Fourier transform of a supergausian

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Discussion Overview

The discussion revolves around the analytic Fourier transform of a Super-Gaussian function, specifically the form Aexp(-(\frac{x}{a})^{2n}), where n is a positive integer. The scope includes theoretical aspects and mathematical reasoning related to Fourier transforms and integral calculus.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant inquires about the analytic Fourier transform of a Super-Gaussian function.
  • Another participant suggests that the term "super-Gaussian" could refer to fat-tailed distributions, noting that different fat-tailed distributions have distinct Fourier transforms.
  • A participant specifies the Super-Gaussian as Aexp(-(\frac{x}{a})^{2n}) and mentions n as a positive integer.
  • Another participant references integrals from Gradshteyn and Ryzhik, indicating that while there is a normalization constant for the distribution, a closed form for the Fourier transform may not exist.
  • This participant proposes expanding the imaginary exponential in a power series and integrating term-by-term to derive a power series representation of the Fourier transform.
  • It is noted that the resulting series may not be a true power series and could be an asymptotic series due to the growth of the Gamma function as m increases.
  • A later reply expresses a preference for numerical transforms over analytical methods.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the existence of a closed form for the Fourier transform of the specified Super-Gaussian function, and multiple approaches and uncertainties are present in the discussion.

Contextual Notes

The discussion highlights limitations regarding the closed form of the Fourier transform and the nature of the series derived from the expansion of the imaginary exponential, which may not converge.

modaniel
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Hi,
I was wondering if anyone might know what the analytic Fourier transform of a Super-Gaussian is?
cheers
 
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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.
 
thanks mute for the reply. The supergaussian i refer to is the Aexp(-(\frac{x}{a})^{2n}) where n is positive an integer and is the order of the supergaussian.
 
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##.
 
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.
 
Hi Mute, thanks for the replies. looks like i might have to stick to numerical transforms!
 

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