# I Problem with this estimation lemma example

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1. Dec 6, 2016

### Jenny short

I have been trying to show that

$$\lim_{U\rightarrow\infty}\int_C \frac{ze^{ikz}}{z^2+a^2}dz = 0$$

Where $$R>2a$$ and $$k>0$$ And C is the curve, defined by $$C = {x+iU | -R\le x\le R}$$

I have tried by using the fact that

$$|\int_C \frac{ze^{ikz}}{z^2+a^2}dz| \le\int_C |\frac{ze^{ikz}}{z^2+a^2}| |dz|$$

I want to use the fact $$|e^{ikz}|=e^{-kU}$$

However I got really stuck after that. I would really appreciate help

2. Dec 6, 2016

### Svein

I do not understand your description of the curve C. Anyhow, $z^{2}+a^{2}=(z-ia)(z+ia)$, so you have poles in ia and -ia. The residue at ia is $\frac{iae^{-ka}}{2ia}=\frac{e^{-ka}}{2}$. Now you just have to calculate the other residue and use the residue theorem...

Last edited: Dec 7, 2016
3. Dec 6, 2016

### mathman

$|\frac{ze^{ikx}}{z^2+a^2}|\leq e^{-kU}|\frac{z}{z^2+a^2}|$

Does that help?

4. Dec 7, 2016

### Jenny short

I've done that, but I'm suck on what to do after that

5. Dec 7, 2016

### Svein

To continue what I said above: As long as C is given by $\vert z\vert=R$ with $R>\vert a \vert$, the value of the integral is given by $2\pi i\sum Res_{\vert z \vert <R}$.

Last edited: Dec 11, 2016
6. Dec 7, 2016

### Svein

More information: If neither ia or -ia is inside C, then the function is analytic there, thus the integral must be 0.