The following signals are shown to be either an energy signal or a power signal
$ \,\!x(t)=e^{-at}u(t) $ for a > 0
solution:
since $ Energy(\infty) = \int_{-\infty}^{\infty} \! |x(t)|^2\ dt $ ,
$ = \int_{0}^{\infty}\!e^{-2at}dt $ $ =\frac{1}{2a} < {\infty} $
therefore x(t) is an energy function because the energy is finite, and not a power function.
A consequence of this is that P=0. If the energy of the signal was infinite, then the power would be found to have a finite value.
