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How can I find the Fourier transform of:

$$f(t) = te^{-t^2}$$

user91500
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Amir Šabanović
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    what have you tried? do you know what is the Fourier transform of $e^{-t^2}$. Then what happens if you multiply by a t? (see 107 in this link: http://en.wikipedia.org/wiki/Fourier_transform#Convolution_theorem under functional relationship) – Lost1 Jun 25 '14 at 15:55
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    Can you tell us what notation do you use for Fourier transform? Is it $\displaystyle F(\omega)=\int_{-\infty}^\infty f(t)e^{-i\omega t}dt$? – Tunk-Fey Jun 25 '14 at 15:55
  • @Lost1 Thanks! Never spotted these down there. – Amir Šabanović Jun 25 '14 at 16:06
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    no worries. it is just a consequence of the fact multiplying ik is the same as differentiating. – Lost1 Jun 25 '14 at 16:11

1 Answers1

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Hint: If using the Fourier transform $\hat{f}(\xi) = \int\limits_{-\infty}^{\infty}f(x)e^{-2\pi i x\xi}dx$,
then $x^nf(x) $ has the Fourier transform $\left(\frac{i}{2\pi}\right)^n\frac{d^n\hat{f}(\xi)}{d\xi}$,
and $f(x) = e^{-\alpha x^2}$ has the Fourier transform $\hat{f}(\xi) = \sqrt{\frac{\pi}{\alpha}}e^{-(\pi\xi)^2/\alpha}$.

What happens if you try to combine them?

fromGiants
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