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Fourier transform of a supergausian

  1. Aug 28, 2012 #1
    I was wondering if anyone might know what the analytic fourier transform of a Super-Gaussian is?
  2. jcsd
  3. Aug 28, 2012 #2


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    That depends on what you mean by "super-Gaussian" distribution. Do you mean fat-tailed distributions? (distributions with tails that fall off rather slowly compared to a Gaussian).

    If so, then there are several kinds of fat tailed distributions, each with its own fourier transform. The fourier transform of the probability density function is just the characteristic function for the distribution, which are usually listed on the wikipedia page for the distribution of interest.
  4. Aug 28, 2012 #3
    thanks mute for the reply. The supergaussian i refer to is the Aexp(-([itex]\frac{x}{a}[/itex])[itex]^{2n}[/itex]) where n is positive an integer and is the order of the supergaussian.
  5. Aug 28, 2012 #4


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    Hm, that looks tricky. Looking in the integral book Gradshteyn and Ryzhik, I found two useful integrals. The first is

    $$\int_0^\infty dx~\exp(-x^\mu) = \frac{1}{\mu}\Gamma\left(\frac{1}{\mu}\right),$$
    which holds for ##\mbox{Re}(\mu) > 0## and can be used to find the normalization constant of your distribution. This is integral 3.326-1 in the sixth edition.

    There does not appear to be an integral for ##\exp(-x^\mu+ix)##, so it's possible there may not be a closed form for the fourier transform. However, you could expand the imaginary exponential in a power series and perform the integral term-by-term to get a power series representation of the fourier transform. In this case, the following integral (3.326-2) is useful:

    $$\int_0^\infty dx~x^m \exp(-\beta x^n) = \frac{\Gamma(\gamma)}{n\beta^\gamma},$$
    where ##\gamma = (m+1)/n##.
  6. Aug 29, 2012 #5


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    I should note that a) the series won't actually be a power series (because the moments aren't powers of anything) and b) the series looks like it will be at best an asymptotic series, as ##\Gamma((m+1)/n)## will grow quite large as ##m## gets large, and so the sum won't actually converge.
  7. Aug 31, 2012 #6
    Hi Mute, thanks for the replies. looks like i might have to stick to numerical transforms!
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