I have been given that

$$\mathcal{F}\{xe^{-x^2}\} = - \frac{i\omega}{2^{3/2}} e^{-\frac{\omega^2}{4}} \tag{1}\label{1}.$$

I am supposed to use (1) and a table of known Fourier transforms to find the function $f$, when

$$xe^{-x^2} = \int_{-\infty}^{\infty} f(v)e^{-2(x-v)^2} dv \tag{2}\label{2}.$$

I have used the definition of the inverse-fourier transform, as well as attempting to Fourier transform both sides of (2), but none of those approaches yielded any results.

Any advice is greatly appreciated!


Fourier transforming both sides should work fine. The trick is that you can separate the resulting double integral: \begin{align} \mathcal{F}\left[\int_{-\infty}^\infty f(v)e^{-2(x-v)^2}dv\right] =& \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty\int_{-\infty}^\infty f(v)e^{-2(x-v)^2}e^{-i\omega x}dvdx \\ =& \frac{1}{\sqrt{2\pi}}\left[\int_{-\infty}^\infty f(v)e^{-i\omega v}dv\right]\left[\int_{-\infty}^\infty e^{-2(x-v)^2}e^{-i\omega (x-v)}dx\right] \\ & = \sqrt{2\pi}\mathcal{F}[f]\mathcal{F}\left[e^{-2x^2}\right]. \end{align} Can you take it from here?

  • $\begingroup$ Thanks! Yeah I believe so. Didn't see that I could separate the double integral! $\endgroup$
    – jakvah
    Jul 29 '19 at 21:04

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