TITLE: The existence of the laplace transform
QUESTION [5 upvotes]: I don't understand why the laplace transform of some function, say f(t), has to be "piecewise continuous" and not "continuous". Is "piecewise continuous" sort of like the minimum requirement? This troubles me because I don't think f(t)=t is piecewise continuous, it's simply continuous...
Also, I don't understand what's so special about a function being of "exponential order". Can someone explain to me why this property is so special and apparently makes the Laplace Transform exist?
Also, my teacher says that f(t) has to be of exponential order from [0,infinity] and other sources say that it at least has to be of exponential order from [T,infinity] where T>0.
Thanks.

REPLY [3 votes]: Any continuous function can be said to be piecewise continuous. Saying only "piecewise continuous" is just a way to use fewer words.
In fact, we can relax the requirement of piecewise continuity, and just require that $f(t)$ be integrable on any finite subset of $[0,\infty)$. An example of such a function which is not continuous on $[0,\infty)$ is $f(t)=t^{-1/2}$.