Re: Integral that Mathematica 5.1 can do and Mathematica 6 cannot do (2)
- To: mathgroup at smc.vnet.net
- Subject: [mg77636] Re: Integral that Mathematica 5.1 can do and Mathematica 6 cannot do (2)
- From: dimitris <dimmechan at yahoo.com>
- Date: Thu, 14 Jun 2007 05:25:21 -0400 (EDT)
- References: <f4okju$6mr$1@smc.vnet.net>
BTW, I think your asumptions could be 0<r&&0<\[Rho] &&0<s.
That is, no infinity is needed!
Dimitris
/ Andrew Moylan :
> Consider the following integral:
>
> Assuming[0 < r < Infinity && 0 < \[Rho] < Infinity && 0 < s <
> Infinity,
> Integrate[(\[Rho]*
> r)/(Sqrt[s^2 + z^2]*(1 + Sqrt[s^2 + z^2]/r)^2), {z, -
> Infinity,
> Infinity}]]
>
> Mathematica 5.1 gives the solution as:
>
> -((1/(45*s^2))*(2*r*\[Rho]*
> (-15*s^2 + 6*r^2*
> Hypergeometric2F1[1, 2,
> 7/2, 1 - r^2/s^2] -
> 6*s^2*Hypergeometric2F1[
> 1, 2, 7/2,
> 1 - r^2/s^2] -
> 3*(r^2 - s^2)*
> HypergeometricPFQ[
> {1, 1, 3/2}, {5/2,
> 7/2}, 1 - r^2/s^2] -
> 8*(r^2 - s^2)*
> HypergeometricPFQ[
> {1, 3/2, 2}, {5/2,
> 7/2}, 1 - r^2/s^2] -
> 4*r^2*HypergeometricPFQ[
> {3/2, 2, 2}, {5/2,
> 7/2}, 1 - r^2/s^2] +
> 4*s^2*HypergeometricPFQ[
> {3/2, 2, 2}, {5/2,
> 7/2}, 1 - r^2/s^2])))
>
> Mathematica 6 instead gives the following conditional solution:
>
> r^3*\[Rho]*If[r < s, ((-Pi)*r + 2*Sqrt[-r^2 + s^2] +
> 2*r*ArcTan[r/Sqrt[-r^2 + s^2]])/(-r^2 + s^2)^(3/2),
> Integrate[1/(Sqrt[s^2 + z^2]*(r + Sqrt[s^2 + z^2])^2),
> {z, -Infinity, Infinity}, Assumptions -> Element[z, Reals] &&
> s > 0 && r >= s && \[Rho] > 0]]