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Math Help - Computing Integral

  1. #1
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    Question Computing Integral

    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

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  2. #2
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    Re: Computing Integral

    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}
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  3. #3
    Grand Panjandrum
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    Re: Computing Integral

    Quote Originally Posted by mathshelpee View Post
    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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