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Fourier transform

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

    Screenshot2012-06-12at23022AM.png



    3. The attempt at a solution

    I don't understand this step. It's got to be some sort of identity that I missed. I also don't understand why the limits of integration change.
     
  2. jcsd
  3. Jun 12, 2012 #2
    Here's an easy way of seeing it:

    Remember that the integral over an even interval of an odd function is zero
    [tex]\int_{-L}^L f(x) dx = 0 [/tex]
    if [itex] f(-x) = -f(x) [/itex].

    You can see fairly easily that [itex] \frac{\sin(\alpha)}{\alpha} [/itex] is an even function and [itex] \sin(\alpha x) [/itex] is an odd function; therefore [itex] \frac{\sin(\alpha) \sin(\alpha x)}{\alpha} [/itex] is odd and it's integral vanishes over an even support interval.
     
  4. Jun 12, 2012 #3
    ok, I understand what you mean, although it took me about 30 minutes to get it. I still understand why the limits of integration change. I also don't understand why 1/pi changes to 2/pi though I think it has something to with the change in the limits of integration.
     
  5. Jun 12, 2012 #4
    For an even function f(-x) = f(x), you can show that [itex] \int_{-L}^L f(x) dx = 2 \int_0^L f(x) dx [/itex]
     
  6. Jun 12, 2012 #5
    cool
     
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