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Incomplete gamma function

  1. Jun 25, 2011 #1
    How do I calculate the integral

    [tex]\int_{ix}^{i\infty} e^{-t} t^{-s-1}dt,[/tex]
    where [itex]x>0[/itex], [itex]s>0[/itex]? Mathematica gives [itex]\Gamma(-s,ix)[/itex], where [itex]\Gamma(\cdot,\cdot)[/itex] is the incomplete gamma function, but I am not sure how to justify this formally.
     
  2. jcsd
  3. Jun 25, 2011 #2

    mathman

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    Substitute u = -it, so the integral is from x to inf.
     
  4. Jun 25, 2011 #3

    Mute

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    The incomplete Gamma function is defined by the integral

    [tex]\Gamma(s,x) = \int_{x}^\infty dt~t^{s-1}e^{-t}.[/tex]

    Replacing x with ix formally gives [itex]\Gamma(s,ix)[/itex]. However, the meaning of the integral with lower bound ix is really just formal, I think. You identify the integral with the incomplete Gamma function, and then you determine the "integral's" value by using the analytic continuation of the incomplete Gamma function for complex arguments.
     
    Last edited: Jun 25, 2011
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