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Integral over the fourier transform

  1. Nov 7, 2008 #1
    It is a well known fact that [tex]\int dk \tilde{F}(k) = F(0)[/tex] where the tilde denotes the Fourier transform. (take or leave some [tex]\pi[/tex]s) Is it possible to show this
    1) without assuming that we know [tex]\int dx e^{ikx} = \delta(k)[/tex]
    and 2) without saying: "well we know what the inverse Fourier transform looks like, so see what you evaluate if you put in a zero". (I think this would burn down to 1 anyways)

    If the prove is impossible, can someone provide a link to a conclusive proof of the delta function property. Sorry if it looks like homework, it really isn't.
  2. jcsd
  3. Nov 7, 2008 #2
    Your "well known" bit isn't true in general. The proper theory of Fourier transforms deals with equivalence classes of functions so that such problems disappear, but if we don't do this then the problem still remains. For the rest of this post we put mathematical rigor to one side. As an example. consider:

    [tex] f(x) = \begin{cases} e^{-x},& x>0 \\
    0,& x\leq 0.[/tex]

    so we have f(0)=0. A quick computation reveals:

    [tex] \hat{f}(k) = \frac{1-ik}{1+k^2} [/tex]

    So by the Fourier inversion theorem, we find:

    [tex] f(0) = \frac{1}{2\pi} \int \frac{\mathrm{d} k}{1+k^2} =\frac{1}{2} [/tex]

    Eek! Note that the result is the average of the original function at the point of discontinuity.
  4. Nov 7, 2008 #3
    Ok I knew that it would not work on functions that are not continuous. It is something that one does expect after learning about Lesbeque measures.
    But I digress. It is true F.A.P.P. (for all practical/physicists' purposes) and there must be something a physicist will take as a proof. Oh I am sorry that would include "it's written in a book" ;) So something that I can accept as a proof of plausibility.
  5. Nov 7, 2008 #4
    Do you happen to know any books or resources that explains Fourier transforms "properly" as you say =-) It's one of those subjects that really interests me, but I simply can't stomach the engineerspeak!
  6. Nov 7, 2008 #5
    I only know one in German. F. Constantinescu "Distributionen und Ihre Anwendung in der Physik" ISBN 3-519-02042-4

    It also gives some Literature recomendations like the original (I think Schwarz invented distributions):
    Schwarz, L "Theorie des Distributions" Paris 1966

    But I think if you search for lecture notes on distributions you'll find a lot online.
  7. Nov 7, 2008 #6
    I found a proof

    I finally found a plausible proof. It goes roughly like this:
    [tex]\int dx\, f(x/\lambda)\tilde{g}(x) = \int dx\, f(x)\lambda \tilde{g}(\lambda x)[/tex]
    [tex]= \int dx\, dy'\, f(x)\lambda g(y') e^{-i\lambda xy'} = \int dx\, dy\, f(x) g(y/\lambda ) e^{-ixy}[/tex]
    [tex]= \int dy\, \tilde{f}(y)g(y/\lambda)[/tex]
    If we let [tex]\lambda \rightarrow \infty [/tex]
    [tex]f(0)\int dx\, \tilde{g}(x)=g(0) \int dx\, \tilde{f}(x)[/tex]
    Now we take a function that we know well for g like a gaussian which will produce some constants which are absorbed in the definition of the Fourier-Transform and
    and we get.
    [tex]f(0) = \int dx\, \tilde{f}(x)[/tex]
    Last edited: Nov 7, 2008
  8. Nov 7, 2008 #7


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    I don't know if you are unaware, or are aware but just omitted it -- but limits and integrals don't always commute like that. For example,

    [tex]\lim_{n \rightarrow +\infty} \int_{(0, 1)}n x^n \, dx
    = \lim_{n \rightarrow +\infty} \frac{n}{n+1} = 1[/tex]

    [tex]\int_{(0, 1)} \lim_{n \rightarrow +\infty} n x^n \, dx
    = \int_{(0, 1)} 0 \, dx = 0[/tex]

