Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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?

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


    User Avatar
    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


    User Avatar
    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook