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Form into a gamma function

  1. Feb 19, 2013 #1
  2. jcsd
  3. Feb 19, 2013 #2
    I don't see how it produced [itex] \Gamma (\frac{4}{5}) [/itex], but I did find one way to solve it:

    The Fourier transform of [itex] f(x) [/itex] in non-unitary, angular frequency form is:

    [itex] \hat f (\upsilon) = \int_{-\infty}^{\infty} f(x) e^{-i \upsilon x} dx [/itex]

    We have [itex] f(x) = (ix)^{-\frac{1}{5}} [/itex] and will just need to evaluate [itex] \hat f (\upsilon) [/itex] at [itex] \upsilon = -1 [/itex] once we find an expression for it:

    You'll note that the bottom entry in this table (http://en.wikipedia.org/wiki/Fourier_transform#Distributions) works for us in this situation, with [itex] \alpha = \frac{1}{5} [/itex]

    At [itex] \upsilon = -1 [/itex] the expression reduces to [itex] \hat f (-1) = \frac{2 \pi}{\Gamma (\frac{1}{5})} = 1.368... [/itex] -- agreeing with wolframalpha's calculated value.
     
    Last edited: Feb 19, 2013
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