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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!

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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?

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  • $\begingroup$ 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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