    You can probably salvage the proof idea by assuming f and g are sufficiently well-behaved, and splitting the integral into 3 or more pieces, such that one piece converges to the desired answer, and the other pieces converge to 0. Or maybe invoke one of the theorems of analysis that talk about this sort of stuff (e.g. maybe by showing the right thing converges uniformly)
  9. Nov 7, 2008 #8
    I think we're over-thinking this one. You shouldn't need the expression in 1), since the basic expression you're starting with already assumes that

    \int e^{ikx} dk \tilde{F}(k) = F(x)

    (up to some [itex]2\pi[/itex] factor anyway), which is just the definition of the Fourier Transform. Now just plug in [itex]x=0[/itex] and you're done, without ever mentioning the Dirac delta. Is there some reason that lacks rigor?
  10. Nov 8, 2008 #9
    I don't see that I assumed I knew anything about the inverse of the Fourier Transform. And if you use the usual integral as the definition you get this (I'll use hats for FT and anti hats for iFT now):

    [tex]\check{\hat{f(x)}}(z)[/tex] Sorry looks like ****. Inverse FT of the FT of f(x)
    [tex] = \int dx\, dy\, e^{izy} e^{-iyx} f(x) = \int dx\, dy\, e^{i(z-x)y} f(x)[/tex]
    now you would use [tex]\int dy\, e^{i(z-x)y} = \delta(z-x)[/tex]
    to get:
    [tex]\check{\hat{f(x)}}(z) =f(z)[/tex]

    If you know another way to show why the inverse FT works. Please enlighten me.

    The critique by Hurkil about the limit is well placed, and I am sure that a close inspections of the requirements for the proof will reveal, in what cases I will not work, non-continuities and the likes...
  11. Nov 8, 2008 #10
    I took one mathematical quantum course last spring, and one proof for the inverse Fourier transform was given there.

    The course web page is this: http://www.helsinki.fi/~jlukkari/courses/mpintro/index_en.html

    The lecture notes seem to be still available there. Download the chapter 6. Inverse Fourier transform is explained there. They are quite messy, scanned, notes, but if you are ready to go through them with time, they should be understandable.
  12. Nov 8, 2008 #11
    Thank you. I just skimmed the lecture notes, and it seems, that they use a very similar trick to the one I mentioned with [tex]\frac{y}{\varepsilon}[/tex] and a similar limit, with a gauss bell as a helper function that you can use to dump your transforms onto.
    Case closed. Back to work.
  13. Nov 11, 2008 #12
    Apologies for not getting back to you sooner 0xDEADBEEF. The following standard argument might help you. Assume [tex]f[/tex] is nice, i.e bounded and continuous with [tex]\|f\|_{L^1}<\infty[/tex]. Then define the mollifier [tex] \phi_\lambda[/tex] by:

    [tex] \phi_\lambda (x) = \int e^{-\lambda |t|} e^{2\pi i t x}\, \mathrm{d} t = \frac{2\lambda}{4\pi^2 x^2 + \lambda^2}. [/tex]

    so clearly [tex]\|\phi_\lambda\|_{L^1} = 1[/tex]. A routine calculation then shows that:

    [tex] f *\phi_\lambda (x) = \int f(y) \phi_\lambda (x-y)\, \mathrm{d}y = \int e^{-\lambda |t|} e^{2\pi i x t}\left( \int f(y) e^{-2\pi i y t}\, \mathrm{d} y\right) \mathrm{d} t = \int e^{-\lambda |t|}\hat{f}(t) e^{2\pi i x t}\, \mathrm{d} t [/tex]

    Showing [tex] f *\phi_\lambda \rightarrow f[/tex] as [tex] \lambda \rightarrow 0[/tex] is routine using [tex] \|\phi_\lambda \|_{L^1}=1[/tex]. You can take limits on the RHS using the dominated convergence theorem. This should provide you with your answer.
  14. Nov 11, 2008 #13
    Last edited by a moderator: May 3, 2017
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