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Inverse Fourier Transformation of a Fourier Transformation

  1. Jan 14, 2012 #1
    I am having problem with the inverse transformation of a fourier transformed function which should give the function itself.

    Let

    [itex]f=f(x)[/itex] and let f be fourier transformable (whatever that implies)

    Let

    [itex]\tilde{f}(k)=∫^{\infty}_{-\infty}dx e^{-ikx}f(x)[/itex] (1)

    then we should have:

    [itex]f(x)=∫^{\infty}_{-\infty}dk e^{ikx}f(k)[/itex] (2)

    This implies:

    [itex]f(x)=∫^{\infty}_{-\infty}dk e^{ikx}∫^{\infty}_{-\infty}dx' e^{-ikx'}f(x')[/itex] (3)

    Note that x'≠x

    My solution is as follows:
    [itex]f(x)=∫^{\infty}_{-\infty}dk e^{ikx}∫^{\infty}_{-\infty}dx' e^{-ikx'}f(x')[/itex] (4)

    [itex]f(x)=∫^{\infty}_{-\infty}dk ∫^{\infty}_{-\infty}dx' e^{-ik(x'-x)}f(x')[/itex] (5)

    [itex]f(x)=∫^{\infty}_{-\infty}dx' ∫^{\infty}_{-\infty}dk e^{-ik(x'-x)}f(x')[/itex] (6)

    [itex]f(x)=∫^{\infty}_{-\infty}dx' δ(x'-x)f(x')[/itex] (7)

    [itex]f(x)=f(x)[/itex] (8)

    Is this correct? Step (6) to (7) bothers me. And what about the change in integration variables? I guess that is correct as well?
     
  2. jcsd
  3. Jan 14, 2012 #2

    marcusl

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    This looks correct to me. In step 6 you can move f(x') from the right integral into the left one, leaving int{ dk exp(-ik(x'-x)) }. This becomes the delta function in 7 as you state, through application of the principle of stationary phase.
     
  4. Jan 14, 2012 #3

    mathman

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    The analysis is essentially correct, except that you need to include 2π in the exponents. That is 2πikx or -2πikx'.
     
  5. Jan 16, 2012 #4

    marcusl

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    Oh, right! Mathman points out that I left off the normalization--sorry! You can solve the problem using the variables he mentions, or you can continue to use k but then the inverse transform has a normalizing constant 1/2π in front like this:

    [tex]f(x)=\frac{1}{2\pi}∫^{\infty}_{-\infty} e^{ikx}\tilde{f}(k)dk.[/tex]
    The key piece of eq. (6) becomes

    [tex]\frac{1}{2\pi}∫^{\infty}_{-\infty} e^{ik(x'-x)}dk[/tex]
    which is evaluated most easily as a limit

    [tex]\frac{1}{2\pi}\lim_{r\rightarrow\infty}∫^{r}_{-r} e^{ik(x'-x)}dk.[/tex]
    The integral becomes

    [tex]\lim_{r\rightarrow\infty}r\frac{\sin r(x'-x)}{r(x'-x)}[/tex]
    which is single-valued only if x'=x. At that point the ratio equals one by L'Hospital's rule, the overall expression becomes infinite, and this expression is a Dirac delta function.
     
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