2
$\begingroup$

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.

$\endgroup$

1 Answer 1

0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.