The Fourier Transform of a spatial variable is no different mathematically from a Fourier Transform of a temporal variable. The mathematics is agnostic to parameter interpretation.
For the Fourier Transform pair for the time-frequency domain are often written
$$F(\omega) = \mathscr{F}(f)(\omega) = \int_{-\infty}^{\infty} f(t) e^{i \omega t} \, dt$$
$$f(t) = \mathscr{F}^{-1}(F)(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty} F(\omega) e^{-i \omega t} \, d\omega$$
while the analogous notation for the spatial-spatial frequency domain are often written
$$F(k) = \mathscr{F}(f)(k) = \int_{-\infty}^{\infty} f(x) e^{i kx} \, dx$$
$$f(x) = \mathscr{F}^{-1}(F)(x) = \frac{1}{2\pi}\int_{-\infty}^{\infty} F(k) e^{-i kx} \, dk$$
Certainly, the only difference between these pairs is symbolic.
However, in physics and engineering, one assigns units to these symbols. For the time-frequency transform pair, units of time and inverse time are assigned to the canonical parameters $t$ and $\omega$, respectively, and hence the reason we have a time-domain-frequency domain pair. For example, units could be in seconds and inverse seconds (i.e. radians/second).
When we move to the spatial Fourier Transform, the canonical units $x$ and $k$ are, for example, meters and inverse meters. The interpretation of inverse meters is that of a "wave number," and represents a spatial frequency for a traveling (or standing) wave.
The interpretation of the spatial Fourier Transform yielding momentum originates in quantum mechanics for which we have the relationship $p=k\hbar$, where $\hbar$ is the Dirac constant or reduced Planck's constant. Then, letting $k=p/\hbar $, we have
$$F(p) = \mathscr{F}(f)(p) = \int_{-\infty}^{\infty} f(x) e^{i px/\hbar } \, dx$$
$$f(x) = \mathscr{F}^{-1}(F)(x) = \frac{1}{2\pi\hbar }\int_{-\infty}^{\infty} F(p) e^{-i px/\hbar } \, dp$$
where $F(p)$ is called the momentum representation of $f(x)$
Best Answer
As you know the Laplace transform is an Operator which transforms a function $f(t)$ into another function $$F(s) =\mathcal{L} \{ f(t) \}$$
We can think of $f(t)$ as the input and $F(s)$ as the output.
Thus the domain of this operator is the set of all functions $f(t)$ for which $$F(s) =\mathcal{L} \{ f(t) \}$$ is convergent and the range or codomain is the set of those well defined transforms $F(s)$