What is the $s$ in the Laplace transform? I know the formula, which is $$F(s) =  \int_0^\infty f(t) e^{-st}\,dt$$ but I don't understand what the $s$ is, and I've been searching everywhere and can't seem to find an answer, or maybe I'm just being really dumb and missing the obvious. Anyway, if anyone could help me with, I would appreciate it tremendously.
 A: Look at it more carefully. That '$s$' is also present on the left hand side. This is the argument of the Laplace transform $F$ of $f$, which itself is a function.
Instead, you could also write anything else in place of $s$, e.g.
$$F(x)=\int_0^\infty f(t)e^{-xt}dt\,.$$
A: Here's a non-rigorous intutive answer.
After watching, 3Blue1Brown's video on the Fourier Transform (do watch it and then read the answer) I visualise 's' as the winding frequency.
Fourier transform, gives us a surge/peak when the winding frequency matches the frequency of the input sinusoid, helping us to identify the frequencies in the input signal.
Laplace transform, in addition to finding sinusoidal frequencies, also identifies the exponentials present in the waveform. So for example, if the input is $$f(x) = e^{at} \sin(ωt)$$ Laplace transform, in addition to having $e^{-iωt}$  (for controlling the 'winding process' like FT) also has $e^{-αt}\, $ (i.e) $  \, e^{-(α+iω)t}$ so instead of taking Fourier transform of $f(x)$, we take the Fourier transform of $f(x)e^{-αt } = e^{(a-α)t} \sin(ωt)$. So now we've got this additional "degree of freedom" ~ $\alpha$.
If $a ≠ α$ the the waveform lasts for a short interval (the exp'l dampens it), thus the peaks of Fourier aren't big. But when $a = α$, $$f(x)=e^{0t} \sin(ωt)$$ therefore we will just be taking the Fourier transform of $f(x)=sin(ωt)$, so the peaks will be huge at $ω$. Thus we've got a way to point out both exponentials and the sines.

Here's another awesome answer on Laplace transform, from Linear Algebra stand point. Laplace transform as a change of basis 
