The Game's Afoot, Watson!

The substitutions $x \,=\, \tan\tfrac12u$ , $y \,=\, \tan\tfrac12v$ , $z \,=\, \tan\tfrac12w$ convert this integral into $\begin{array}{rcl} I & = & \displaystyle 8\int_0^\infty\int_0^\infty \int_0^\infty \frac{dx\,dy\,dz}{(1+x^2)(1+y^2)(1+z^2) - (1-x^2)(1-y^2)(1-z^2)} \\ & = & \displaystyle 4\int_0^\infty \int_0^\infty \int_0^\infty \frac{dx\,dy\,dz}{x^2 + y^2 + z^2 + x^2y^2z^2} \end{array}$ Introducing spherical polar coordinates $x = r\sin\theta\cos\phi$ , $y = r\sin\theta\sin\phi$ , $z = r\cos\theta$ converts the integral to $\begin{array}{rcl} I & = & \displaystyle 4\int_0^\infty dr \int_0^{\frac12\pi}d\theta \int_0^{\frac12\pi}d\phi \frac{r^2\sin\theta}{r^2 + r^6\sin^4\theta\cos^2\theta\cos^2\phi\sin^2\phi} \\ & = & \displaystyle 4\int_0^\infty dr \int_0^{\frac12\pi}d\theta \int_0^{\frac12\pi}d\phi \frac{\sin\theta}{1 + r^4\sin^4\theta\cos^2\theta\cos^2\phi\sin^2\phi} \\ & = & \displaystyle 4\int_0^\infty dr \int_0^{\frac12\pi}d\theta \int_0^{\frac12\pi}d\phi \frac{\sin\theta}{1 + \frac14r^4\sin^4\theta\cos^2\theta\sin^22\phi} \\ & = & \displaystyle 2\int_0^\infty dr \int_0^{\frac12\pi}d\theta \int_0^{\pi}d\phi \frac{\sin\theta}{1 + \frac14r^4\sin^4\theta\cos^2\theta\sin^2\phi} \end{array}$ Since the substitution $x = t^4$ shows us that $\int_0^\infty \frac{dt}{1+t^4} \; = \; \tfrac14B\big(\tfrac14,\tfrac34\big) \; = \; \frac{\Gamma(\frac14)\Gamma(\frac34)}{4\Gamma(1)} \; =\; \frac{\pi}{2\sqrt{2}}$ we can perform the integration with respect to $r$ to deduce that $\begin{array}{rcl} I & = & \displaystyle 2\int_0^{\frac12\pi}d\theta \int_0^{\pi}d\phi \frac{\pi\sin\theta}{2\sqrt{2}\left(\frac14\sin^4\theta\cos^2\theta\sin^2\phi\right)^{\frac14}} \\ & = & \displaystyle \pi \int_0^{\frac12\pi} \frac{d\theta}{\sqrt{\cos\theta}} \int_0^\pi \frac{d\phi}{\sqrt{\sin\phi}} \end{array}$ Using symmetry properties of $\sin$ and $\cos$ , we deduce that $\begin{array}{rcl} I & = & \displaystyle 2\pi\left(\int_0^{\frac12\pi}\frac{d\theta}{\sqrt{\sin\theta}}\right)^2 \; = \; \tfrac12\pi B\big(\tfrac14,\tfrac12\big)^2 \; = \; \tfrac12\pi\left(\frac{\Gamma(\frac14)\Gamma(\frac12)}{\Gamma(\frac34)}\right)^2 \; = \; \tfrac14\Gamma\big(\tfrac14\big)^4 \end{array}$ which makes the answer $1+4+4+4 = 13$ .

Note: The proof of this integral was first published by G.H.Watson, but the result had been previously obtained by W.F. van Peype (when he was a graduate student).