Do decaying exponential signals have finite energy? 
"A signal that decays exponentially has finite energy, so, it is also
  an energy signal."
http://www.songho.ca/dsp/signal/signals.html


I don't quite get how that can be true. Energy of a signal is defined as:
$$
E_s = \langle x(t),x(t)\rangle = \int_{-\infty}^\infty |x(t)|^2\,dt.
$$
According to Wolfram Alpha, this integral does not converge for $e^{-2x}$, so how can this qualify as a finite energy signal?
 A: Probably they mean to consider the integral starting from a particular point in time (as opposed to all the way to $-\infty$), in which case the integral is finite.
A: Following up on the comment discussion:
Courses are written by humans, who are at times imperfect.  An expert in signal analysis might see the term "signal that decays exponentially" and immediately picture a function like $$x(t) = \begin{cases}e^{-t},& t\ge 0 \\ 0, &t < 0,\end{cases}$$
or $x(t) = e^{-|t|}$ as Ian hinted at.  This interpretation would be entirely natural to an expert because


*

*they have probably seen such "decaying pulse" signals many times before;

*it occurs in a context immediately following discussion of a single square wave pulse (which is finitely supported);

*the alternative interpretation of a signal that grows unbounded in one direction would be a signal of infinite power and doesn't make sense in this context.


It would be very easy for such a human being, especially someone writing without feedback from an coauthor/editor/referee, to overlook the possibility that a beginner will see the term "signal that decays exponentially" and picture the function $x(t) = e^{-t}, -\infty < t < \infty$.
In short, yes, decaying exponential signals have finite energy, but $e^{-t}$ by itself is not a decaying exponential signal.
