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Math Help - Gamma Distribution

  1. #1
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    Gamma Distribution

    Expected Value

    \displaystyle\text{E}[X]=\frac{1}{\Gamma\left(\frac{2}{5}\right)20^{2/5}}\lim_{b\to\infty}\int_0^bx^{2/5}e^{-x/20}dx

    \displaystyle\Gamma\left(\frac{2}{5}\right)=\lim_{  b\to\infty}\int_0^bx^{-3/5}e^{-x}dx

    How do I solve this Gamma function?
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  2. #2
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    Have a go with this,

    It is known that

    \displaystyle\Gamma\left(\frac{1}{5}\right)\approx 4.591

    And \displaystyle\frac{\Gamma (x)\Gamma (y)}{\Gamma (x+y)}= \int_0^1 t^{x-1}(1-t)^{y-1}~dt

    so \displaystyle \Gamma \left(\frac{2}{5}\right) = \Gamma \left(\frac{1}{5}+\frac{1}{5}\right)=\frac{\Gamma \left(\frac{1}{5}\right)\Gamma \left(\frac{1}{5}\right)}{ \int_0^1 t^{\frac{-4}{5}}(1-t)^{\frac{-4}{5}}~dt}

    Not sure how valid this will be given my time away from the properties governing this function.
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  3. #3
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    Quote Originally Posted by dwsmith View Post
    Expected Value

    \displaystyle\text{E}[X]=\frac{1}{\Gamma\left(\frac{2}{5}\right)20^{2/5}}\lim_{b\to\infty}\int_0^bx^{2/5}e^{-x/20}dx

    \displaystyle\Gamma\left(\frac{2}{5}\right)=\lim_{  b\to\infty}\int_0^bx^{-3/5}e^{-x}dx

    How do I solve this Gamma function?
    Why can't you leave it as \Gamma \left(\frac{2}{5}\right)?
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  4. #4
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    Quote Originally Posted by mr fantastic View Post
    Why can't you leave it as \Gamma \left(\frac{2}{5}\right)?
    I don't know if I can. I was just going to solve everything if I could.
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  5. #5
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