Beginner question about existence of Laplace transform

Question

I am having problems understanding why a Laplace transform exists or not. Here is my math and logic, hopefully someone can point out where I am wrong.

$$f(t)=e^{at} \implies ℒ[e^{at}] = F(s)=\int_0^{\infty}\exp([a-s]t)dt$$

From this point I am at a loss both mathematically and notationally, but here I go anyways:

$$\implies \frac{1}{a-s}\lim_{t \to\infty}\exp([a-s]t)$$

According to Schaum's "Laplace transform":

The Laplace transform of f(t) is said to exist if the integral converges for some value of s; otherwise it does not exist.

What does this mean for my case? The way I understand it is that $s>a$, because only then the integral will converge. Is this correct? And if it is, why can we just assume that $s>a$?

I am really frustrated, I have tried to look it up on the internet and here, but I just do not understand it.

Edited from $s>t$ to $s>a$.

Answer 1

Taking $s$ and $a$ to be real, the integral defining $F(s)$ in your case converges provided $s>a$. This specifies the domain of $F$, which is the interval $(a,\infty)$. When $s$ is real, the transform is only intended to be defined for $s$ large enough to make the integral converge.

To understand this it may help to note that the Laplace transform produces another function, which may in fact have another domain.

Answer 2

$s$ is complex of the form $s=\sigma +j\omega$ where $\sigma,\omega\in\mathbb{R}$ and $j=\sqrt{-1}$, so for convergence, the real part of $s$ has to be greater than zero when we have $e^{-st}$. Now, let's take your example, we have that an exponential will decay or blow up. The argument of the exponential is $a-s < 0$ for $t\in(0,\infty)$ so $a<s$ or the integral will converge when $\text{Re}\{s\} > a$.

On the Wikipedia page, you will see the formal definition and that $s\in\mathbb{C}$.