Meaning of opposite sign in Fourier's exponents I'm currently exploring the meaning, intuition and differences of the various types of Fourier transforms. I can't say everything is ok, but I think I got most of them (hopefully, Dunning Kruger is not involved).
I know that Fourier transform is a symmetrical transformation, meaning that $$\mathscr{F}\{f\} = F$$$$\mathscr{F}\{F\} = f$$ Given: $$F(\omega)=\int_{-\infty}^{\infty}{f(x)e^{i2\pi x\omega}dx}$$$$f(x)=\int_{-\infty}^{\infty}{F(\omega)e^{-i2\pi x\omega}d\omega}$$However, I also know that, to make the thing work, there are two caveats:

*

*The sign of the exponents in the integral of the transform must be opposites

*The scaling factors of the two must be equal, when multiplied together, to a factor of $1/N$.

I'm interested in the first one. Why is it that? Is it jus to "make the math work"? Why? If so, is there an intuition behind it?
 A: Summary: I don't think there's anything super profound about the sign inversion. The minus sign comes from complex conjugation that arises when taking the (hermitian) inner product between $F(\omega)$ and the complex exponentials.
One way to think about it is in terms of linear algebra. The following is not rigorous, but is aimed at giving you some intuition.
$F$ and $\omega \mapsto e^{2i\pi x\omega}$ can be thought of as two infinite dimensional vectors indexed by $\omega$.
Using the integral Fourier formula, $F$ is a linear combination of complex exponential functions (with the positive sign) $\{\omega \mapsto e^{2i\pi x\omega}\}_{x\in\mathbb R}$. Here $x$ plays the role of the index in the linear combination sum, and $f(x)$ is the "weight" (coefficient)  of $e^{2i\pi x\omega}$ in the linear combination.  And in a certain sense (again, not rigorous), the family of exponential functions is an orthonormal family.
So to reconstruct the weight vector $f$ from vector $F$, you only need to take the inner product with the exponential functionss
$$f(x) = \langle \omega \mapsto e^{2i\pi x\omega} , F\rangle = \int_{\mathbb R} \overline{e^{2i\pi x\omega}} F(\omega)d\omega = \int_{\mathbb R} e^{-2i\pi x\omega} F(\omega)d\omega$$
That explains the minus sign.
All of the above can be made rigorous using the theory of distribution. But to see it more simply, you can look at a finite discrete signal $\{f_k\}_{0\leq k < N}$ and compute its Discrete Fourier Transform
$$F_j = \sum_{k=0}^{N-1} f_ke^{2i\pi k j}$$
Then the family $\{ k\mapsto \frac 1 {\sqrt N} e^{2i\pi k j} \}_{0\leq j < N}$ of vectors is orthonormal. And the algebra from above with the hermitian inner product inverting the sign rigorously applies.
