# How to calculate the following fourier transform?

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?

• 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. Sep 24 '15 at 9:39
• @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 $x^2 e^{-x^2}$? Sep 24 '15 at 9:45
• 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 Sep 24 '15 at 12:32
• @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. Sep 24 '15 at 12:54

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