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Fourier transform of t, 1/t and t^n

  1. Nov 24, 2011 #1
    I would like to know how one finds the Fourier transforms of





    with the definition of the fourier transform as

    [tex]\mathscr{F}\{f(t)\}=\mathcal{F}\{f(t)\}=\frac{1}{ \sqrt{2\pi} }\int\limits_{-\infty}^{\infty}{e}^{-i\omega t}f(t)\mbox{d}t[/tex]

    I have tried the definition of a fourier transform and I got some weird limits. Laplace transforms are so much easier!

    Thanks in advance.
  2. jcsd
  3. Nov 24, 2011 #2


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    Science Advisor

    In the usual definition of Fourier transform, f(t) is usually presumed to be integrable, or square integrable. None of your functions satisfy this requirement.
  4. Nov 25, 2011 #3


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    The functions do, however, have Fourier transforms in terms of distributions. Consider

    [tex]\sqrt{2\pi}\delta(\omega) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty dt~e^{i\omega t}.[/tex]

    Now, take a derivative of both sides with respect to the frequency:

    [tex]\sqrt{2\pi}\delta'(\omega) = \frac{i}{\sqrt{2\pi}} \int_{-\infty}^\infty dt~t e^{i\omega t}.[/tex]

    You can keep taking derivatives to get the Fourier transform of tn. For 1/t, the fourier transform will be proportional to the [itex]\mbox{sgn}(\omega)[/itex] function, where sgn(x) returns the sign of x.
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