MathGroup Archive 2003

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

Search the Archive

Re: Integrate Problem

  • To: mathgroup at smc.vnet.net
  • Subject: [mg40527] Re: [mg40502] Integrate Problem
  • From: Dr Bob <majort at cox-internet.com>
  • Date: Wed, 9 Apr 2003 01:33:22 -0400 (EDT)
  • References: <200304080705.DAA23549@smc.vnet.net>
  • Reply-to: majort at cox-internet.com
  • Sender: owner-wri-mathgroup at wolfram.com

ComplexExpand@Integrate[1/Pi^2 1/(1 + x^2 + y^2 + z^2)^2,
 {z, -Infinity, Infinity}, Assumptions -> {Im[x] == 0, Im[y] == 0, Im[z] == 
0}]

Bobby

On Tue, 8 Apr 2003 03:05:18 -0400 (EDT), Stewart Mandell 
<stewart at rentec.com> wrote:

> When I run
>
> Integrate[1/Pi^2  1/(1 + x^2 + y^2 + z^2)^2, {z, -Infinity, Infinity},
> Assumptions -> {Im[x] == 0, Im[y] == 0, Im[z] == 0}]
>
> I get
> (I*(Log[-(I/Sqrt[1 + x^2 + y^2])] -
> Log[I/Sqrt[1 + x^2 + y^2]]))/
> (2*Pi^2*(1 + x^2 + y^2)^(3/2))
>
> I would like
>
> 1/(2*Pi ) 1/(1 + x^2 + y^2)^3/2
>
> for an answer. How do I get Mathematica to forego the complex
> answer?
>
> thanks, Stewart
>
>
>
>



-- 
majort at cox-internet.com
Bobby R. Treat



  • Prev by Date: RE: How do you get polynomials listed in order of decreasing exponent?
  • Next by Date: Re: Integrate Problem
  • Previous by thread: Integrate Problem
  • Next by thread: Re: Integrate Problem