Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2012 / S1 / 3 / c / Solution

The Gamma function identity $\Gamma (1/2)=\sqrt{\pi }$ gives the Bromwich integral of an inverse square root directly:Equivalently, put $x=u^2$ in the real integral. It becomes $2{\int }_0^{\infty }{e}^{-u^2}du=\sqrt{\pi }$ by the Gaussian integral.