1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook