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 / b / Solution

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.