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How do I calculate the fourier transform of $x^2 * e^{-x^2}$

If I let $f(x) = x^2$ and $g(x) = e^{-x^2}$, my attempt has been to calculate the fourier transform of $f$ and $g$ seperately and plug them into the convolution theorem formula

$transform(f*g) = \sqrt{2\pi} * transform(f) * transform(g)$

Is this the correct way to go about it, am I using the convolution theorem correctly?

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  • $\begingroup$ The problem is that the Fourier transform of $f $ only exists in the sense of tempered distributions.If you know about these, your idea will work. One thing about which I am not sure is if you are using the convolution theorem correctly, since you use the same symbol for multiplication and convolution. $\endgroup$
    – PhoemueX
    Sep 24 '15 at 9:39
  • $\begingroup$ @PhoemueX I am not entirely sure about the use of symbols, I definately meant multiplication. That is to calculate the fourier transform of the product $x^2 e^{-x^2}$ can I calculate the fourier transform of each of those functions seperately and plug them into the formula given by the product $\sqrt{2\pi}\cdot transform(f) \cdot transform(g)$ and get the fourier transform for $x^2 e^{-x^2}$? $\endgroup$ Sep 24 '15 at 9:45
  • $\begingroup$ I answered yesterday a question concerning the Fourier transform, that contains the answer to this question. I think you might be interested in reading it, since you asked the question. Any feedback on the answer (comment, acceptance) would also be greatly appreciated. Thanks $\endgroup$
    – Tom-Tom
    Sep 24 '15 at 12:32
  • $\begingroup$ @ArnoldDoveman: No, you have (up to multiplies of $2\pi $) that $F (fg)= F (f)\ast F (g) $, where $F $ is Fourier transform and $\ast $ is convolution. $\endgroup$
    – PhoemueX
    Sep 24 '15 at 12:54
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For any good enough function $\phi$ (like $e^{-x^2}$) $$ \int_{\mathbb{R}}x\,\phi(x)\,e^{ix\xi}\,dx=\frac1i\int_{\mathbb{R}}\phi(x)\,\frac{d}{d\xi}\,e^{ix\xi}\,dx=\frac1i\,\frac{d}{d\xi}\int_{\mathbb{R}}\phi(x)\,e^{ix\xi}\,dx. $$ Use twice.

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    $\begingroup$ ... or use the fact that $x^2\mathrm e^{-x^2}=\left.\partial_\lambda\mathrm e^{-\lambda x^2}\right\rvert_{\lambda=1}$ $\endgroup$
    – Tom-Tom
    Sep 24 '15 at 12:45

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