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Dirac Delta Function - unfamiliar definition

  1. Aug 8, 2008 #1


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

    [tex]\int_{-\infty}^{\infty} g(x)\delta(x-a)\,dx=g(a)[/tex]

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

    [tex]\int e^{ix(k'-k)}d^3x=\delta^3(k'-k)[/tex]

    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. jcsd
  3. Aug 8, 2008 #2


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    yes, the dirac delta function is proportional to the fourier transform of the constant function equal to 1. You actually get [itex](2\pi)^3\delta^3(k)[/itex].
    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 [itex]\hat f[/itex]
    \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
    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 [itex]\int f(x)\delta(x)\,dx = f(0)[/itex].

    This is a bit sketchy, because the integral of [itex]e^{ixy}[/itex] 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
  4. Aug 8, 2008 #3


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    Check out this thread for a very non-rigorous argument.
  5. Aug 8, 2008 #4
    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.
  6. Aug 10, 2008 #5


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    Lol small world indeed
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