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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You will end up with some kind of convolution – Graham Hesketh May 25 '18 at 16:17
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The IFT will be $$\frac{d}{dt}\int_{-\infty}^\infty f(t-t')g(t'),dt'=\frac{d}{dt}\int_{-\infty}^\infty f(t')g(t-t'),dt'$$ – Mark Viola May 25 '18 at 16:19
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Allright. Thank's a lot – 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:
$\mathcal{F}^{-1}\left(i\omega \hat{f}(\omega)\hat{g}(\omega)\right)(t)=\frac{d}{dt}\mathcal{F}^{-1}\left(\hat{f}(\omega)\hat{g}(\omega)\right)(t)=\frac{d}{dt}\mathcal{F}^{-1}\left(\widehat{(f*g)}(\omega)\right)(t)=\frac{d}{dt}(f*g)(t)$
Just remember (and never forget) that convolution turns into multiplication after applying the Fourier transform and vice versa.
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