= Solution
Put $x=r\cos\theta$, $y=r\sin\theta$, and $s=|\sin\theta|$. From the exterior transform, the inverse integral has the <amplitude>
$$
f(k)=-\frac{iC_0\sinh(\gamma(k)b)}{2\pi L^+(k)(k+k_0)}
$$
multiplying $e^{-ikr\cos\theta-\gamma rs}$. Apply the supplied <method of steepest descent> formula at $k_s=k_0\cos\theta$. On the outgoing sheet, $\gamma(k_s)=ik_0s$, so $-i\sinh(ik_0bs)=\sin(k_0bs)$. The far-field scattered potential is
$$
\boxed{\phi_{\rm sc}^{\rm out}(r,\theta)\sim
C_0\sqrt{\frac{k_0}{2\pi r}}\,
\frac{s\sin(k_0bs)}{L^+(k_0\cos\theta)(k_0\cos\theta+k_0)}
\,e^{-ik_0r+i\pi/4}.}
$$
Restore $e^{i\omega t}$ 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.
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