroup$ How do you hunt a non-trivial integral with closed forms like this? Anyway, I love your integrals! (+1) $\endgroup$ – Sangchul Lee May 24 '13 at 5:01
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    $\begingroup$ @sos440 I am experimenting a lot. $\endgroup$ – Vladimir Reshetnikov May 24 '13 at 14:16

Edited It turns out this integral is done in http://arxiv.org/abs/0801.0891, which O.L. linked to in a comment to another question. This isn't much of an answer, really, as for all the closed forms I refer you to that paper.

Write the integral (by change of variable) as $$ \frac12 \int_0^\infty K(\sqrt{-1-2\cosh \theta})\,d\theta, $$ then expand this into a double integral from the definition of $K$, and integrate over $\theta$ to get $$ \int_0^1 \frac{dt}{\sqrt{1-t^2}\sqrt{1+3t^2}} K\left( \sqrt{\frac{1-t^2}{1+3t^2}} \right). $$ Another change of variable $$ t^2 = \frac{1}{v^2+1} $$ will bring it to the form $$ \frac1\pi\overline V_3(1,1,1) = \int_0^\infty \frac{dv}{\sqrt{(v^2+4)(v^2+1)}}K\left(\sqrt{\frac{v^2}{v^2+4}}\right), $$ which is equation (32) in that paper. It is evaluated by considering an integral of a product of three modified Bessel functions, and that integral was done in two papers by W. N. Bailey in 1934 ("Some indefinite integrals involving Bessel functions"). As a consequence, the integral equals $$ \Re\left( K(e^{-\pi i/6})^2\right). $$ In that paper, (equation 39) they also give the equivalent form (I don't know a proof why this is equivalent to the above) $$ \frac12 K(\sin(\pi/12))K(\cos(\pi/12)) = \frac{\sqrt3}2 K(\sin(\pi/12))^2, $$ and the closed form exists because $\sin\frac\pi{12}$ is a singular value (see mathworld) of the elliptic integral: $$ \frac{3\Gamma(\frac13)^6}{32\pi^2 2^{2/3}}. $$

  • $\begingroup$ Actually, this question is positively settled by your another recent answer. $\endgroup$ – Vladimir Reshetnikov May 29 '13 at 0:04
  • $\begingroup$ I did them in the wrong order. $\endgroup$ – Kirill May 29 '13 at 0:16
  • $\begingroup$ Kirill, thank you for all your answers, they are amazing! I learn from them a lot. $\endgroup$ – Vladimir Reshetnikov Jun 4 '13 at 21:49
  • $\begingroup$ I'm happy to help. $\endgroup$ – Kirill Jun 5 '13 at 2:11

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