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

$\displaystyle f: \mathbb{R}^n \rightarrow \mathbb{R}$

$\displaystyle f(x)=e^{-\pi a |x|^2}$

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

Thank you.

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- Oct 18th 2008, 02:26 PMmariannefourier transform
I need to find the Fourier transform for the following function:

$\displaystyle f: \mathbb{R}^n \rightarrow \mathbb{R}$

$\displaystyle f(x)=e^{-\pi a |x|^2}$

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

Thank you. - Oct 19th 2008, 02:53 AMLaurent
First note that, by Fubini theorem, the Fourier transform equals: for every $\displaystyle \xi\in\mathbb{R}^n$,

$\displaystyle F(\xi)=\int e^{-2i\pi x\cdot \xi}e^{-\pi a |x|^2}dx = \prod_{k=1}^n \int e^{-2i\pi x_k \xi_k} e^{-\pi a x_k^2} dx_k$

(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 $\displaystyle F$ of the function $\displaystyle x\mapsto e^{-\pi a x^2}$ from $\displaystyle \mathbb{R}$ to $\displaystyle \mathbb{R}$.

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 $\displaystyle F$ (with respect to $\displaystyle \xi$) and integrate by parts (by integrating $\displaystyle x e^{-\pi a x^2}$ and dividing $\displaystyle e^{-2 i\pi \xi x}$). You get $\displaystyle F'(\xi)=-\frac{2\pi\xi}{a} F(\xi)$.

Hence there is $\displaystyle C$ such that $\displaystyle F(\xi)=C e^{-\frac{\pi}{a}\xi^2}$.

To find the value of $\displaystyle C$, remark that $\displaystyle C=F(0)=\int e^{-\pi a x^2}dx=\frac{1}{\sqrt{\pi a}}\int e^{-u^2}du$. And $\displaystyle \int e^{-x^2}dx = \sqrt{\pi}$ (this can be proved by polar change of variable if you don't know that). As a consequence, $\displaystyle C=\frac{1}{\sqrt{a}}$. (There may be mistakes here)

Notice that if $\displaystyle a=1$, the function is equal to its own Fourier transform.