# Dirac Delta Function - unfamiliar definition

1. Aug 8, 2008

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?

2. Aug 8, 2008

### 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: Aug 8, 2008
3. Aug 8, 2008

### Fredrik

Staff Emeritus
Check out this thread for a very non-rigorous argument.

4. Aug 8, 2008

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.

5. Aug 10, 2008

### nicksauce

Lol small world indeed