Past exam of the short options paper of the physics Prelims of the University of Oxford / 2021 / S01 / 3 / d / Solution

On the small clockwise contour $z=ϵe^{i\theta }$, $\theta$ decreases from $\pi$ to $-\pi$. Uniformly on that arc, $e^{z-\sqrt{z}}=1+O(\sqrt{ϵ})$, and henceThe large arcs also vanish. To make this precise for the displayed geometry, take them on the left semicircle $z=\gamma +R{e}^{i\theta }$, $\pi /2\le \theta \le 3\pi /2$, split at the cut. On the principal holomorphic branch, $\mathrm{Re}\sqrt{z}\ge 0$, $|z|\ge R-\gamma$, and $\mathrm{Re}z=\gamma +R\cos \theta$. The estimation lemma givesFor the last bound, symmetry and $\sin u\ge 2u/\pi$ on $0\le u\le \pi /2$ imply that the angular integral is at most $\pi /R$. The exact choice of large arcs does not affect the limiting result. The vertical improper integral converges absolutely because $e^{-\mathrm{Re}\sqrt{\gamma +iy}}/|\gamma +iy|$ decays as $O(e^{-c\sqrt{|y|}}/|y|)$.

Taking limits in the closed contour integral now yields $2\pi iJ+2iK-2\pi i=0$, where $K={\int }_0^{\infty }{e}^{-x}\sin (\sqrt{x})dx/x$. ThusAs an optional explicit evaluation, put $x=t^2$ and define $K(a)=2{\int }_0^{\infty }{e}^{-t^2}\sin (at)dt/t$. Differentiation under the integral sign and the Gaussian integral give $K^{\prime }(a)=\sqrt{\pi }{e}^{-a^2/4}$ and $K(0)=0$. Therefore $K=K(1)=\pi \mathrm{erf}(1/2)$ and $J=\mathrm{erfc}(1/2)\simeq 0.479500$ in terms of the error function and complementary error function.