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It's long known that the first Watson triple integral evaluates to,

$$I_1 = \frac{1}{\pi^3} \int_{0}^{\pi} \int_{0}^{\pi} \int_{0}^{\pi} \frac{dx \, dy \, dz}{1-\cos x \cos y \cos z} = \frac{\Gamma^4\left(\tfrac{1}{4}\right)}{4\pi^3} = 1.3932039\dots$$

Apparently, it is also equivalent to the simple infinite series,

$$I_1= \frac{4}{3} \sum_{n = 0}^{\infty} \frac{(4n)!}{n!^4} \frac{1}{(2 \cdot 18^2)^n} $$

which is connected to the Ramanujan-type pi formula,

$$\frac{1}{\pi}= \frac{2}{9} \sum_{n = 0}^{\infty} \frac{(4n)!}{n!^4} \frac{7n+1}{(2 \cdot 18^2)^n} $$

I found such series for $I_1$, $I_2$, and $I_3$. Using Mathematica, it is easy to see that they agree to an arbitrary number of decimal digits. But if you can prove them rigorously, it would be interesting to know how. More details are at,

http://sites.google.com/site/tpiezas/0025

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2 Answers 2

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The sum

$$ \frac{4}{3} \sum_{n \ge 0} \frac{(4 n)!}{(n!)^4} z^4 = \frac{4}{3} \sum_{n \ge 0} \frac{ (\frac{1}{4})_n (\frac{3}{4})_n (\frac{1}{2})_n }{ (1)_n (1)_n} \frac{(256 z)^n}{n!} = \frac{4}{3} {}_3F_2\left( \left. \begin{array}{c} \frac{1}{4}, \frac{1}{2}, \frac{3}{4} \\ 1, 1 \end{array} \right| 256 z \right) $$

This particular hypergeometric function can be expressed in term of complete elliptic integral:

$$ {}_3F_2\left( \left. \begin{array}{c} \frac{1}{4}, \frac{1}{2}, \frac{3}{4} \\ 1, 1 \end{array} \right| w \right) = \frac{4 \sqrt{2} }{ \pi^2 } \frac{ K\left( \frac{1}{2}- { \frac{1}{\sqrt{2(1+ \sqrt{1-w} )} } } \right)^2}{ \sqrt{(1 + \sqrt{1 - w}) } } $$

For $z =\frac{1}{2 \cdot 18^2}$, $w =\frac{32}{81}$. Then, it results that

$$ I_1 = \frac{4 \sqrt{2}}{\pi^2} \left(K\left(\frac{1}{2} - \frac{3 \sqrt{2}}{8}\right)\right)^2 $$ Now elliptic integrals for certain moduli are known to related to gamma functions. These are known as singular values of elliptic integrals.

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  • $\begingroup$ (+1) Looks cute. $\endgroup$ Commented Dec 11, 2019 at 17:53
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A beautiful well-explained solution may be found in the fantastic book, Inside Interesting Integrals by the author Paul J. Nahin, pages $206$-$212$.

The transformation presented in the solution turns the initial integral into a product of three integrals as you may see below, which looks fascinating,

$$\int_0^{\pi}\int_0^{\pi}\int_0^{\pi}\frac{\textrm{d}u \textrm{d}v \textrm{d}w}{1-\cos(u)\cos(v)\cos(w)}$$ $$=4\sqrt{2}\int_0^{\infty} \frac{\textrm{d}t}{1+t^4}\int_0^{\pi/2} \frac{\textrm{d}\theta}{\sqrt{\cos(\theta)}}\int_0^{\pi/2} \frac{\textrm{d}\psi}{\sqrt{\sin(\psi)}}.$$

Finding new solutions to the Watson triple integrals is always a very exciting mission.

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