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Definite integral using complex analysis

  1. Oct 23, 2011 #1
    1. The problem statement, all variables and given/known data
    Considering the appropriate complex integral along a semi-circular contour on the upper half plan of z, show that
    [itex] \int^{\infty}_{\infty} \frac{cos(ax)}{x^2 + b^2} dx = \frac{\pi}{b}e^{-ab} (a>0, b>0) [/itex] ​


    2. Relevant equations
    [itex] \int_{C} = 0 [/itex] For C is a semi-circle of infinite radius in the upper half of the complex plane.
    [itex] \oint f(z)dz = 0 [/itex] for analytic functions
    [itex]\oint f(z)dz = 2\pi i \sum Res[f(z_{0})] [/itex] Where z0 is a singularity and you sum over all the residues within the contour.
    [itex] z^2 + b^2 = (z+ib)(z-ib) [/itex]



    3. The attempt at a solution
    I've been trying to solve this by using [itex] \int \frac{e^{iaz}}{z^2 + b^2}dz [/itex]. Which must be solved using Cauchy's Formula (the residue thing above), since f(z) is not analytic over all space. There is a singularity of order one at [itex]\pm ib[/itex].
    Is there a better contour integral I could use? Right now the algebra is getting messy.
     
  2. jcsd
  3. Oct 23, 2011 #2

    vela

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    Show us your work. You must be doing something wrong because that contour integral is pretty straightforward to evaluate.
     
  4. Oct 24, 2011 #3
    I tried using the contour integral [itex]\int^{\infty}_{\infty} \frac{e^{-iaz}}{z^2+b^2}dz[/itex] and found that it was equal to [itex]\frac{\pi}{b}e^{-ab}[/itex]. Now I think I have to consider that a sum of different integrals. We have the top hemisphere, the x-axis and the circular contour about z0 and the lines leading to +ib. I'm finding that must of those are zero and I think I can pull out the cosine function from the integral along the x-axis...

    I think before I was just forgetting how to factor...
     
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