Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2005
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*November
*December
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 2005

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: Hypergeometric integral looks wrong ?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg55481] Re: Hypergeometric integral looks wrong ?
  • From: Maxim <ab_def at prontomail.com>
  • Date: Fri, 25 Mar 2005 05:48:22 -0500 (EST)
  • Organization: MTU-Intel ISP
  • References: <d1tv31$rk0$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

On Thu, 24 Mar 2005 08:51:45 +0000 (UTC), luigi <junk1 at lafaena.com> wrote:

> 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.
>
>

Yes, it is possible to convert one form of the answer to another, except  
that b*v (and not bv) should be put in the numerator. Mathematica's answer  
contains Hypergeometric2F1[..., E^(v*t)/z] rather than  
Hypergeometric2F1[..., z*E^(-v*t)], so we need an appropriate  
transformation identity. Using the formula found at  
http://functions.wolfram.com/07.23.17.0057.01 , we obtain (in version 5.1)

In[1]:=
Assuming[a < 0 && b > 0 && v > 0 && z < 1 && t > 0,
   b*v*Integrate[E^(-b*v*s)*(1 - z*E^(-v*s))^(-a), {s, 0, t}] /.
         Beta[z_, a_, b_] ->
           z^a/a*Hypergeometric2F1[a, 1 - b, a + 1, z] /.
       Hypergeometric2F1[a_, b_, c_, z_] ->
         Gamma[b - a]*Gamma[c]/(Gamma[b]*Gamma[c - a])*
             Hypergeometric2F1[a, a - c + 1, a - b + 1, 1/z]/(-z)^a +
           Gamma[a - b]*Gamma[c]/(Gamma[a]*Gamma[c - b])*
             Hypergeometric2F1[b, b - c + 1, b - a + 1, 1/z]/(-z)^b //
     FullSimplify[#, ExcludedForms -> _Hypergeometric2F1]&]

Out[1]=
Hypergeometric2F1[a, b, 1 + b, z] -
   Hypergeometric2F1[a, b, 1 + b, z/E^(t*v)]/E^(b*t*v)

as requested.

Maxim Rytin
m.r at inbox.ru


  • Prev by Date: symbolic quaternionic analysis
  • Next by Date: MapAt changes its response towards different expressions
  • Previous by thread: Hypergeometric integral looks wrong ?
  • Next by thread: Re: Hypergeometric integral looks wrong ?