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

The square-root branch points are zero and infinity. At zero the local behavior is $z^{-1/2}$; in the coordinate $u=1/z$ near infinity it is $u^{1/2}$. In either case one circuit reverses the sign and two restore it. Thus the ramification index is $2$: these are two-sheeted algebraic branch points, sometimes described as branch order one when order means ramification index minus one. The decay to zero at infinity does not remove the branching.

TODO: the original closed-contour drawing is absent from the source. The following independent Bromwich contour deformation establishes the required discontinuity relation without asserting the missing drawing's exact geometry. Orient the vertical line upward, as required by the positive imaginary answer in the next part, and use the principal square root, $-\pi <\arg z<\pi$.

Close a truncated vertical line to the left with a rectangle, slit it along the negative real axis, and exclude a small circle about zero. The integrand is holomorphic inside that slit domain, so the boundary contour integral is zero. Horizontal boundary integrals are $O(R^{-1/2})$: integrating $e^x$ along $-R\le x\le 1$ gives a bounded factor while $|z{|}^{-1/2}\le R^{-1/2}$. The left boundary contributes $O(e^{-R}{R}^{1/2})$, and the small circle contributes $O(\sqrt{\epsilon })$. All vanish in the indicated limits.

At $z=-x$ with $x>0$, the upper and lower bank values are respectively $-i{e}^{-x}/\sqrt{x}$ and $+i{e}^{-x}/\sqrt{x}$. The boundary follows the upper bank toward zero and the lower bank away from zero. Their combined contribution is $-2i{\int }_0^{\infty }{e}^{-x}{x}^{-1/2}dx$. HenceThe vertically improper integral converges conditionally; these estimates also justify the contour limit. Reversing the vertical orientation reverses its sign.

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.