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Math Help - Integration proof

  1. #1
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    Integration proof

    How would i prove this integral
    for all integrals m,n >0
    Attached Thumbnails Attached Thumbnails Integration proof-problem.jpg  
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  2. #2
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    m,n\in\mathbb{Z} \ \mbox{or} \ m,n\in\mathbb{R}

    If integers, I would think by induction.
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  3. #3
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    for all integers m,n greater or equal to 0
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  4. #4
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    Proof: by induction
    Show P(1) is true
    Assume P(k) is true
    Prove P(k+1) is true.
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  5. #5
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    Quote Originally Posted by Monster32432421 View Post
    How would i prove this integral
    for all integrals m,n >0
    Use repeated integration by parts:

    u = (1 - x)^n \Rightarrow du = -n (1 - x)^{n-1} dx

    dv = x^m dx \Rightarrow v = \frac{1}{m+1} x^{m+1}

    Then

    \displaystyle I(m, n) = \frac{n}{m+1} \int_0^1 x^{m+1} (1 - x)^{n-1} \, dx
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    \displaystyle = \frac{n(n-1)(n-2) .... (n - [t-1])}{(m+1)(m+2)(m+3) .... (m+t)}\int_0^1 x^{m+t}(1 - x)^{n-t} \, dx


    Substitute t = n and it's left for you to finish.
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  6. #6
    MHF Contributor harish21's Avatar
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    Since

    \displaystyle \int_0^1 x^{\alpha - 1} \;(1-x)^{\beta -1} dx is a Beta function with B(\alpha,\beta) = \dfrac{\Gamma(\alpha) \; \Gamma(\beta)}{\Gamma(\alpha + \beta)}

    Also, \Gamma(\alpha) = (\alpha-1)!

    therefore,

    you can write:

    \displaystyle \int_0^1 x^m \; (1-x)^n \;dx = \int_0^1 x^{(m+1)-1} \; (1-x)^{(n+1)-1} \;dx,

    which is a beta function B(m+1, n+1)

    B(m+1, n+1) = \dfrac{\Gamma(m+1) \; \Gamma(n+1)}{\Gamma(m+n+2)}

    Now finish it off!
    Last edited by harish21; November 6th 2010 at 11:45 PM.
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  7. #7
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    Quote Originally Posted by harish21 View Post
    Since

    \displaystyle \int_0^1 x^{\alpha - 1} \;(1-x)^{\beta -1} dx is a Beta function with B(\alpha,\beta) = \dfrac{\Gamma(\alpha) \; \Gamma(\beta)}{\Gamma(\alpha + \beta)}

    Also, \Gamma(\alpha) = (\alpha-1)!

    therefore,

    you can write:

    \displaystyle \int_0^1 x^m \; (1-x)^n \;dx = \int_0^1 x^{(m+1)-1} \; (1-x)^{(n+1)-1} \;dx,

    which is a beta function B(m+1, n+1)

    B(m+1, n+1) = \dfrac{\Gamma(m+1) \; \Gamma(n+1)}{\Gamma(m+n+2)}

    Now finish it off!
    I don't think that quoting, more or less, a standard result constitutes a proof in the sense that the OP has requested.
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