Consider the following function:
$F(\omega )=\frac{2 \sin \left(a \omega ^2\right) \sin (b \omega )}{\omega }=\frac{\cos \left(a \omega ^2-b \omega \right)-\cos \left(a \omega ^2+b \omega \right)}{\omega }$
Mathematica calculates the same inverse Fourier for the two functions in the right-hand side, namely,
$ \mathcal{F}_{\omega }^{-1}\left[\frac{\cos \left(a \omega ^2+b \omega \right)}{\omega }\right](t)=\frac{\log (-i (b+t))-\log (i (b+t))-\log (i t-i b)+\log (i b-i t)}{2 \sqrt{2 \pi }} $
$ \mathcal{F}_{\omega }^{-1}\left[\frac{\cos \left(a \omega ^2-b \omega \right)}{\omega }\right](t)=\frac{\log (-i (b+t))-\log (i (b+t))-\log (i t-i b)+\log (i b-i t)}{2 \sqrt{2 \pi }} $
Therefore, Mathematica yields zero value for $\mathcal{F}_{\omega }^{-1}\left[\frac{2 \sin (a \omega ) \sin \left(b \omega ^2\right)}{\omega }\right](t)$, which is incorrect. How can I solve such problem in Mathematica?