A Laplace Transform Example
$ x(t)=-e^{-2t}u(-t) $
Therefore the Laplace Transform is:
$ X(s) = \int_{-\infty}^{\infty} x(t)e^{-st}dt $
$ = \int_{-\infty}^{\infty} -e^{-2t}u(-t)e^{-st}dt $
$ = \int_{-\infty}^{0}-e^{-2t}e^{-st}dt $
$ = \int_{-\infty}^{0}-e^{-2t}e^{-(a+j\omega )t}dt $
$ = \int_{-\infty}^{0}-e^{-(2+a)t}e^{-j\omega t}dt $
$ = \frac{-e^{-(2+a)t}e^{-j\omega t}}{-(2+a+j\omega )}|_{-\infty}^{0} $
If 2+a<=0 then integral diverges
else:
$ X(s) = \frac{-1}{-(2+a+j\omega)} - 0 $
$ = \frac{1}{2+s} $
With a ROC: a<-2
