# Proving a series for the Watson Triple Integrals?

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,

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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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