- #1

modaniel

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I was wondering if anyone might know what the analytic Fourier transform of a Super-Gaussian is?

cheers

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- Thread starter modaniel
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- #1

modaniel

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I was wondering if anyone might know what the analytic Fourier transform of a Super-Gaussian is?

cheers

- #2

Mute

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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.

- #3

modaniel

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- #4

Mute

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$$\int_0^\infty dx~\exp(-x^\mu) = \frac{1}{\mu}\Gamma\left(\frac{1}{\mu}\right),$$

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##.

- #5

Mute

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- #6

modaniel

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Hi Mute, thanks for the replies. looks like i might have to stick to numerical transforms!

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