I'm looking into the proof behind why the Fourier transform and inverse Fourier transform are inverse operations of each other. However, I'm having trouble understanding the following integral:
$\int_{-\infty}^{\infty}e^{i\omega(\tau-x)}d\omega=2\pi\delta(\tau-x)$.
Mainly, it is the relation of the $sinc$ function to the delta function that confuses me.
Not an answer, but too long for a comment:
It should be noted that the integral $\int_{-\infty }^{\infty } e^{i \omega (\tau -x)} \, d\omega$ does not converge, so it has no representation in terms of classical functions. The Dirac $\delta$ generalized function is used to give meaning to the value of the integral, but this is no longer a function in the normal sense, i.e. a map $f:\mathbb R\to\mathbb R$ assigning to each $x\in\mathbb R$ a value $f(x)\in\mathbb R$.
Instead, we can define $\delta$ as a measure on the Lebesgue $\sigma$-algebra of $\mathbb R$ with $\delta(A) = \mathsf 1_{A}(0)$. Then the Lebesgue integral with respect to the measure $\delta$ satisfies $$ \int_{-\infty}^\infty f(x)\ \mathsf d\ \! \delta(x) = f(0). $$ Note that $\delta$ is not absolutely continuous with respect to the Lebesgue measure $m$, as e.g. $m(\{0\})=0$ while $\delta(\{0\})=1$ (in fact it is a singular measure), and therefore there does not exist a Radon-Nikodym derivative $g:\mathbb R\to\mathbb R$ for which $$ \int_{-\infty}^\infty g(x)\delta(x)\ \mathsf dx = g(0) $$ holds. So note that any expression involving an integral of the form above is a valid Lebesgue integral, but instead an abuse of notation.
