Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 78 / 2 / i / Solution

Put $x=r\cos \theta$, $y=r\sin \theta$, and $s=|\sin \theta |$. From the exterior transform, the inverse integral has the amplitudemultiplying $e^{-ikr\cos \theta -\gamma rs}$. Apply the supplied method of steepest descent formula at $k_s=k_0\cos \theta$. On the outgoing sheet, $\gamma (k_s)=i{k}_{0}s$, so $-i\sinh (i{k}_{0}bs)=\sin (k_{0}bs)$. The far-field scattered potential isRestore $e^{i\omega t}$ for the physical time-harmonic potential. This is cylindrical outgoing spreading, with angular directivity determined by the kernel factor and aperture. The ordinary saddle formula applies for a fixed non-grazing direction, away from a coalescing pole or branch point. Its finite upstream angular limit is consistent with the separate branch-point calculation in the root solution; at grazing directions that separate calculation supplies the appropriate justification.