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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$\begingroup$ You should tell us what is known about $\alpha$ and $A$. Are they positive? In a given range? $\endgroup$– Andrej BauerCommented Aug 21, 2013 at 10:59
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$\begingroup$ Yes. $A$ and $\alpha$ are both positive. $\endgroup$– user38925Commented Aug 22, 2013 at 2:47
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$\begingroup$ you will also need A<1 to make the integral converge $\endgroup$– Carlo BeenakkerCommented Aug 22, 2013 at 7:08
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1 Answer
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If $\Re(A)<1$ and $\Re(\alpha)>0$ Mathematica gives:
$$\int_0^1 \mbox{ }x^{\alpha-1} \,(1-x) ^{-A}\, {}_2F_1 (1-A, \alpha -1-A; \alpha -A; x) \mbox{ }dx$$ $$\mbox{ }=\frac{\Gamma(\alpha)\Gamma(1-A)}{\Gamma(1+\alpha-A)}\mbox{ } \mbox{}_3F_2(\alpha,1-A,\alpha-A-1;\alpha-A,1+\alpha-A;1)$$
answered Aug 21, 2013 at 11:52
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1$\begingroup$ This is what you get by integrating termwise, using the beta integral. $\endgroup$ Commented Aug 21, 2013 at 12:44
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$\begingroup$ @IraGessel By termwise, I assume you mean "integrating the hypergeometric series termwise"?! $\endgroup$ Commented Aug 21, 2013 at 17:34
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$\begingroup$ Yes, I mean integrating $x^{\alpha -1}(1-x)^{-A}$ times each term of the hypergeometric series. $\endgroup$ Commented Aug 21, 2013 at 19:10