# Dirac Delta Function - unfamiliar definition

## Main Question or Discussion Point

Given:

$$f(x)=\delta(x-a)$$

Other than the standard definitions where f(x) equals zero everywhere except at a, where it's infinity, and that:

$$\int_{-\infty}^{\infty} g(x)\delta(x-a)\,dx=g(a)$$

Is there some kind of other definition involving exponentials, like:

$$\int e^{ix(k'-k)}d^3x=\delta^3(k'-k)$$

I remember learning something about this, but can't find a proof of it in any textbook or online at the moment, and I don't trust my memory enough to know if this is precise. Any thoughts?

gel
yes, the dirac delta function is proportional to the fourier transform of the constant function equal to 1. You actually get $(2\pi)^3\delta^3(k)$.
To be rigorous, the fourier transform is defined in terms of distributions and not just standard functions. It is not a standard Riemann integral, although physicists often treat it as such.

The proof could go like the following. If f is a fourier transformable function with transform $\hat f$
\begin{align*} \int f(x) \left(\int e^{ixy}\,dy\right)\,dx &= \int \left(\int e^{ixy}f(x)\,dx\right)\,dy\\ &= \int \hat f(y) dy \end{align*}
Using z=0, the last line is
$$\int \hat f(y)e^{-iyz}\,dy = 2\pi f(z)=2\pi f(0)$$
-- using the inverse fourier transform. Compare this to $\int f(x)\delta(x)\,dx = f(0)$.

This is a bit sketchy, because the integral of $e^{ixy}$ doesn't make sense using the Riemann integral.

Also, the delta function is the derivative of the Heaviside function f(x)=1{x>0}, in the sense of distributions.

Last edited:
Fredrik
Staff Emeritus
Gold Member
Check out this thread for a very non-rigorous argument.

Wow... Irrelevant, but that nicksauce guy who started the other thread is a good friend of mine in RL. I'm gonna call him now!

Ha, small world.

Oh, and I understood the (false) justifications, thanks.

nicksauce