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I was wondering if someone could possibly tell me what this integral is or possibly help me figure out an upper and lower bound for this integral (in terms of $X,Y,\alpha$):

$\int_{X}^{Y} v^{\alpha} (1-v)^{\alpha} dv$, where $-1< \alpha <0$ and $0< X < Y <1$. $\alpha, X, Y$ are all real numbers.

Thank you!

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Google "Beta Function". It is very beautiful stuff. –  DonAntonio Nov 11 '12 at 3:39
    
Thank you very much! –  J Kasahara Nov 11 '12 at 18:44
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1 Answer

up vote 0 down vote accepted

You can have a closed form formula in terms of the hypergeometric function

$$\int_{x}^{y} v^{\alpha} (1-v)^{\alpha} dv= -{\frac {{x}^{1+\alpha}{ _2F_1(-\alpha,1+\alpha;\,2+\alpha;\,x)}-{y}^{1+ \alpha}{_2F_1(-\alpha,1+\alpha;\,2+\alpha;\,y)}}{1+\alpha}},$$ where $_2F_1$ $$ \,_2F_1(a,b;c;z) = \sum_{n=0}^\infty {(a)_n (b)_n \over (c)_n} \, {z^n \over n!} $$

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Thank you very much! –  J Kasahara Nov 11 '12 at 18:44
    
@JKasahara: You are welcome. Glad to assist. –  Mhenni Benghorbal Nov 11 '12 at 20:36
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