Easy Description but Hard Answer

Let $\theta$ be the angle with the horizontal. Equate the change in potential energy to the kinetic energy:

$m g R \, sin \theta = \frac{1}{2} m R^2 \Big(\frac{d \theta}{dt} \Big)^2$

Re-arranging gives:

$dt = \sqrt{\frac{R}{ 2 g \, sin \theta}} \, d \theta$

Total elapsed time from start to end:

$T = \sqrt{\frac{R}{2 g}} \int_0^{\pi/2} \frac{1}{\sqrt{sin \theta}} d \theta \\ = \sqrt{\frac{R}{2 g}} \int_0^{\pi/2} sin^{2 (1/4) -1} \, \theta \,\, cos^{2 (1/2) -1} \, \theta \, d \theta \\ = \frac{1}{2} \, \sqrt{\frac{R}{2 g}} \, \frac{\Gamma(1/4) \, \Gamma(1/2)}{\Gamma(3/4)} \\ = \frac{1}{2} \, \sqrt{\frac{R}{2 g}} \, \frac{1}{\sqrt{2 \pi}} \Gamma(1/4)^2 \\ = \sqrt{\frac{R}{g}} \, \frac{1}{4 \sqrt{\pi}} \, \Gamma(1/4)^2$