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Math Help - Integral

  1. #1
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    Integral

    does anyone know how to evaluate this integral

     \Phi(X) = C \cdot (b-X)^{\alpha-1}(X-a)^{\beta-1}  dX

    where C, a,b,\alpha and  \beta are constants
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  2. #2
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    Quote Originally Posted by chogo View Post
    does anyone know how to evaluate this integral

     \Phi(X) = C \cdot (b-X)^{\alpha-1}(X-a)^{\beta-1}  dX

    where C, a,b,\alpha and  \beta are constants
    You fail to mention the limits of integration.
    ---
    This is what I will assume:
    1) b>a
    2)The limits of integration are from x=a \mbox{ to }x=b.
    3)And \alpha,\beta >0.
    ---
    We can ignore the constant C and instead consider:
    \int_a^b (b-x)^{\alpha - 1}(x-a)^{\beta - 1} dx
    Use the functional substitution s=x-a:

    \int_0^{b-a} (b-a -s)^{\alpha - 1}\cdot s^{\beta -1} ds = (b-a)^{\alpha - 1} \int_0^{b-a} \left( 1 - \frac{s}{b-a} \right)^{\alpha - 1}\cdot s^{\beta - 1} ds

    Now use the substitution t=\frac{s}{b-a}:

    (b-a)(b-a)^{\alpha - 1}\int_0^1 (1-t)^{\alpha - 1}\cdot t^{\beta - 1} \cdot (b-a)^{\beta - 1} dt = (b-a)^{\alpha + \beta - 1} \int_0^1 (1-t)^{\alpha - 1} t^{\beta-1} dt

    Realize that this is the Beta Integral.

    Thus,
    (b-a)^{ \alpha + \beta - 1} \cdot \bold{B} (\alpha , \beta )

    We can convert this to the Gamma function.

    \boxed{ (b-a)^{\alpha + \beta - 1} \cdot \frac{\Gamma (\alpha )\cdot \Gamma (\beta )}{\Gamma (\alpha + \beta) }}

    So if \alpha,\beta are positive integers we end up with a nice looking result in the end.

    This is Mine 6th Post!!!
    Last edited by ThePerfectHacker; June 13th 2007 at 07:52 PM.
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  3. #3
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    thanks alot. I stuck it into the wolfram integrator and got a formula using the hypergeometric funtion. This makes more sense. The limits are however

     x=t, x=s

    where  s>t

    thanks and congrats on the 6000 post
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