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Getting a delta function from an indefinite integral

  1. Nov 26, 2008 #1
    Hey everybody,

    One question that I've had for a week or so now is how the following integral can equal a Dirac delta function:

    [tex] \frac{1}{2\pi} \int_{-\infty}^{\infty}{dt} \:e^{i(\omega - \omega^{'})t}\: = \: \delta(\omega - \omega^{'})[/tex]

    A text that I was reading discusses Fourier transforms and eventually arrives at the above equation through the use of definitions. Since the book was taking an inverse Fourier transform, the solution is already known just to be f(t) (the function first operated on by the regular Fourier transform.)

    We know the definition of the Dirac delta function:
    [tex]f(t)\: = \: \int_{-\infty}^{\infty}{d\tau f(\tau) \delta(t - \tau)[/tex]

    And here we have a function f(t) being transformed and inverse transformed (with a dummy variable tau):

    [tex]f(t)\: = \: \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}{d\omega\:e^{i\omega t}} \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} d\tau \: e^{-i\omega \tau} f(\tau)[/tex]

    which can be manipulated to become

    [tex]f(t)\: = \: \int_{-\infty}^{\infty}{d\tau f(\tau) [ \frac{1}{2\pi} \int_{-\infty}^{\infty}{d\omega} \:e^{i(t - \tau)\omega}][/tex]

    Together these equations produce:

    [tex] \frac{1}{2\pi} \int_{-\infty}^{\infty}{dt} \:e^{i(t - \tau)\omega}\: = \: \delta(t - \tau)[/tex]

    (which is essentially the first equation)

    I see how a delta function operating within an integral applies to the above case to show how the first equation works out, however, I wasn't sure if there was a direct mathematical formulation that would get from the original indefinite integral of e^(w-w') to the dirac delta function.

    Obviously if I try and integrate this function and evaluate the solution at infinity and negative infinity the function diverges.

    Let me know if there are any extra ways to look at this integral (or evaluate it for that matter) that would lead to the solution of a delta function.

  2. jcsd
  3. Nov 28, 2008 #2
    One way is to not integrate directly to infinities, but to only over some interval [tex][-L,L][/tex], and delay the limit [tex]L\to\infty[/tex] to the end. First check this:

    \int\limits_{-L}^L dt\; e^{i(\omega-\omega')t} = \frac{2\sin((\omega-\omega')L)}{\omega-\omega'}

    Then suppose [tex]f[/tex] is some test function. In the following integration the change of variable [tex]u=(\omega-\omega')L[/tex], [tex]du=L d\omega[/tex], is used.

    \lim_{L\to\infty} \int\limits_{-\infty}^{\infty} d\omega\;\Big( f(\omega) \int\limits_{-L}^L dt\; e^{i(\omega-\omega')t}\Big)
    = \lim_{L\to\infty} \int\limits_{-\infty}^{\infty} d\omega\; \frac{2f(\omega)\sin((\omega-\omega')L)}{\omega-\omega'}
    = \lim_{L\to\infty} \int\limits_{-\infty}^{\infty} du\; \frac{2f(\frac{u}{L} + \omega') \sin(u)}{u}
    = 2f(\omega') \int\limits_{-\infty}^{\infty} du\;\frac{\sin(u)}{u} = 2\pi f(\omega')


    \lim_{L\to\infty} \int\limits_{-L}^L dt\; e^{i(\omega-\omega')t} = \lim_{L\to\infty} \frac{2\sin((\omega-\omega')L)}{\omega-\omega'} = 2\pi \delta(\omega - \omega')

    in the usual sense, that the limit should be taken after integration over omega.

    I have two comments to this:

    (1) What I showed here, is not yet a rigor proof of the Fourier inverse transformation formula. There is a problem with the step, where the order of integration and limit is changed. The function [tex]\frac{\sin(u)}{u}[/tex] is not Lebesgue integrable, and thus the standard dominated convergence theorem cannot be used. I have so far been unable to complete this step properly. I don't know any other way to justify this properly than first proving the inverse transformation formula in some other way, and then getting the delta-function identity backwards, like you had already done. Anyway, this calculation is useful for pedagogical and heuristic purposes even without proper justification of this one step.

    (2) I have heard some people claiming, that representations of delta function always approach infinity at origo, and zero elsewhere. The claim is wrong, and right here we have a counterexample, since this collection of functions don't converge towards zero anywhere, and still they behave as a delta function. Can you see what is actually happening on the limit [tex]L\to\infty[/tex]?
    Last edited: Nov 28, 2008
  4. Nov 28, 2008 #3
    I just remembered that there is another way too. This:

    \lim_{\epsilon\to 0^+} \int\limits_{-\infty}^{\infty} dt\; e^{-\epsilon t^2 + i(\omega - \omega')t}

    But I have not spent much time thinking about this, so I think I'll not write anything about this too quickly now.
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