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Homework Help: Fourier sine transform of 1

  1. Jan 11, 2010 #1
    1. The problem statement, all variables and given/known data
    I'm looking to determine the Fourier sine transfom of 1.


    2. Relevant equations
    One this site http://mechse.illinois.edu/research/dstn/teaching_files2/fouriertransforms.pdf [Broken] (page 2) it gives the sine transform as

    [tex]\frac{2}{\pi \omega}[/tex]

    3. The attempt at a solution
    However, since the Fourier sine fransform of 1 is defined via,

    [tex]\frac{2}{\pi} \int_0^\infty \sin (\omega x) dx ,[/tex]

    I figure that its value should be,

    [tex] \frac{2}{\pi \omega} -\lim_{L\rightarrow \infty } \frac{2}{\pi \omega} \cos (r L) .[/tex]

    It seems like they've just thrown the cosine term away, but is this legal? If so why?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jan 11, 2010 #2

    LCKurtz

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    The usual condition for any fourier transform is

    [tex]\int_{-\infty}^\infty |f(x)|\ dx < \infty[/tex]

    which f(x) = 1 doesn't satisfy. The sine transform doesn't exist, and the integral for it diverges as you have observed.
     
  4. Jan 11, 2010 #3
    Excellent. Thanks, LCKrutz.
     
  5. Jan 11, 2010 #4

    vela

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    The straightforward integral diverges, so what they probably did was throw in an integrating factor [itex]e^{-\lambda x}[/itex] to make the integral converge, and then take the limit as [itex]\lambda\rightarrow0^+[/itex]. Try that and see what you get.
     
  6. Jan 11, 2010 #5
    That's a good trick. I'll have to remember that one.

    Cheers.
     
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