Computing Integral

**mathshelpee** · September 2nd 2011, 07:17 AM

Hey,

I've done this problem, but towards the end, I get a term with "i" (imaginary) in it. However I assume the question only asks for the real part. Therefore, since I have only one term at the end...will the final answer be zero?

Thanks

http://i53.tinypic.com/so31c6.jpg

**Random Variable** · September 2nd 2011, 08:04 AM

Actually it appears that the integral is real-valued.

Let $f(z) = \frac{e^{iz}}{(z-\pi)^{2}+a^{2}}$

Let the contour go from (-R,0) to (R,0) and then back to (-R,0) along an arc in the upper half plane. f(z) has a simple pole in the upper half plane at $\pi + ia$ . If we let R go to infinity, the integral will evaluate to zero along the arc because of Jordan's lemma.

therefore $\int_{-\infty}^{\infty} \frac{e^{ix}}{(x-\pi)^{2}+a^{2}} \ dx = 2 \pi i \ \text{Res} \ [f,\pi +ia]$

$= 2 \pi i \lim_{z \to \pi + ia} (z- \pi - ia) \frac{e^{iz}}{(z-\pi)^{2}+a^{2}}$

$= 2 \pi i \lim_{z \to \pi + ia} \frac{e^{iz} + i(z-\pi-ia)e^{iz}}{2(z-\pi)}$

$= 2 \pi i \frac{e^{i\pi -a}}{2ia} = \frac{\pi}{a}e^{\i \pi}e^{-a} = -\frac{\pi}{a} e^{-a}$

**CaptainBlack** · September 2nd 2011, 09:03 AM

Originally Posted by mathshelpee

Hey,

I've done this problem, but towards the end, I get a term with "i" (imaginary) in it. However I assume the question only asks for the real part. Therefore, since I have only one term at the end...will the final answer be zero?

Thanks

http://i53.tinypic.com/so31c6.jpg

You are interested in:

$I=\int_{\mathbb{R}} \frac{e^{ix}}{(x-\pi)^2+a^2}\; dx$

Change the variable to $u=x-\pi$ to get:

$I=\int_{\mathbb{R}} \frac{e^{i(u+\pi)}}{u^2+a^2}\; du=e^{i\pi}\int_{\mathbb{R}} \frac{e^{iu}}{u^2+a^2}\; du$

Now the real part of the integrand of the right most integral is symmetric and the imaginary part is anti-symmetric, and as for $a \ne 0$ the integral exists the imaginary part is zero. So:

$I=-\int_{\mathbb{R}} \frac{\cos(u)}{u^2+a^2}\; du=-2\int_{0}^{\infty} \frac{\cos(u)}{u^2+a^2}\; du$

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