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What is the inverse Fourier transform of $i\omega f(\omega)g(\omega)$?

is it just $\frac{d}{dt}(f(t)\cdot g(t))$ or will I end up with some kind of convolution?

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  • $\begingroup$ You will end up with some kind of convolution $\endgroup$ – Graham Hesketh May 25 '18 at 16:17
  • $\begingroup$ The IFT will be $$\frac{d}{dt}\int_{-\infty}^\infty f(t-t')g(t')\,dt'=\frac{d}{dt}\int_{-\infty}^\infty f(t')g(t-t')\,dt'$$ $\endgroup$ – Mark Viola May 25 '18 at 16:19
  • $\begingroup$ Allright. Thank's a lot $\endgroup$ – OD IUM May 25 '18 at 16:20
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It will be a convolution by the convolution theorem for inverse Fourier transforms:

$\mathcal{F}^{-1}\left(i\omega \hat{f}(\omega)\hat{g}(\omega)\right)(t)=\frac{d}{dt}\mathcal{F}^{-1}\left(\hat{f}(\omega)\hat{g}(\omega)\right)(t)=\frac{d}{dt}\mathcal{F}^{-1}\left(\widehat{(f*g)}(\omega)\right)(t)=(f*g)(t)$

Just remember (and never forget) that convolution turns into multiplication after applying the Fourier transform and vice versa.

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