# Choosing between a semicircular contour and a rectangular contour

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?

• 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