Entry 29

The first Watson triple integral was discussed in Enry 26. It turns out all three integrals can be expressed simply by the Dedekind eta function $\eta(\tau)$ $$\begin{aligned}I_1 &= \frac{1}{\pi^3}\int_0^\pi \int_0^\pi \int_0^\pi \frac{dx\, dy\, dz}{1-\cos x\cos y\cos z}=4\,\eta^4(i)\\I_2 &= \frac{1}{\pi^3}\int_0^\pi \int_0^\pi \int_0^\pi \frac{dx\, dy\, dz}{3-\cos x\cos y-\cos x\cos z-\cos y\cos z}=4^{1/3}\sqrt{3}\,\eta^4(\sqrt{-3})\\I_3 &= \frac{1}{\pi^3}\int_0^\pi \int_0^\pi \int_0^\pi \frac{dx\, dy\, dz}{1-\cos x-\cos y-\cos z}=2\sqrt{6}\,(1+\sqrt{2})^{1/3}\eta^4(\sqrt{-6})\end{aligned}$$ In terms of the gamma function $$\begin{aligned}I_1&= \frac{\Gamma^4(\frac{1}{4})}{4\pi^3}\\I_2&= \frac{3\Gamma^6(\frac{1}{3})}{2^{14/3}\pi^4}\\I_3&= \frac{\Gamma(\frac{1}{24})\Gamma(\frac{5}{24})\Gamma(\frac{7}{24})\Gamma(\frac{11}{24})}{16\sqrt{6}\,\pi^3} \end{aligned}$$ . There are infinitely many Ramanujan-type formulas for the $I_i$ a few of which are $$\begin{aligned}\frac{(1+\sqrt{2})\Gamma\big(\tfrac{3}{8}\big)^4}{(2\pi)^{5/2}}\sqrt{I_1} &= \sum_{n=0}^\infty \frac{(2n)!^3}{n!^6} \frac{1}{(-2^6)^n}\\ \frac{1}{\sqrt{2}}\,I_1 &=\sum_{n=0}^\infty \frac{(2n)!^3}{n!^6} \frac{1}{(-2^9)^n}\\ \frac{4}{\sqrt{3}}\,I_2 &=\sum_{n=0}^\infty \frac{(2n)!^3}{n!^6} \frac{1}{(2^8)^n}\\ \frac{1+\sqrt{2}+\sqrt{6}}{\sqrt{6}}\,I_3 &=\sum_{n=0}^\infty \frac{(2n)!^3}{n!^6} \frac{1}{\big(2^3(1+\sqrt{2}+\sqrt{6})^3(1+\sqrt{2})\big)^n}\end{aligned}$$ There is one other formula that belongs to the family with denominators as powers of $2$ $$16\,\eta(\sqrt{-7})^4 =\sum_{n=0}^\infty \frac{(2n)!^3}{n!^6} \frac{1}{(2^{12})^n}$$ but I don't know if this has an equivalent and analogous integral.