# Keyhole integration

Ok so my lecturer gave us this powerful lemma for doing contour integrals over a semi-circle. The lemma is:

Let $C_{R}$ be the contour defined as $\{z \in \mathbb{C} \mid z = R e^{i \theta } , \theta \in [0, \pi] \}$.

If $|f(z)| \le \frac{k}{ |z| ^2}$ for some $k \in \mathbb{R}$ , $\Im(z) > 0$ and $|z|$ large enough, then:

$$\lim_{R \to 0} \int_{C_{R}} f(z) dz = 0$$

So in a lecture we did this problem:

$$\int_0^{\infty} \dfrac{dx}{ \sqrt{x} (1 + x) }$$

He then defined $\Gamma _{R} = C_R \cup [\epsilon,R] \cup (-C_{\epsilon}) \cup [R,\epsilon]$

Where $C_R = \{z \in \mathbb{C} \mid |z| = R\}$, $C_{\epsilon} = \{z \in \mathbb{C} \mid |z| = \epsilon \}$

He somehow used the lemma to show that:

$$\lim_{R \to 0} \int_{C_{R}} f(z) dz = 0$$

But he can't do that can he?

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Is this really all that you remember about the step that he "somehow" used it? He didn't maybe change variables? – Phira May 21 '11 at 6:50
Yes, he can. $\phantom{c}$ – Raskolnikov May 21 '11 at 6:53

The statment of the powerful lemma has $|f(z)|\le k|z|^{-2}$ as a condition, but it holds if $|f(z)|\le k|z|^{-\alpha}$ with $\alpha>1$. The proof is the same. He is using this with $\alpha=3/2$.