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]]

```

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