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Homework Help: Finding Fourier transform

  1. Jan 12, 2010 #1
    1. The problem statement, all variables and given/known data
    Find the Fourier transform of
    [tex]f(x)=\frac{1}{(x^2+a^2)^2},\ a>0[/tex], and show by direct calculation that with inverse Fourier transform you'll get the original function [tex]f(x)[/tex]!


    2. Relevant equations
    Fourier transform and it's inverse:
    [tex]F(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty f(x)e^{i k x}dx[/tex]
    [tex]f(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty F(k)e^{-ikx}dk[/tex]

    3. The attempt at a solution
    I have put my function into the transform and I got:
    [tex]F(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \frac{e^{ikx}}{(x^2+a^2)^2}dx[/tex]

    I have transformed it to complex integral and since k is arbitrary I have two possible paths of integration, for k>0 in the upper plane and for k<0 in the lower plane. My resudues are:

    [tex]Res(f,ia)=-i\frac{e^{-ka}}{4a^3}(1+ka)[/tex] and

    [tex]Res(f,-ia)=i\frac{e^ka}{4a^3}(1-ka)[/tex].

    So for k>0 my integral (without the [tex](2\pi)^{-1/2}[/tex]) is:
    [tex]I=\pi \frac{e^{-ka}}{2a^3}(1+ka)[/tex]
    And for k<0:
    [tex]I=\pi\frac{e^{ka}}{2a^3}(1-ka)[/tex]

    Now the problem is: which one do I use? Mathematica gives me this:
    [tex]\frac{1}{2a^3}\cdot\ \sqrt{\frac{2}{{\pi}}}e^{-ak}\left((ak+1)\theta(k)-e^{2ak}(ak-1)\theta(-k)\right)[/tex]
    Where [tex]\theta(x)[/tex] is Heaviside Step function.

    I have noticed that the solutions of those integral are similar but how do I implement the step function? Plus how would I integrate it? I know that the derivative of step function is Delta function, but I don't know what the integral of it is?

    Thanks!
     
  2. jcsd
  3. Jan 12, 2010 #2

    Matterwave

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    The integral of a step function is much simpler to deal with than the derivative. It's simply the area under the step!
     
  4. Jan 12, 2010 #3

    vela

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    Just break the integral up into two pieces, from [itex]-\infty[/itex] to 0 and from 0 to [itex]\infty[/itex]. One step function will equal 0, and the other will equal 1 in each integral.
     
  5. Jan 13, 2010 #4
    Oh, ok that's fine. But still how did I get the step in the first place? When I was computing the transform I got 2 solutions for [tex]k\lessgtr 0[/tex], do I need to combine them to get the step or? And If I do how to do that?
     
  6. Jan 13, 2010 #5

    vela

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    Yes, the step function just lets you indicate for which part of the domain the function it multiplies should apply. The [itex]\theta(k)[/itex] says the [itex](ak+1)[/tex] part only contributes for [itex]k>0[/itex] because [itex]\theta(k)[/itex] is zero for [itex]k<0[/itex] and one for [itex]k>0[/itex]. Similarly, the [itex]\theta(-k)[/itex] says the [itex]e^{2ak}(ak-1)[/itex] term contributes only for [itex]k<0[/itex]. Using the step function allows you to combine the two solutions you got for [itex]k>0[/itex] and [itex]k<0[/itex] into one expression.
     
  7. Jan 13, 2010 #6
    Thanks for clarification!
     
  8. Jan 18, 2010 #7
    Hi I have a question about this problem... again :(

    I know why I get these results with step function, that's all fine, but now I need to find the inverse (the original function).

    So I have:
    [tex]\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty F(k)e^{-i kx}dk[/tex]. I put my previous solution up there, step function 'kills' the bounds in the integral (for [tex]\theta(k)[/tex] my integral becomes: [tex]\int_{-\infty}^\infty \theta(k)dk=\int_0^\infty[/tex], right?) So I end up with 2 integrals which could be solved easily. But I have problem with that. If I try to solve this integral:
    [tex]\int_0^\infty e^{-ik(x-ia)}(1+ka)dk[/tex]
    I have problem with that upper bound. I have [tex]e^{i\cdot\infty}[/tex], and what should I make of that? If I try to solve this with residues, I have a difficulty, because I don't have [tex]\int_{-\infty}^\infty[/tex] or even function so that I could make [tex]\int_0^\infty=\frac{1}{2}\int_{-\infty}^\infty[/tex]. So how to solve it?
     
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