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Math Help - Integration Problem

  1. #1
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    Integration Problem

    Question: "The factorial n! of a non-negative integer n is the product of all positive integers less than or equal to n. The gamma function is an extension of the factorial function to real numbers. It is defined by:

    \Gamma(x)=\int_0^\infty{e^{-t}}{t^{x-1}}{dt}, {x>0}

    (a) Evaluate \Gamma(1)

    Assuming the limits represent t, if I evaluate \Gamma(1), does the function become \int_0^\infty{e^{-t}{dt}? In which case when I integrate do I get: {-e^{-t}} with limits between 0 and infinity, which equals 1?

    Part (b):

    "Show that for x>0:

    \Gamma(x+1)={x\Gamma(x)}

    So I thought that \Gamma(x+1)=\int_0^\infty{e^{-t}}{t^x}{dt}. If I let u=t^x, then u'={xt^{x-1}}, and if I let {v'=e^{-t}}, then {v=-e^{-t}}. Then I can use 'by parts' to show this. Do I treat x as some constant, like n?

    Finally:

    "(c) Show that for n=1,2,3...

    \Gamma(n)=(n-1)!

    That is, the gamma function generalizes the factorial, with its argument shifted down by 1"

    I am entirely unsure about how to work this one out. Any tips/hints? Many thanks!
    CP
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  2. #2
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    Re: Integration Problem

    Quote Originally Posted by CrispyPlanet View Post
    Assuming the limits represent t, if I evaluate \Gamma(1), does the function become \int_0^\infty{e^{-t}{dt}? In which case when I integrate do I get: {-e^{-t}} with limits between 0 and infinity, which equals 1?
    Yes

    Quote Originally Posted by CrispyPlanet View Post
    Then I can use 'by parts' to show this. Do I treat x as some constant, like n?
    Also yes.

    Quote Originally Posted by CrispyPlanet View Post
    That is, the gamma function generalizes the factorial, with its argument shifted down by 1"

    I am entirely unsure about how to work this one out. Any tips/hints? Many thanks!
    CP
    Try induction along with your result from part (b).
    Thanks from CrispyPlanet
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  3. #3
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    Re: Integration Problem

    Ok thanks, will give that a go.
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  4. #4
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    Re: Integration Problem

    Quote Originally Posted by SlipEternal View Post


    Try induction along with your result from part (b).
    Ok I've done the base step, and shown that both sides are equal to 1. For the induction step I am stuck. Using part b: \Gamma(k+1)=k\Gamma(k). Where do I go from here?
    CP

    EDIT: I'm rubbish at math induction!
    Last edited by CrispyPlanet; October 22nd 2013 at 01:01 PM.
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  5. #5
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    Re: Integration Problem

    By the induction hypothesis, k\Gamma(k) = k(k-1)! = k! = ((k+1)-1)!.
    Thanks from CrispyPlanet
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  6. #6
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    Re: Integration Problem

    Note: \displaystyle \begin{align*} \Gamma \left( n \right) = (n - 1)! \end{align*} if n is a positive integer, so \displaystyle \begin{align*} \Gamma (1) = 0! = 1 \end{align*}.
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