The Fourier transform of $ f(x)=xe^{{-x}^2} $

Question

I know how to calculate the Fourier Transform using the main formula and integrals calculations. But i want to calculate the Fourier Transform of

$ f(x)=xe^{{-x}^2} $

using the following formula :

for $f(x)=e^{{-ax}^2} (a>0) $ then $ \mathbb{F}(f)=\frac{1}{\sqrt{2a}} e^{{{-\omega}^2}/4a} $

How can i do that?

Note: Should I also use the rule of $ \mathbb{F}\{f'(x)\}=i\omega\mathbb{F} \{f(x)\} $ ?

Answer 1

You can use the derivative formula of the Fourier transform.

Assume $\mathcal{F}\{f(x)\}=F(\omega)$. then $$\mathcal{F}\{f'(x)\}=(i\omega)F(w)$$

For $f(x)=e^{-ax^2}$ and $F(\omega)=\frac{1}{\sqrt{2a}} e^{{{-\omega}^2}/4a}$ we have $f'(x)=(-2ax)e^{-ax^2}$ . Hence, $$\mathcal{F}\{(-2ax)e^{-ax^2}\}=\frac{i\omega}{\sqrt{2a}} e^{{{-\omega}^2}/4a}$$ Therefore, choosing $a=1$ and simplifying we have

$$\boxed{\mathcal{F}\{xe^{-x^2}\}=\frac{i\omega}{-2\sqrt{2}} e^{{{-\omega}^2}/4}}$$

Answer 2

$\displaystyle\mathcal{F}(xe^{-x^2} )(\omega)= \frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^{+\infty} xe^{-x^2}e^{-i\omega x} dx =$

$\displaystyle=\left(-\frac{1}{\sqrt{2\pi}}\frac{1}{2} e^{-x^2}e^{-i\omega x}\right)|_{-\infty}^{+\infty} -\frac{i\omega}{2\sqrt{2\pi}}\int\limits_{-\infty}^{+\infty}e^{-x^2}e^{-i\omega x} dx$

$\displaystyle=0-\frac{i\omega}{2}\mathcal{F}(e^{-x^2})(\omega)=-\frac{i\omega}{2\sqrt{2}} e^{-\omega^2/4} $