       Re: Hypergeometric integral looks wrong ?

• To: mathgroup at smc.vnet.net
• Subject: [mg55500] Re: Hypergeometric integral looks wrong ?
• From: "luis serven" <junk1 at lafaena.com>
• Date: Sat, 26 Mar 2005 02:39:35 -0500 (EST)
• References: <d1tv31\$rk0\$1@smc.vnet.net> <d20r1q\$baq\$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```Thanks for your reply. However, the transformation you mention involves real
powers of (-z) ands is defined for z outside (0,1), while in my case I need
to allow for 1> z >=0.

Luigi

"Maxim" <ab_def at prontomail.com> wrote in message
news:d20r1q\$baq\$1 at smc.vnet.net...
> 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:=
> 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=
> 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
>

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