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MathGroup Archive 2007

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Integral that Mathematica 5.1 can do and Mathematica 6 cannot do

  • To: mathgroup at smc.vnet.net
  • Subject: [mg77562] Integral that Mathematica 5.1 can do and Mathematica 6 cannot do
  • From: Andrew Moylan <andrew.j.moylan at gmail.com>
  • Date: Wed, 13 Jun 2007 07:24:04 -0400 (EDT)

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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