Evaluate the integral, $$\int_a^\infty dx \frac{1}{x-b-i0}\exp\big(-\sqrt{x^2-a^2}\big),$$ where $0<a<b<\infty$, and $i0$ is a positive infinitesimal imaginary number, shifting the pole slightly above the real axis. To use residue theorem, I tried taking as my contour, the perimeter of the sector on the first quadrant, which includes the real axis, the imaginary axis at $a$ that is $x=a(1+iy)$ with $y>0$, and the arc at infinity. But, on the imaginary axis, the argument of the exponential becomes $$-\sqrt{(a+iy)^2-a^2}=-\sqrt{2iay-y^2}\xrightarrow{y\to\infty} -iy,$$where $x=a(1+iy)$. To get an approximation, I used the asymptotic form of the exponential at large $y$ throughout, $$\int_0^\infty dy \frac{1}{iay-(b-a)}\exp(-iya)=-i\int_0^\infty dz \frac{1}{z+i(b-a)}\exp(-iz)=\frac{-i}{2}e^{-(b-a)}(-i\pi-2\mathrm{Ci}(i(b-a))-2\mathrm{Shi}((b-a)),$$ Here $\mathrm{Ci}$ and $\mathrm{Shi}$ are the Cos integral and Sinh integral, respectively. Now, there's some confusion regarding the contribution at along the arc at infinity. For the "upper" part of the arc, where the real part of $x$ is finite but the imaginary part goes to infinite, using $x=\mathrm{Re}x+iy$ where $y=\mathrm{Im}x$ as before, $$|\exp\big(-\sqrt{(\text{Re}x+iy)^2-a^2}\big)|\to|\exp(-\sqrt{-2i\mathrm{Re}xy+(\mathrm{Re}x^2-a^2-y^2)})|\xrightarrow{y\to\infty}|\exp(-iy)|=1$$ remains finite. Hence that the integral over the arc seems to diverge.
Is there a better contour, or a different method to evaluate this integral?