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Fourier transform of sin(3pix/L)

  1. May 11, 2016 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations


    3. The attempt at a solution
    So we want sine in terms of the exponentials when we take the fourier transform [tex]F(k)=\int_{-\infty}^{\infty}f(x)e^{-ikx}dx[/tex] where [itex]f(x)=\sin(3\pi x/L)[/itex]. Let a=3pi/L. Then [tex]\sin(ax)=\frac{e^{iax}-e^{-iax}}{2i}[/tex].
    (Is this correct?)
    Then we can take the fourier transform:
    [tex]F(k)=\int_{-\infty}^{\infty}\frac{e^{iax}-e^{-iax}}{2i}e^{-ikx}dx[/tex]. Rearranging gives [tex]\frac{1}{2i}[\delta(K+a)-\delta(K-a)][/tex]. But my notes says there is [itex]\sqrt{2\pi}[/itex] in front and I'm not sure where it came from?
    Any help will be appreciated.
     
  2. jcsd
  3. May 11, 2016 #2

    Ray Vickson

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    Google is your friend; look up "Dirac delta function".
     
  4. May 12, 2016 #3
    I don't understand how to apply the dirac delta function here? I just used the integral representation of delta to get to the last line.
     
  5. May 12, 2016 #4
    Oh I am missing 2pi when I integrate. But where is the square root coming from?
     
  6. May 12, 2016 #5

    vela

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    You're probably using a different convention for the Fourier transform compared to what was done in your notes (or your notes are wrong). I think you also made a sign error.
     
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