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So far I have $B(x,y+1) = \int_{0}^{1} t^{x-1}(1-t)^{(y+1)-1} dt =\int_{0}^{1} t^{x-1}(1-t)^{y} dt $ for $x,y>0$. I tried doing integration by parts by letting $dv = (1-t)^y$ and $u=t^{x-1}$ but it didn't really take me anywhere. I have to find this identity starting with the beta definition I used above. I know there is a gamma identity I can use to derive this, but like I said, my instructions are to start with the beta function I used above. Any recommendations? Note I am an undergraduate learning bessel functions so I don't have detailed background on beta functions and their properties as the graduate school level. I am looking for mathematical manipulations rather than theoretical explanations that wouldn't make sense to me. Thanks.

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If you want to do this by integration by parts, try showing that

$$B(x+1,y+1)=\int_0^\infty\frac{s^{x}}{(1+s)^{x+y+2}}ds$$

which can be done by making the substitution $t=s^2/(1-s^2)$. Once you've shown the above, you'll see that integration by parts will give you what you want.

As you know this can also be derived from a broader identity:

$$\Gamma(x)\Gamma(y)=\Gamma(x+y)B(x,y)$$

whose proof can be found here. Indeed, we get that:

$$B(x,y+1)=\frac{\Gamma(x)\Gamma(y+1)}{\Gamma(x+y+1)}=\frac{\Gamma(x)\Gamma(y)y}{\Gamma(x+y)(x+y)}=\frac{y}{x+y}B(x,y)$$

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"I know there is a gamma identity I can use to derive this, but like I said, my instructions are to start with the beta function I used above. Any recommendations?" That's why I deleted my comment. –  Pedro Tamaroff Oct 7 '12 at 22:50
    
@Peter Tamaroff: corrected. –  Alex R. Oct 8 '12 at 1:09

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