Let the signal $ X(\omega) $ be equal to:
$ X(\omega) = \delta(\omega) + \delta(\omega - 2) - \delta(\omega - 3) \, $
The Inverse Fourier Transform of a signal in Continuous Time is:
$ x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}X(\omega)e^{j\omega t}d\omega \, $
Using this, we obtain:
$ x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}(\delta(\omega)e^{j\omega t} + \delta(\omega - 2)e^{j\omega t} - \delta(\omega - 3)e^{j\omega t}) d\omega \, $
$ x(t) = \frac{1}{2\pi}(e^{j\omega t} +e^{j2\omega t} - e^{j3\omega t}) \, $
