The problem
Find the Laplace Transform of the following: $ x(t)=e^{-2t}u(t) $
Solution
Now, from other classes we have become used to simply looking up a simple problem like this in a Laplace table. This may or may not be valid on exam 3 and the final, so we have to know how to solve the problem using the definition of a Laplace Transform.
$ X(s)=\int_{-\infty}^{+\infty}x(t)e^{-st}dt $
$ =\int_{0}^{\infty}e^{-2t}e^{-st}dt $
$ =\int_{0}^{\infty}e^{-(2+s)t}dt $
$ =\int_{0}^{\infty}e^{-(2+a+j\omega)t}dt $
We can now see that if if $ 2+a\leq 0 $, the integral diverges and if $ 2+a>0 $:
$ X(s)=\frac{e^{-(2+a)t}e^{-j\omega t}}{-(2+s)} \bigg| _0^{\infty} $
$ =0-\frac{1}{-2-s} $ $ =\frac{1}{2+s} $
