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

Use the principal complex square root, $\sqrt{z}=\sqrt{r}{e}^{i\theta /2}$ with $-\pi <\theta <\pi$. For $F(z)=e^{z-\sqrt{z}}/z$, the local expansion on this holomorphic branch isThe fractional power prevents a single-valued Laurent series in a punctured disc. In particular, zero is a branch point, not an isolated singularity classified as a pole. Merely observing that $zF(z)\to 1$ on the slit plane would not establish a simple pole: $zF(z)$ does not extend as a holomorphic function through a whole disc around zero.

The negative real half-line is a chosen branch cut. Its upper and lower boundary values differ because $\sqrt{-x+i0}=i\sqrt{x}$ and $\sqrt{-x-i0}=-i\sqrt{x}$ for $x>0$. Continuation once around zero changes the sign of the complex square root, and hence changes $F$ in general. Other branch cuts are possible; this one defines the principal holomorphic branch.

Excise a small disc about zero and a thin slit along the negative real axis, using the orientations shown in the source: $C_1$ upwards, $C_3$ towards zero above the slit, $C_4$ clockwise around zero, and $C_5$ away from zero below the slit. On the resulting region $F$ is a single-valued holomorphic function; zero lies in the excluded disc and there are no poles in the region. Cauchy's integral theorem givesHere the paths on the cut mean limiting paths from its two sides. Apply Cauchy's integral theorem before taking those limits, when the entire contour lies inside the slit domain.

For $x>0$, the upper and lower boundary values on the branch cut areWrite $x=A_R$ for the outer endpoint distance on the cut. The upper path runs from $x=A_R$ to $x=ϵ$, while the lower path runs back from $ϵ$ to $A_R$; in both cases $dz=-dx$. ThereforeandThe source's limits zero and infinity require $ϵ\to 0$ and $R\to \infty$; they are not an exact identity on a finite contour. For a concrete counterexample take $ϵ=1/4$ and $A_R=1$: the finite real integral is at most ${\int }_{1/4}^1{x}^{-1/2}dx=1$, whereas the full-range integral is $\pi \mathrm{erf}(1/2)\simeq 1.6352$, as evaluated in the next part. Near zero the combined integrand is $O(x^{-1/2})$, so the combined improper integral converges, although the individual cut integrals have logarithmic divergences. At infinity the factor $e^{-x}$ ensures convergence. Consequently

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.