In a paper I came across (page 10, section 7), the authors state that $\int_{-\infty}^{\infty} \frac{dx}{(b^{2}+x^{2})\cosh ax} $ can be evaluated by "closing the real axis with a semi-circle centered at the origin located in the upper half-plane. An elementary estimate shows that the integral over the circular boundary vanishes as the radius goes to infinity."
But I don't think that the integral over the circular boundary is going to vanish if one simply lets the radius go to infinity in a continuous manner. Furthermore, wouldn't it be easier to use a rectangular contour?
You're right, the argument in the paper isn't rigorous because one would have to consider what happens when $|\cosh ax|\lt1$. Continously taking either a semicircle or a rectangle to infinity would cross the poles. You're also right that showing how to avoid this complication is easier for a rectangle than for a semicircle. Since $\cosh (x+\mathrm iy)=\cosh x\cos y+\mathrm i\sinh x\sin y$, rectangles at arbitrary distances from the origin can be chosen such that $|\cosh ax|\ge1$, and then the factor $b^2+x^2$ in the denominator ensures convergence.
Of course, since the integrals along the rectangular contours converge, the integrals along the semicircular contours also converge, since they're equal if they enclose the same poles; but this would be more difficult to show.