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The Incomplete Gamma Function

  1. Mar 5, 2009 #1
    1. The problem statement, all variables and given/known data

    I read in a paper that:
    [tex]\Gamma\left(c,\,d\frac{x+e}{x-y}\right) = (c-1)!\,exp\left[-d\frac{y+e}{x-y}\right]\,exp[-d]\,\sum_{k=0}^{c-1}\,\sum_{l=0}^k \frac{d^k}{k!}{k\choose l}\left(\frac{y+e}{x-y}\right)^l [/tex]

    2. Relevant equations

    But the incomplete gamma function defined in the book of table of integrals and series as:
    [tex]\Gamma(1+n,x) = n!\,exp[-x]\,\sum_{k=0}^n \frac{x^m}{m!}[/tex]

    3. The attempt at a solution

    Applying this we get:

    [tex]\Gamma\left(c,\,d\frac{x+e}{x-y}\right) = (c-1)!\, exp\left[-d\frac{x+e}{x-y}\right]\,\sum_{k=0}^{c-1} \frac{d^k}{k!}\,\left(\frac{x+e}{x-y}\right)^k \neq (c-1)!\,exp\left[-d\frac{y+e}{x-y}\right]\,exp[-d]\,\sum_{k=0}^{c-1}\,\sum_{l=0}^k \frac{d^k}{k!}{k\choose l}\left(\frac{y+e}{x-y}\right)^l[/tex]


    How did the authors get their result?

    Regards
     
    Last edited: Mar 5, 2009
  2. jcsd
  3. Mar 5, 2009 #2

    Avodyne

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    Science Advisor

    Try simplifying the would-be identity as much as possible, and see how far you can get.
     
  4. Mar 5, 2009 #3
    Execuse me, what is the would-be identity? Can you help me a little to start?
     
  5. Mar 5, 2009 #4

    Avodyne

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    Science Advisor

    The would-be identity is

    [tex](c-1)!\,\exp\!\!\left(-d\,\frac{x+e}{x-y}\right)\sum_{k=0}^{c-1} \frac{d^k}{k!}\!\left(\frac{x+e}{x-y}\right)^{\!\!k} = (c-1)!\,\exp\!\!\left(-d\,\frac{y+e}{x-y}\right)\exp(-d)\sum_{k=0}^{c-1}\sum_{l=0}^k \frac{d^k}{k!}{k\choose l}\!\!\left(\frac{y+e}{x-y}\right)^{\!\!l}[/tex]

    Try to simplify this, and then show that the simplified equation is true. You can start with the obvious step of canceling the [itex](c{-}1)![/itex] on each side. Then, put all the exponentials on one side and see if you can simplify them.
     
  6. Mar 6, 2009 #5
    Ok, now I know what they did, they just changed the form of one equation, which is [tex]\left(\frac{x+e}{x-y}\right)[/tex] ,to the more convenient form for mathematical manipulation [tex]\left(\frac{y+e}{x-y}+1\right)[/tex].
    Thank you very much Avodyne, you helped me to figure it out.
    Regards
     
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