Laplace transform of floor$(x)$ 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!
 A: 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,
$$
F(s) = \frac1{s(e^s-1)}.
$$
A: There is an alternative way of computing $\mathcal{L}\left\{\left\lfloor\dfrac{t}{a}\right\rfloor\right\}$.  
Consider the sawtooth wave of period $a$.  Let us define one period of this function by $f_a(t) = \dfrac{t}{a}(u(t) - u(t-a))$ (several periods of $f(t)$ are shown below).

We now can define $\left\lfloor\dfrac{t}{a}\right\rfloor = \dfrac{t}{a} - f(t)$ for any $t\geq 0$ (which results in the familiar graph below).

Therefore, $$\mathcal{L}\left\{\left\lfloor\frac{t}{a}\right\rfloor\right\} = \mathcal{L}\left\{\frac{t}{a}\right\} - \mathcal{L}\{f(t)\} = \frac{1}{as^2} - \mathcal{L}\{f(t)\}.$$
To compute the Laplace transform of $f(t)$, you'll need to recall the Laplace transform formula for periodic functions; that is, if $g(t)$ is piecewise continuous for $t\geq 0$ and is $p$-periodic, then 
$$\mathcal{L}\{g(t)\} = \frac{1}{1-e^{-ps}}\mathcal{L}\{g_p(t)\} = \frac{1}{1-e^{-ps}}\int_0^p e^{-st}g_p(t)\,dt$$
where, again, $g_p(t)$ is one period of $g(t)$.
To get that formula, though, you'd have to rewrite the improper integral (from the integral definition) as an infinite series and do some straightforward computations.  For now, let's assume that we have the luxury of using this formula.  Also, to help us compute $\mathcal{L}\{f_a(t)\}$, recall the derivative formula $\mathcal{L}\{tf_a(t)\} = -\dfrac{d}{ds}\mathcal{L}\{f_a(t)\}$. 
We now see that 
$$\begin{aligned} \mathcal{L}\{f_a(t)\} &= \frac{1}{a}\mathcal{L}\{t(u(t) - u(t-a))\} \\ &= -\frac{1}{a}\frac{d}{ds}\left[\mathcal{L}\{u(t)\} - \mathcal{L}\{u(t-a)\}\right]\\ &= -\frac{1}{a}\frac{d}{ds}\left(\frac{1}{s} - \frac{e^{-as}}{s}\right)\\ &= -\frac{1}{a}\left(-\frac{1}{s^2} - \frac{-ase^{-as} - e^{-as}}{s^2}\right)\\ &= \frac{1-e^{-as}-ase^{-as}}{as^2}\end{aligned}$$ 
Thus,
$$\mathcal{L}\{f(t)\} = \frac{1}{1-e^{-as}} \mathcal{L}\{f_a(t)\} = \frac{1-e^{-as}-ase^{-as}}{as^2(1-e^{-as})}$$
and therefore, we have that
$$\mathcal{L}\left\{\left\lfloor\frac{t}{a} \right\rfloor\right\} = \mathcal{L}\left\{\frac{t}{a}\right\} - \mathcal{L}\{f(t)\} = \frac{1}{as^2} - \frac{1-e^{-as}-ase^{-as}}{as^2(1-e^{-as})} = \frac{e^{-as}}{s(1-e^{-as})}.$$
