What is the Fourier transform of $f(x) = x^2 e^{-x^2}$?

Question

I have a hand-in for my math course, and don't really know how to attack it. I've seen examples of $x e^{-x^2}$, but can't quite seem to do it with $x^2$.

Answer 1

For $f(x)=x^2e^{-x^2}$ let $g(x)=e^{-x^2}$ with it's Fourier transform is ${\cal F}(g)=\dfrac{1}{\sqrt{2}}e^{-\frac{w^2}{4}}$, then \begin{align} g(x) &=e^{-x^2} \\ g'(x) &=-2xe^{-x^2} \\ g''(x) &=4x^2e^{-x^2}-2e^{-x^2} \\ &=4f(x)-2g(x) \\ {\cal F}(g'')&=4{\cal F}(f)-2{\cal F}(g)\\ -w^2{\cal F}(g)&=4{\cal F}(f)-2{\cal F}(g)\\ {\cal F}(f) &= \dfrac{2-w^2}{4}{\cal F}(g)\\ {\cal F}(f) &= \dfrac{2-w^2}{4}\dfrac{1}{\sqrt{2}}e^{-\frac{w^2}{4}} \end{align}

Answer 2

Hint:

You can use the fact that the Fourier Transform of $x^nf(x)$ is $$ i^n\frac{d^nF(\omega)}{d \omega^n} $$

where $F$ is the Fourier transform of $f$ ( and $\omega$ is the angular frequency)

Answer 3

Hint: Another method that is more laborious but easier to remember is to use the Taylor expansion:

$$x^2\exp(-x^2)=x^2\sum_{n=0}^{\infty}\frac{(-x^2)^n}{n!}=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+2}}{n!}.$$

This the problem boils down to determining Fourier transforms of powers. This method might be not the best for this problem but sometimes it is very useful. Especially for Laplace transform this method is very powerful as powers of $x$ or $t$ can be determined very easily for the Laplace transform.