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    • Thread Starter

    Would just like a hint on how to get it because I'm clearly not doing it right, I did the following:
    \int_{0}^{1}x\frac{\Gamma(\alpha  +\beta)}{\Gamma(\alpha)\Gamma(\b  eta)}x^{\alpha-1}(1-x)^{\beta - 1}dx
    =\int_{0}^{1}\frac{\Gamma(\alpha  +\beta)}{\Gamma(\alpha)\Gamma(\b  eta)}x^{\alpha}(1-x)^{\beta - 1}dx
    then I thought I could just change it to this
    =\int_{0}^{1}\frac{\Gamma(\alpha  +\beta)}{\Gamma(\alpha)\Gamma(\b  eta)}\frac{\Gamma(\alpha+1)\Gamm  a(\beta)}{\Gamma((\alpha+1)+ \beta)}dx
    and then cancel from there but it obviously is not the answer I'm looking for as I end up with
    =\int_{0}^{1}\frac{\Gamma(\alpha  +\beta)}{\Gamma(\alpha)}\frac{\G  amma(\alpha+1)}{\Gamma((\alpha+1  )+ \beta)}dx
    Which obviously will not give me the answer I'm looking for,can anyone tell me what I should have done instead?
    (I know the latex isn't perfect but can't see where the errors are, obviously the lpha is an alpha and the eta is beta)

    This is all correct. From here, note that the integral is not needed (everything under the integral sign is a constant). Also, use the fact that Gamma(a+1)=(a)(Gamma(a)) (or prove this using the definition of the gamma function and integration by parts).
    • Thread Starter

    Yeah, I previously got rid of the integral just didn't type it up. Wouldn't that all just cancel to 1 though if changed it to aGamma(a)? I know the mean is supposed to be \frac_{\alpha+\beta}{\alpha}

    No. Gamma(a+1)/Gamma(a) = a and Gamma ((a+b)+1)/Gamma(a+b) = a+b.
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Updated: November 9, 2008
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