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Inverse Fourier Transform using complex variables

  1. Jan 12, 2012 #1
    1. The problem statement, all variables and given/known data

    For α > 0, determine u(x) by the inverse Fourier transform

    [tex]u(x) = \frac{1}{2\pi}\int_{-\infty}^{\infty}\ \frac{e^{ikx}}{ik+\alpha}\ dk [/tex]

    2. Relevant equations

    3. The attempt at a solution

    This seemed like a relatively simple residue problem. You just note that the denominator has a pole at αi, rewrite the integrand by multiplying through by -i (in order to get the denominator into the form k - αi), and use the residue theorem. As mentioned, the only pole is a simple one, so

    [tex]\mbox{Res}_{k=ai}\ \frac{-ie^{ikx}}{k-αi} = -ie^{i(iα)x} = -ie^{-αx}[/tex]

    Since the inverse Fourier transform has that factor of 1/2∏, but the residue theorem requires multiplication by 2∏i, I get

    [tex]u(x) = e^{-αx}[/tex]

    But that seems wrong. Mathematica gives me

    [tex]\sqrt{2\pi}e^{\alpha x}\ H(x)[/tex]

    Now I remember from days gone by that the square root prefactor is more of a conventional thing, so I'm not so concerned about that. I'm mildly concerned about the sign of α and much, much more concerned about the appearance of the Heaviside function. Where did that come from?

  2. jcsd
  3. Jan 12, 2012 #2

    I like Serena

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    Hi tjackson3! :smile:

    Fourier's theorem has a precondition.
    The function that is transformed has to be integrable.

    The function you got is not integrable from ##-\infty## to ##+\infty##.
    Btw, your result of Mathematica is not integrable either (assuming alpha > 0).
    For positive alpha it should be ##e^{\alpha x}H(-x)##.
    This also explains the change in sign of alpha.

    As it is, this function is integrable, and if you calculate its Fourier transform, you find your original function.
    This Fourier transform can be calculated directly from the definition.
  4. Jan 12, 2012 #3
    Hi Serena! Thanks for your response!

    Ah, see, Mathematica did give me H(-x), but I thought that was just a Mathematica quirk - since my braindead self thought for a moment H(x) = H(-x) - and decided not to include it (of course, even if that were true, I imagine that would have messed up the rest of the formula). That was very thoughtless and indicative of the studying overload.

    This is for a Complex Variables prelim, so I haven't actually had any experience with Fourier transforms, so please excuse my silliness here. So you're saying that when you perform an inverse Fourier transform like this, if the integral doesn't converge, you add a Heaviside function in order to ensure that it does?

    Thanks again!
  5. Jan 12, 2012 #4

    I like Serena

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    You're welcome!

    I'm not aware of any rule for this.
    As far as I know it's trial and error.
    If there are possible problems with integrability, you need to verify your result.
  6. Jan 12, 2012 #5

    Ray Vickson

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    In order to apply residues you need to "complete the contour". Look at exp(i*k*x). For k = kr + i*ki we have exp(-ki*x + i*kr*x), so for x > 0 we need to complete the contour in the upper k-plane (ki > 0); this makes exp(-ki*x) convergent as ki --> infinity. The contour is the real axis plus an infinite semicircle in the upper k-plane; it includes the pole at k = iα, so we get something nonzero. However, for x < 0 we need to complete the contour in the lower k-plane (ki < 0); that contour does not include the pole, so you get zero.

  7. Jan 12, 2012 #6

    I like Serena

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    I just realized that I approached your problem from the Fourier theorem, showing your result could not be right.

    In this case, your actual question would be why the residue theorem did not appear to work.
    As it is, to use the residue theorem you defined a curve around the pole consisting of a straight path over the real line, and an arc through the positive imaginary plane.
    You assumed the arc integral would approach zero at infinity, but it doesn't.
    It only approaches zero at the part where x is positive.

    So you need to define your arc integral differently to get the proper result.
  8. Jan 13, 2012 #7
    That makes total sense. Thank you both!
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