Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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

1 Answer 1

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 $\alpha=3/2$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.