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MathGroup Archive 2005

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Hypergeometric integral looks wrong ?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg55455] Hypergeometric integral looks wrong ?
  • From: "luigi" <junk1 at lafaena.com>
  • Date: Thu, 24 Mar 2005 03:41:54 -0500 (EST)
  • Sender: owner-wri-mathgroup at wolfram.com

I have to compute the following integral:

(1/ bv) Integrate[Exp[- b v s](1 - z Exp[-v s])^(-a), {s, 0, Infinity},
Assumptions -> {Re[b] > 0, Re[v] > 0, Re[b v] > 0}]

In my case, z runs over (-Infinity,1), and a, b v are real ( a < 0).
Now if the limits of integration were (0, Infinity) then this is just the
hypergeometric function 2F1[a, b, b+1, z]. But since I am integrating over
the interval (0, t), direct calculation yields

2F1[a, b, b+1, z] - Exp[-b v t] 2F1[a, b, b+1, z Exp [- v t]]

But Mathematica 5.0 instead yields

(1-z)^(-a) 2F1[a, a-b, 1+a-b , 1/z] ((z-1)/z)^(a) - Exp[-b v t] (1-Exp[v
t]/z)^a (1 - Exp[-v t] z)^(-a) 2F1[a, a-b, 1+a-b , Exp[v t]/z ]

This looks quite different and, moreover, does not seem to make sense for
the range of z I am allowing. Is mathematica's result right ? Am I missing
some transformation that links both results ?

Thanks for any help.



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