I Problem with this estimation lemma example

[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

 

[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...

 

[mathman]

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

Does that help?

 

[Jenny short]

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

Does that help?

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

 

[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}.

 

[Svein]

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