Signal
$ x(t) = e^{-2t} u(t) + e^{-5t} u(t-4) \! $
Fourier Transform
$ X(\omega) = \int_{-\infty}^{\infty} x(t) e^{-j\omega t} dt \! $
$ X(\omega) = \int_{0}^{\infty} e^{-2t} e^{-j\omega t} dt + \int_{4}^{\infty} e^{-5t} e^{-j\omega t} dt \! $
$ X(\omega) = \int_{0}^{\infty} e^{-(2+j\omega )t} dt + \int_{4}^{\infty} e^{-(5+j\omega )t} dt \! $
$ X(\omega) = {\left. \frac{e^{-(2+j\omega )t}}{-(2+j\omega )} \right]^{\infty}_0 } + {\left. \frac{e^{-(5 + j\omega )t}}{-(5 +j\omega )} \right]^{\infty}_4 }\, $
$ X(\omega) = \frac{1}{2+j\omega } + \frac{e^{-(20+4j\omega )}}{5+j\omega } \! $
