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Fourier Transform and Infinity Problem

  1. May 2, 2010 #1
    1. The problem statement, all variables and given/known data
    Hi, I need to find the Fourier Transform of: [tex]g(t)=\frac{1}{x}e^{\frac{-\pi t^2}{x^2}}[/tex]



    2. Relevant equations

    [tex]G(f)=\int_{-\infty}^{\infty }g(t)e^{-j2\pi ft}dt
    \therefore
    G(f)=\int_{-\infty}^{\infty }\frac{1}{x}e^{\frac{-\pi t^2}{x^2}-j2\pi ft}dt[/tex]



    3. The attempt at a solution
    [tex]G(f)=\frac{1}{x}[\frac{x^2e^{-t(\frac{\pi t}{x^2}+j2\pi f)}}{-2\pi t-j2\pi fx^2}] (limits:t=\infty,t=-\infty)[/tex]

    If you sub in t=infinity or t=-infinity, both will give you 0 because t is on the denominator.

    Thanks for the help in advance! :)
     
    Last edited: May 3, 2010
  2. jcsd
  3. May 2, 2010 #2

    gabbagabbahey

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    Gold Member

    Check your antiderivative; you are essentially claiming that

    [tex]\frac{d}{dt}\left(\frac{\rm{e}^{h(t)}}{h'(t)}\right)=\rm{e}^{h(t)}[/tex]

    which isn't true.
     
  4. May 3, 2010 #3
    Hi, I had a - insead of a + in the last exponential.

    Basically to integrate the exponential, I took the derivative of the power and divided the exponential by that derivative. Isn't this the right method?

    So the derivative of [tex]\frac{-\pi t^2}{x^2}-j2\pi ft=\frac{-2\pi t}{x^2}-j2\pi f=\frac{-2\pi t-j2\pi fx^2}{x^2}
    [/tex]
    [tex]G(f)=\int_{-\infty}^{\infty }\frac{1}{x}e^{\frac{-\pi t^2}{x^2}-j2\pi ft}dt=\frac{1}{x}[\frac{e^{\frac{-\pi t^2}{x^2}-j2\pi ft}}{\frac{-2\pi t-j2\pi fx^2}{x^2}}=\frac{1}{x}[\frac{x^2e^{-t(\frac{\pi t}{x^2}+j2\pi f)}}{-2\pi t-j2\pi fx^2}][/tex]
     
    Last edited: May 3, 2010
  5. May 3, 2010 #4

    gabbagabbahey

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    No, for example, [itex]\int e^{-t^2}dt\neq-\frac{1}{2t}e^{-t^2}+C[/itex]. Take the derivative of it to convince yourself:

    [tex]\frac{d}{dt}\left(-\frac{1}{2t}e^{-t^2}+C\right)=\left(\frac{1}{2t^2}+1\right)e^{-t^2}[/tex]

    via the product rule.

    In fact, there is no nice antiderivative for [itex]e^{-t^2}[/itex], or your integrand (Unlees you consider the error function to be nice). You can however integrate it over the interval [itex](-\infty,\infty)[/itex] easily by completing the square on the exponent, and using the fact that [itex]\int_{-\infty}^{\infty}e^{-x^2}dx=\int_{-\infty}^{\infty}e^{-y^2}dy[/itex], and hence

    [tex]\int_{-\infty}^{\infty}e^{-x^2}dx=\left(\int_{-\infty}^{\infty}e^{-x^2}dx\int_{-\infty}^{\infty}e^{-y^2}dy\right)^{1/2}=\left(\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2+y^2)}dxdy\right)^{1/2}[/tex]

    Which you can integrate by switching to polar coordinates.
     
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