Your question is a bit ambiguous, since you don't state what you mean by fourier integral and fourier transform.
One possible source of confusion is that, while the fourier transform is indeed a linear isometry on $L^2$, the integral $$
\int_{-\infty}^\infty f(t)e^{-i2\pi\omega t} \,dt
$$
does not converge for every $f \in L^2$. It does, however converge for every $f \in L^1 \cap L^2$, and the fourier transform on the full space $L^2$ can therefore be defined as the unique extension of the transform defined by the integral on $L^1 \cap L^2$. The result is then sometimes called the Fourier-Plancherel-Transform, but sometimes also simply the fourier transform on $L^2$.
Or you could simply be referring to the difference between the integral one uses to compute the coefficients of a fourier series and the integral used to define the fourier transform (on $L^2 \cap L^2$).
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
Very roughly speaking: you can think of the difference in terms of the Heisenberg Uncertainty Principle, one version of which says that "bandwidth" (frequency spread) and "duration" (temporal spread) cannot be both made arbitrarily small.
The classical Fourier transform of a function allows you to make a measurement with 0 bandwidth: the evaluation $\hat{f}(k)$ tells us precisely the size of the component of frequency $k$. But by doing so you lose all control on spatial duration: you do not know when in time the signal is sounded. This is the limiting case of the Uncertainty Principle: absolute precision on frequency and zero control on temporal spread. (Whereas the original signal, when measured at a fixed time, gives you only absolute precision on the amplitude at that fixed time, but zero information about the frequency spectrum of the signal, and represents the other extreme of the Uncertainty Principle.)
The wavelet transform take advantage of the intermediate cases of the Uncertainty Principle. Each wavelet measurement (the wavelet transform corresponding to a fixed parameter) tells you something about the temporal extent of the signal, as well as something about the frequency spectrum of the signal. That is to say, from the parameter $w$ (which is the analogue of the frequency parameter $k$ for the Fourier transform), we can derive a characteristic frequency $k(w)$ and a characteristic time $t(w)$, and say that our initial function includes a signal of "roughly frequency $k(w)$" that happened at "roughly time $t(w)$".
How is this helpful? Let us say we are looking at the signal of the light emitted from a traffic light. So for some time it will be red, and for some time it will be green (ignore the yellow for now). If we take the Fourier transform of the observed frequency, we can say that
But a functioning traffic light would have either red or green shown at a time, and not both. And if the traffic light malfunctions and shows both lights at the same time, we would still see from the Fourier transform
But if we take the wavelet transform we can sacrifice frequency precision to gain temporal information. So with the wavelet transform done on the working traffic light we may see
This would tell us that not only can the traffic light show both red and green lights, that at least at around 1 o'clock the light is working properly and only showing one light.