Past exam of the mathematics course of the University of Cambridge / 2008 / ib / Paper 3 / 5C (Complex Methods)

For $t>0$, use the Bromwich inversion formula with a vertical contour $\mathrm{Re}s=c>0$:Close the contour to the left, where the exponential decays. The large-arc contribution tends to zero by the usual exponential arc estimate, and the two simple poles are at $s=ia$ and $s=-ia$. The residues of the complete integrand are $e^{iat}/2$ and $e^{-iat}/2$. The residue theorem therefore givesThe right limit at zero is one. The sign of nonzero real $a$ has no effect on the answer.

Take the branch of the complex logarithm with $-\pi <\arg s<\pi$ and write $s^{-1/2}=\exp [-\log s/2]$. Deform the Bromwich contour into a Hankel contour about the negative real axis. The small circle at zero contributes $O(r^{1/2})$ and vanishes; the large closing arcs vanish by the exponential estimates for $t>0$. On the upper lip $s=-r+i0$, $s^{-1/2}=-i/\sqrt{r}$, while on the lower lip it is $i/\sqrt{r}$. Keeping the lip orientations in the Bromwich inversion with a square-root branch cut givesWith $r=u^2$, the supplied Gaussian integral evaluates this as $2{\pi }^{-1}{\int }_0^{\infty }{e}^{-t{u}^2}du$. HenceThe inverse is singular but locally integrable at zero, which is compatible with existence of its Laplace transform.