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?

  • $\begingroup$ 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

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

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