How to take the Laplace transform of floor(t)

Question

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:)

Answer 1

I think I would add to vonbrand answer

$\begin{align*} S(s) &= \sum_{n \ge 0} e^{-n s} \\ &= \frac{1}{1 - e^{-s}} \end{align*}$

How does he get that?

Let's start with

$$ S(s) = \sum_{n \ge 0} e^{-n s} $$

That's how we define it.

Let's multiply both side by $e^{- s}$

$$ e^{- s} S(s) =e^{- s} \sum_{n \ge 0} e^{-n s} $$

$$= \sum_{n \ge 0} e^{-(n+1) s} $$

$$= \sum_{n \ge 1} e^{-n s} $$

Notice that multiplying with $e^{- s}$ we still have the same sum but losing it's first term. Hence,

$$ (1-e^{- s}) S(s)= \sum_{n \ge 0} e^{-n s}- \sum_{n \ge 1} e^{-n s} = e^{-0 s}=1 $$

Dividing both sides with $(1-e^{- s})$ we got

$$ S(s) = \sum_{n \ge 0} e^{-n s} =\frac{1}{1-e^{-s}}$$

Answer 2

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., $s > 0$.