So I am trying to take the Laplace transform of $\text{floor}(t)$, and I am stuck.
First of all $\text{floor}(t)$ is defined to be the highest integer $<=t$
With that in mind, here is what I tried to do:
$$f(t)=\text{floor}(t)$$
$$f(t)=u(t-1)+u(t-2)+u(t-3)+…$$
Where $~u(t)~$ is the unit step function.
By using the definition of the Laplace transform:
$$ \mathcal{L}{f}=\int_0^\infty e^{-st}\sum_{i=1}^\infty u(t-i)dt $$
$$ \mathcal{L}{f}= \int_1^2e^{-st}1dt + \int_2^3 e^{-st}2dt + \int_3^4 e^{-st}3dt + … $$
$$ \mathcal{L}{f} = \sum_{i=1}^{\infty} n\int_n ^{n+1}e^{-st} dt$$
$$ \mathcal{L}{f} = \frac{1}{s}(1-e^{-s}) \sum_{i=1}^{\infty} ne^{-sn} $$
Since I don't know how to compute this infinite series I don't know how to go forward from here. If anyone can help me with that it would be much appreciated. Also if there is a "better" way to solve this or I am doing something wrong, which is likely, I would appreciate if you point it out:)
Best Answer
Consider the sum:
$\begin{align*} S(s) &= \sum_{n \ge 0} e^{-n s} \\ &= \frac{1}{1 - e^{-s}} \end{align*}$
So your mystery sum is just:
$\begin{align*} -S'(s) &= \sum_{n \ge 0} n e^{- n s} \\ &= \frac{e^s}{(1 - e^s)^2} \end{align*}$
The frobbing is valid as long as $e^{-s} < 1$, i.e., $s > 0$.