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I've been trying to find an expression for the Fourier transform of the function:

$e^{-|x|^n}$

I believe it doesn't have a closed form, but someone did get an expression in terms of an infinite series:

The Fourier transform of $ e^{-|x|^\alpha}, \alpha>0. $

It's a nice solution, but I was looking for something I could evaluate numerically, so I can't use that expression since it's an infinite sum. I tried cutting the series up to the 100th term, but after a certain number of terms my computer cannot longer handle the expression. I was wondering if someone could suggest me a way of getting an approximate expression that doesn't require an infinite sum.

Since I'm dealing with a negative exponential, I would expect that my integral converges to something. I just don't know what approach to take.

During my search I also found this post where they talk about something called the "Method of Steepest Descent", but I'm not familiarized with that method, so I'm unsure if it will help me at all.

Asymptotic behaviours from Fourier transforms

I've been scratching my head with this, but I can't find a way around this.

Thanks.

Alb
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  • The main asymptotic term comes directly. Given that $x^n\to 0$ at $x<1$ and $x^n\to \infty$ at $x>1$ $$\int_{-\infty}^\infty e^{-|x|^n}e^{2\pi ik x}dx=2\int_0^\infty e^{-|x|^n}\cos(2\pi k x)dx=2\int_0^1\cos(2\pi k x)dx+o(1)$$ – Svyatoslav Dec 09 '23 at 12:11
  • Thanks for trying to answer the question! – Alb Dec 12 '23 at 01:20
  • That's correct but only if you consider $n \rightarrow \infty$ right? if $n$ is a value somewhere between 1 and 2, should we still consider this argument, correct? I believe not, because that would be saying that the integral of a gaussian is the same as the integral of $$e^{-|x|}$$ I was hoping to get something that would give me a distinction between Fourier Transforms depending on the $n$ exponent. – Alb Dec 12 '23 at 01:26
  • Yes, it is valid for $n\to\infty$; for just $n>1$ there will be correction terms. I'm not sure that we can get these corrections in a nice form... – Svyatoslav Dec 12 '23 at 17:02
  • I see. Thank you. I appreciate your answers. – Alb Dec 13 '23 at 22:54

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