fourier transform

**marianne** · October 18th 2008, 02:26 PM

I need to find the Fourier transform for the following function:

$f: \mathbb{R}^n \rightarrow \mathbb{R}$
$f(x)=e^{-\pi a |x|^2}$

I'm a newbie when it comes to Fourier transforms so any help would be great.

Thank you.

**Laurent** · October 19th 2008, 02:53 AM

Originally Posted by marianne

I need to find the Fourier transform for the following function:

$f: \mathbb{R}^n \rightarrow \mathbb{R}$
$f(x)=e^{-\pi a |x|^2}$ .

First note that, by Fubini theorem, the Fourier transform equals: for every $\xi\in\mathbb{R}^n$ ,
$F(\xi)=\int e^{-2i\pi x\cdot \xi}e^{-\pi a |x|^2}dx = \prod_{k=1}^n \int e^{-2i\pi x_k \xi_k} e^{-\pi a x_k^2} dx_k$
(you may have a slightly different definition of the Fourier transform; there are several conventions)

This shows that it suffices to find the Fourier transform $F$ of the function $x\mapsto e^{-\pi a x^2}$ from $\mathbb{R}$ to $\mathbb{R}$ .
There are various proofs for this, and I think you should have been given hints to find it.
A possibility is to show that the Fourier transform satisfies a first order differential equation. Here are the steps:
Differentiate $F$ (with respect to $\xi$ ) and integrate by parts (by integrating $x e^{-\pi a x^2}$ and dividing $e^{-2 i\pi \xi x}$ ). You get $F'(\xi)=-\frac{2\pi\xi}{a} F(\xi)$ .
Hence there is $C$ such that $F(\xi)=C e^{-\frac{\pi}{a}\xi^2}$ .
To find the value of $C$ , remark that $C=F(0)=\int e^{-\pi a x^2}dx=\frac{1}{\sqrt{\pi a}}\int e^{-u^2}du$ . And $\int e^{-x^2}dx = \sqrt{\pi}$ (this can be proved by polar change of variable if you don't know that). As a consequence, $C=\frac{1}{\sqrt{a}}$ . (There may be mistakes here)
Notice that if $a=1$ , the function is equal to its own Fourier transform.

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