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I obtained the following integral when looking for a probability density function: $$\int_0^1 x^{\alpha-1} \,(1-x) ^{-A}\, {}_2F_1 (1-A, \alpha -1-A, \alpha -A, x) \,dx$$ Can anyone please give me some hints of evaluating the value of the integral?

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You should tell us what is known about $\alpha$ and $A$. Are they positive? In a given range? –  Andrej Bauer Aug 21 '13 at 10:59
    
Yes. $A$ and $\alpha$ are both positive. –  user38925 Aug 22 '13 at 2:47
    
you will also need A<1 to make the integral converge –  Carlo Beenakker Aug 22 '13 at 7:08

1 Answer 1

If $\Re(A)<1$ and $\Re(\alpha)>0$ Mathematica gives:

$$\int_0^1 \mbox{ }x^{\alpha-1} \,(1-x) ^{-A}\, {}_2F_1 (1-A, \alpha -1-A; \alpha -A; x) \mbox{ }dx$$ $$\mbox{ }=\frac{\Gamma(\alpha)\Gamma(1-A)}{\Gamma(1+\alpha-A)}\mbox{ } \mbox{}_3F_2(\alpha,1-A,\alpha-A-1;\alpha-A,1+\alpha-A;1)$$

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This is what you get by integrating termwise, using the beta integral. –  Ira Gessel Aug 21 '13 at 12:44
    
@IraGessel By termwise, I assume you mean "integrating the hypergeometric series termwise"?! –  Igor Rivin Aug 21 '13 at 17:34
    
Yes, I mean integrating $x^{\alpha -1}(1-x)^{-A}$ times each term of the hypergeometric series. –  Ira Gessel Aug 21 '13 at 19:10

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