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Fourier Sine Transform

  1. Feb 26, 2013 #1
    I am having a hard time understanding how the Fourier sine transform works. I understand that you input a function and you get a function as an output, but I have no idea how certain inputs are even valid. Here is what the book gives for the transform:
    [itex]F(\omega )=\frac{2}{\pi}\int_{0}^{\infty}f(t)\displaystyle\sin (\omega t)dt[/itex]
    For my example, I need to transform an IC in a PDE problem. [itex]u(x,0)=H(1-x)[/itex] where [itex]H[/itex] is the Heaviside function. Basically, [itex]u(x,0)=1[/itex] when [itex]0\le x<1[/itex] and [itex]u(x,0)=0[/itex] when [itex]x\ge 1[/itex]. I assume then that the transform works out in the following manner:
    [itex]F(\omega )=\frac{2}{\pi}\int_{0}^{1}\displaystyle\sin (\omega t)dt=\frac{2}{\pi \omega}(1-\displaystyle\cos (\omega)[/itex]
    What would happen in the case that [itex]f(t)=1[/itex]? The integral would not converge, yet, mathematica tells me there is an answer.
     
  2. jcsd
  3. Feb 26, 2013 #2

    jbunniii

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    If ##f(t) = 1## for all ##t \in \mathbb{R}##, you are of course correct. The integral ##\int_{0}^{\infty} \sin(\omega t) dt## does not converge for any ##\omega \neq 0##. What "answer" did Mathematica give you?
     
  4. Feb 27, 2013 #3
    I was getting [itex]\sqrt{\frac{2}{\pi}}/\omega[/itex]
     
    Last edited: Feb 27, 2013
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