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Inverse fourier transform of gaussian

  1. Oct 10, 2011 #1
    well, i have to prove that the inv. fourier transform of a gaussian (e^(-(k^2/2)) is a gaussian, i know some elementary complex analysis(never actually taken a class in it), not well enough, it seems, to find the solution to this. I tried to integrate over a circular contour, and let the radius of the circle go to infinity, i couldn't solve the integral that i obtained (it was pretty complicated). Also, i don't quite understand this, the integrand is complex analytic everywhere, so if i integrate it over a circular contour, wouldn't i get 0, by cauchy's theorem?
    Any help much appreciated
     
  2. jcsd
  3. Oct 10, 2011 #2

    diazona

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    Sure, but why are you integrating over a circular contour in the first place?

    Start by writing out the integral you have to do to find the inverse Fourier transform.
     
  4. Oct 10, 2011 #3
    well its the integral (e(-k^2)e^ikx)dk over the entire complex plane right?, so unless im wrong there are two ways to integrate over the complex plane, a circle of infinite radius, or k=x+iy, and x,y go to infinity
    riight??
     
  5. Oct 10, 2011 #4

    diazona

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    For one thing, you can't integrate over the entire complex plane. Well, you can, but it has to be a double integral, which is not what you have here. There's only one differential (dk), so you only get to integrate in one dimension.

    Check your references if you need to, in order to find the correct limits of integration for the integral
    [tex]\int_?^? e^{-k^2/2}e^{ikx}\mathrm{d}k[/tex]
    which is involved in the Fourier transform.
     
  6. Oct 10, 2011 #5
    im fairly certain that the limits are -inf. to inf . well yes, i get that you only integrate in one variable, but isn't it true that a complex number a+bi, with arbitrary a and b can span the entire complex plane? so you would have a line integral in da and idb, with both a and b (-∞,∞) or equivalently circle of radius of r, a=rcosθ ; b= rsinθ with r going to ∞.
    sorry if im being slow btw
     
  7. Oct 11, 2011 #6

    diazona

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    Yes, that's correct.
    Not just one variable, but one dimension. A single integral with a single differential dk is a one-dimensional integral. A dimension corresponds to one real variable.
    Yes, but if you have both da and db, then it's not a line integral, it's a (two-dimensional) surface integral.
     
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