I obtained the following integral when looking for a probability density function: $$\int_0^1 x^{\alpha1} \,(1x) ^{A}\, {}_2F_1 (1A, \alpha 1A, \alpha A, x) \,dx$$ Can anyone please give me some hints of evaluating the value of the integral?
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$\begingroup$ You should tell us what is known about $\alpha$ and $A$. Are they positive? In a given range? $\endgroup$ – Andrej Bauer Aug 21 '13 at 10:59

$\begingroup$ Yes. $A$ and $\alpha$ are both positive. $\endgroup$ – user38925 Aug 22 '13 at 2:47

$\begingroup$ you will also need A<1 to make the integral converge $\endgroup$ – Carlo Beenakker Aug 22 '13 at 7:08
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If $\Re(A)<1$ and $\Re(\alpha)>0$ Mathematica gives:
$$\int_0^1 \mbox{ }x^{\alpha1} \,(1x) ^{A}\, {}_2F_1 (1A, \alpha 1A; \alpha A; x) \mbox{ }dx$$ $$\mbox{ }=\frac{\Gamma(\alpha)\Gamma(1A)}{\Gamma(1+\alphaA)}\mbox{ } \mbox{}_3F_2(\alpha,1A,\alphaA1;\alphaA,1+\alphaA;1)$$

1$\begingroup$ This is what you get by integrating termwise, using the beta integral. $\endgroup$ – Ira Gessel Aug 21 '13 at 12:44

$\begingroup$ @IraGessel By termwise, I assume you mean "integrating the hypergeometric series termwise"?! $\endgroup$ – Igor Rivin Aug 21 '13 at 17:34

$\begingroup$ Yes, I mean integrating $x^{\alpha 1}(1x)^{A}$ times each term of the hypergeometric series. $\endgroup$ – Ira Gessel Aug 21 '13 at 19:10