# the proper contour

In the following paper (page 10, section 7), the authors state that $\displaystyle\int_{-\infty}^{\infty} \frac{1}{(b^{2}+x^{2})\cosh ax} \ dx$ 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."

The integrals in Gradshteyn and Ryzhik. Part 21: Hyperbolic functions

Does the integral over the circular boundary really evaluate to zero when you take the limit? I don't think it does. Wouldn't a rectangle be the appropriate contour for this problem?

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The top half of the circle tends to zero. The real part does not. A semicircle is the simplest contour to use for this integral. –  Argon Apr 6 '12 at 19:54

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.
Say $b=1$ and $a = \pi$. My idea would be to use a rectangle with vertices at $N, N+iN,-N+iN,$ and $-N$ where $N$ is an integer greater than $1$ so that I'm not going through a pole. Then showing the integral tends to zero around the three other sides is an easy application of the ML inequality. But if that works, then I guess I could instead use a circle with a radius of $N$. –  Random Variable Apr 6 '12 at 20:48
@RandomVariable: Yes, the modulus of $\cosh az$ increases exponentially with the real part of $z$. And yes, the ML inequality is the basis for all this; but to easily apply it we need $|\cosh ax|\ge1$, which we can ensure easily at the top of the rectangle but not as easily along the semi-circle. –  joriki Apr 6 '12 at 22:24