For the signal:
$ X(\omega)= 2\pi \delta(\omega) + 3\pi \delta(\omega - 3\pi) - 4\pi \delta(\omega + 5\pi) $
$ x(t) = \frac{1}{2\pi} \int_{-\infty}^\infty (2\pi \delta(\omega) + 3\pi \delta(\omega - 3\pi) - 4\pi \delta(\omega + 5\pi)) e^{j\omega t} \mathrm{d}\omega $
$ = \int_{-\infty}^\infty ( \delta(\omega) + \frac{3}{2} \delta(\omega - 3\pi) - 2 \delta(\omega + 5\pi)) e^{j\omega t} \mathrm{d}\omega $
$ x(t) = 1 + \frac{3}{2}e^{j3\pi t} - 2e^{-5\pi t} $
