If $f(x,t) = g(u)$, how do I relate $\iint{fdxdt}$ to $\int{gdu}$?

Question

I'm learning about Fourier Transforms in the context of travelling waves on a dispersive medium, and my textbook sort of handwaves a simplification in which

$$ \textbf{F}(k,\omega) = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(x,t)e^{-i(kx-\omega t)}dxdt $$

becomes

$$ \textbf{F}(k) = \int_{-\infty}^{\infty}f(x,0)e^{-ikx}dx $$

Which, in conjunction with a dispersion relation $\omega =\omega (k)$, allows the Fourier series for the travelling wave to be expressed as a single integral with respect to k.

In trying to wrap my head around the subtleties of why this works, I ran into a more general problem that I think is at the crux of my confusion: if $f(x,t)=g(u)$ where $u$ is a function of $x$ and $t$, is there a way to relate $\iint{fdxdt}$ to $\int{gdu}$? Of course $du=\frac{\partial u}{\partial x}dx + \frac{\partial u}{\partial t}dt$, but I have no clue how to use that to turn a single integral into two. I was also having trouble finding anything about this online since I don't know if there's a name for this simplification.

Many thanks in advance for any and all help.

Edit:

I've managed to grasp a intuitive understanding for the Fourier simplification, but I'm still confused on how to mathematically justify it. Essentially, if you have a formula for the waveform at a specific time, then you can do a spatial FT to get the wavenumbers k that comprise it. Then, if you know the dispersion relation, essentially knowing the wavespeed associated with each wavenumber, then you can imagine pushing all the component waves of the spatial FT at their associated wavespeed, giving

$$ f(x,t)=\frac{1}{2\pi} \int_{-\infty}^\infty \textbf{F}(k)e^{i(kx-\omega (k)t)}dk $$ With $\textbf{F}(k)$ being what I had written above.

While I have a clear enough picture in my head as to why this works, I'm still having trouble justifying it algebraically. Hopefully the following picture makes my question clear:

Answer 1

I will give an answer to (a re-interpretation of) the title question below, although this will likely not answer the Fourier question.

For the latter I seem to need more context to be sure, say a reference to the statement in your book including assumptions. It is good practice to provide this, and to be clear about what the symbols mean and which one question you actually ask.

1. However, I suspect that one integrates the $t$-variable out which somehow results in this 'simplification'. This would equate integrals of multiple variables to less variables. An example would be that $f$ is some traveling wave as in your context: $$f(x,t) = f(x,0)e^{-i\omega t} \delta(t)$$ (E.g. a solution to some separable wave-equation, where the x and t dependencies are split into two functions. I also added a contrived delta-distribution to make the integral work. Yes, better examples exist.) We will substitute this into the integral:

Because the complex exponents cancel, the integral reads: $$F(k,\omega) = \int\int f(x,t)e^{-i(kx-\omega t)}\delta(t) dxdt = \int\int f(x,0)e^{-i(kx)}\delta(t)dxdt = \int f(x,0) e^{-i(kx)} dx. $$

This most likely is just a contrived illustration of how integration out a variable would work.

2. For a re-interpretation of your title question, I changed the $u$ to output 2 numbers instead of 1:

There is a change of variables theorem for multi-variate integrals, which in 2D roughly (i.e. ignoring differentiability and integrability properties) reads as follows:

if $f$ and $g$ are functions from 2 variables to 1 real number, and $$f(x,t) = g(u(x,t))$$ for some (injective) function $$u(x,t) = (u_1, u_2)$$ that outputs 2 real numbers, then $$\int\int_{u(\mathbb{R}^2)} g(u_1,u_2)du_1du_2 = \int\int_{\mathbb{R}^2} g(u(x,t)) |det(Ju)(x,t)| dxdt $$ $$= \int\int_{\mathbb{R}^2} f(x,t) |det(Ju)(x,t)| dxdt$$

Here $|det(Ju)(x,t)|$ is the absolute value of the determinant of the 2-by-2 Jacobian of $u$, i.e. the matrix of partial derivatives of $u$. and $u(\mathbb{R}^2$) is the image of the plane after applying $u$. Multivariate analysis courses would teach this.