Fourier Transform in space and time Solving the 3D wave equation, we use the (inverse) Fourier Transform like that:
\begin{equation}
 f(\mathbf{x},t) \propto\int_\mathbb{R}\int_{\mathbb{R}^3} e^{i(\mathbf k \cdot \mathbf x - \omega t)} \widehat f(\mathbf k,\omega)\ \mathrm d^3k\,\mathrm d\omega
\end{equation}
where $f$ is such that the (inverse) FT is well-defined and $\widehat f$ is the Fourier transformed of $f$ in both space and time.
Why can I not use $i(\mathbf k \cdot \mathbf x + \omega t)$ in the exponent instead?
A physical answer would be that in special relativity, we use the Minkowski metric $g=\mathrm{diag}(-1,1,1,1) $ and integrate over the four-dimensional Minkowski spacetime, so the exponent comes from the Minkowski inner product:
\begin{equation}
i(x^\mu k_\mu) = i(x_1k_1+x_2k_2+x_3k_3-\omega t)
\end{equation}
This is equivalent to our exponent.
However, people did Fourier integrals like this one before special relativity was a thing. What was their (more mathematical?) reason to choose the signs like that?
Thanks in advance!
 A: I think I figured it out myself.
Mathematically, it does not matter if we use $i(\mathbf k \cdot \mathbf x -\omega t)$ or $i(\mathbf k \cdot \mathbf x +\omega t)$.
But it is indeed true that people used a different relative sign for the $\mathbf k$ and $\omega$ terms in the exponent before special relativity and the Minkowski metric. But they still knew that looking at a physical wave propagating in one direction with velocity $c$ at time $t=t_0$ is equivalent to looking at a wave shifted in the opposite direction by $ct_0$  for $t=0$.
\begin{equation}
r = r_0+ct_0 \iff r - ct_0 = r_0 \iff \tilde r = r_0 +c\tilde t_0
\end{equation}
where $\tilde r := r-ct_0$ and $\tilde t_0 = 0$.
So while using the exponent $i(\mathbf k \cdot \mathbf x +\omega t)$ is giving a mathematically valid solution, it is not a solution for physical time $t$, but for a universe where time goes backwards. The direction of time is also included in the Minkowski metric, which is why the explanation using Minkowski space works, too.
I hope that makes sense to anyone having the same question.
