I'm trying to understand the statement of the Final Value Theorem for Laplace transforms. Unfortunately I don't own an authoritative reference, so I'm resorting to Wikipedia.
On this wikipedia page, it is stated that if $f(t)$ is a function on $(0, \infty)$ and $F(s)$ is its Laplace transform, then $\lim_{t \to \infty} f(t)$ exists and is equal to $\lim_{s\to 0} sF(s)$ on the condition that all of the poles of $sF(s)$ lie in the left half-plane.
However, this wikipedia page states it differently. It says that $f(t)$ is a bounded function, and if it is known that $\lim_{t \to \infty} f(t)$ exists and is finite, then $\lim_{t \to \infty}f(t) = \lim_{s \to 0} sF(s)$.
So my question is the following:
Is it true that if the poles of $sF(s)$ are all the left half-plane, then $f(t)$ is guaranteed to be bounded and to have a finite limit as $t \to \infty$? If so, why?
I can see a little bit of the intuition. A pole at $s = \alpha$ contributes a residue whose $t$-dependence is "similar to" $e^{\alpha t}$, and if ${\rm Re}(\alpha) < 0$, then $e^{\alpha t}$ tends to zero at infinity. Furthermore, a simple pole at $s = 0$ contributes a residue independent of $t$, and $\lim_{s \to 0} sF(s)$ is precisely the value of this residue. This residue is then the constant contribution that "persists" in the limit as $t \to \infty$. But this isn't a rigorous argument. For example, what happens if there are infinitely many poles in the left half-plane? And what exactly do I mean by "similar to"?
[By the way, it's also annoying that Wikipedia doesn't say what kind of function $f$ is. Presumably it's a function in $L^1(0, \infty)$?]
They are both correct when $F$ is a rational function or the product of a rational with an exponential. The original statement of the theorem postulates that $f$ is bounded and $\lim_{t\to\infty} f(t)$ exists. The question is how we can know that given $F$.
For $f$ to be bounded and for the limit $\lim_{t\to\infty} f(t)$ to exist, the (rational) function $F$ must have its poles in the open left half plane.
Moreover, $F$ is guaranteed to exist because $f$ is bounded. There is a little caveat here: We have
$$ \int_0^\infty f(t)e^{-st}\mathrm{d}t \leq \int_0^\infty |f(t)|e^{-st}\mathrm{d}t \leq \int_0^\infty Me^{-st}\mathrm{d}t , $$
which converges for all $s \in \mathbb{C}$ with $\Re(s) > 0$. However, we must assume that $f(t)e^{-st}$ is integrable. It suffices to assume that $f$ has a finite number of discontinuities, or assume that the integral is meant in the Lebesgue sense.