TITLE: Laplace transform of floor$(x)$
QUESTION [5 upvotes]: One way to compute the Laplace transform of floor$(x) = \left \lfloor{x}\right \rfloor$ (defined as the greatest integer $\leq x$) for $x$ positive is to use the definition of the transform.  Write out the integral, compute the telescoping series, and get the answer 
$$\mathcal{L}\left(\left\lfloor\frac{t}{a}\right\rfloor\right) = \frac{e^{-as}}{s(1-e^{-as})}$$
Is there an alternate/more clever way to compute this?  Using the definition is too much brute force for my liking!

REPLY [4 votes]: Late response but there's actually a very straightforward way. In plain English, the floor function plus the Heaviside function, all delayed by 1, is actually the floor function. So if $F(s)$ is the Laplace transform of the floor function, it must satisfy the functional equation:
$$
\left(F(s)+\frac1s\right) e^{-s} = F(s),
$$
therefore,
$$
F(s) = \frac1{s(e^s-1)}.
$$