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I'm reading about residue calculus. I read that the standard procedure for integrating $\int_{-\infty}^\infty R(x)dx$ where $R(x)$ is a rational function whose denominator has degree at least 2 units greater than the numerator and no poles on the real line, is to integrate the complex function $R(z)$ over a closed curve consisting of the segment $(-r,r)$ and the semicircle from $r$ to $-r$ in the upper half plane, so that if $r$ is large enough, the curve encloses all poles in the upper half plane.

I see this works because the integral over this closed curve can be found by summing the residues, which may be easier. Since I'm only interested in the integral along the real line, I would hope that the integral over the semicircle goes to $0$ as $r\to\infty$. My text says this is true by "obvious estimates."

Can someone please explain more explicitly how one sees the integral over the semicircle goes to $0$? The obvious estimates is cryptic to me. The text is Ahlfors' Complex Analysis, page 156, if it's helpful. Many thanks.

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Intuitively, let $-n$ be the degree of $R(z)$ (that is, $n = [\text{degree of the denominator}] - [\text{degree of the numerator}]$). Then as $r \to \infty$ and $|z| = R$, we have approximately $|R(z)| \approx C r^{-n}$, where $C$ is a positive constant depending on $R$. But the integration is performed on the semicircle, so this attributes to the factor $\pi r$, corresponding to the arc-length of the semicircle. So the size of the integral is bounded by $\pi C r^{-(n-1)}$, which goes to zero if $n \geq 2$. You may elaborate this argument to obtain a rigorous proof. – Sangchul Lee Mar 18 '12 at 7:19
Thanks @sos440, that does make it easier to see intuitively what is going on. – Dedede Mar 18 '12 at 7:22
up vote 3 down vote accepted

Consider the integral on the the semicircle $C$, where $|z|$ is large. In that limit $R(z)$ goes like $1/z^2$. Let $z = r e^{i\theta}$. Notice that $dz = i r e^{i\theta} d\theta$. We find $$\int_C d z R(z) \sim \int_C \frac{dz}{z^2} = \frac{i}{r} \int_0^\pi d \theta \ e^{-i\theta} = \frac{2}{r}$$ Thus, in the limit the integral on the semicircle vanishes.

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Thanks oenamen! – Dedede Apr 13 '12 at 4:59
@Dedede: Sure thing, Dedede. Glad to help. – user26872 Apr 13 '12 at 5:40

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