# inverse Fourier transform of product of two functions

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?

• You will end up with some kind of convolution – Graham Hesketh May 25 '18 at 16:17
• 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
• Allright. Thank's a lot – OD IUM May 25 '18 at 16:20

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