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Step in fourier transform derivation

  1. Jul 4, 2009 #1
    Looking at how the fourier transform comes about from the fourier series when the period goes to infinity, they make the following step

    [tex] h \left( x \right) = \frac{2}{\pi}\int^{\infty}_{0} \left( dk \right) \left[ \int^{\infty}_{0} h \left( \varsigma \right) sin \left( k \varsigma \right) sin \left( x \varsigma \right) d\varsigma \right] [/tex]

    and the limits of integration can be changed to

    [tex] h \left( x \right) = \frac{1}{2 \pi}\int^{\infty}_{-\infty} \left( dk \right) \left[ \int^{\infty}_{-\infty} h \left( \varsigma \right) sin \left( k \varsigma \right) sin \left( x \varsigma \right) d\varsigma \right] [/tex]

    I'm trying to understand why this factor of 1/4 arises...

    My understand is this
    - in the first case, one integral gives us a fourier coefficent [tex] A_{k} [/tex] the other integration is effectively doing this; [tex]\sum^{\infty}_{k=0} A_{k} sin \left( kx \right)[/tex]

    - in the second case, one integral gives [tex] 2 A_{k} [/tex], because it integrates across 2 periods, but it can't matter which period you integrate across to find the fourier coefficent. The other integral is effectively [tex]2 \sum^{\infty}_{k=0} A_{k} sin \left( kx \right)[/tex] since it is summing over 2 periods instead of 1.

    Is this a correct way of looking at it? I'm more confident I know whats going on after writing this post and having to articulate it...
     
  2. jcsd
  3. Jul 5, 2009 #2

    HallsofIvy

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    The simplest way to see it is to note that in the first equation you cite, the integrals are both going form 0 to [itex]\infty[/itex] while in the second equation, they are going from [itex]-\infty[/itex] to [itex]\infty[/itex].
    [tex]\int_{-\infty}^\infty} f(x)dx= \int_{-\infty}^0 f(x)dx= \int_0^\infty f(x)dx[/tex]

    and, since both integrands are symmetric,
    [tex]\int_{-\infty}^0 f(x)dx= \int_0^\infty f(x)dx[/tex]
    which gives
    [tex]\int_{-\infty}^\infty f(x)dx= 2\int_0^\infty f(x)dx[/tex]
    or
    [tex]\int_0^\infty f(x)dx= \frac{1}{2}\int_{-\infty}^\infty f(x) dx[/tex]
    for each integral.

    It has nothing to do with "periods" because, in the Fourier transform, the functions are not necessarily periodic.
     
  4. Jul 5, 2009 #3

    uart

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    The factor of 4 is easy enough to explain. I'm a little more concerned with where the :

    [tex] \frac{2}{\pi} \int^{\infty}_{0} \left( dk \right) [/tex]

    bit comes from. That's bit is clearly infinite.

    BTW. I'm reasonably familiar with the development of the Fourier transform from the Fourier series but I haven't seen that method before. Do you have a either a link or the equations/development that precedes the lines shown above?
     
  5. Jul 5, 2009 #4
    This all comes from Morse and Feshbach's "Methods of Theoretical Physics". They make a substitution, take the limit of the period going to infnity and the summation becomes an integral. Its the best description I've found but its taking me a while to assimulate it. Most explanations online are "heres the formula - magnificent!".

    Also that whole thing could be rewritten [tex] \frac{2}{\pi} \int^{infty}_{0} \int^{infty}_{0} \left( \right) d\varsigma dk [/tex]

    So its not infinity. Its just a double integral.

    edit: also I made a mistake in those formulas in the first post but it shouldn't matter now.
     
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