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Re: Integrate Problem

  • To: mathgroup at smc.vnet.net
  • Subject: [mg40512] Re: [mg40502] Integrate Problem
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Wed, 9 Apr 2003 01:29:53 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

First of all, you are just wasting effort typing in the Assumptions etc 
part here. It makes no difference at all:


Integrate[1/Pi^2  1/(1 + x^2 + y^2 + z^2)^2, {z, -Infinity, Infinity}]

(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))

The Assumptions mechanism in Integrate does not work the way you think. 
Basically it is just a way of avoiding getting conditional answers in 
Integrate so you should first try Integrate without any Assumptions. If 
the output begins with If ... then you can run it again with suitable 
Assumptions.  But actually it is usually easier to just apply Simplify 
with assumptions to the output. (Besides, Assumptions in Integrate are 
intended to ensure convergence and not to determine whether the answer 
is real or not).

As for your desired answer you can get it simply by applying 
ComplexExpand:


ComplexExpand[Integrate[1/Pi^2  1/(1 + x^2 + y^2 +
         z^2)^2, {z, -Infinity, Infinity}]]



1/(2*Pi*(1 + x^2 + y^2)^(3/2))


Alternatively, use Simplify with the Assumption that x and y are both 
real:


Simplify[Integrate[1/Pi^2  1/(1 + x^2 + y^2 + z^2)^2, {z, -Infinity,
                Infinity}],Element[x|y,Reals]]



1/(2*Pi*(1 + x^2 + y^2)^(3/2))

You should not confuse Assumptions in Integrate and in Simplify, they 
use a different mechanism and serve a different purpose.

Andrzej Kozlowski
Yokohama, Japan
http://www.mimuw.edu.pl/~akoz/
http://platon.c.u-tokyo.ac.jp/andrzej/



On Tuesday, April 8, 2003, at 04:05  pm, Stewart Mandell 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
>
>
>
>
>


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From: bobhanlon at aol.com (Bob Hanlon)
To: mathgroup at smc.vnet.net
References: <b6tshi$n0o$1 at smc.vnet.net>
Organization: AOL http://www.aol.com
Subject: [mg40512] Re: arrows at both ends of a line?

Needs["Graphics`Arrow`"];

twoHeadedArrow[pt1_, pt2_, opts___] := 
  {Arrow[pt1, pt2, opts], Arrow[pt2, pt1, opts]}

twoHeadedArrow[{twoHeadedArrow[{,0},{1,1}],Hue[0],
        twoHeadedArrow[{.75,.25},{.25,.75}]}]];


Bob Hanlon

In article <b6tshi$n0o$1 at smc.vnet.net>, Daniel Nettels
<Daniel.Nettels at unifr.ch> wrote:

<< 
Subject:	arrows at both ends of a line?
From:		Daniel Nettels <Daniel.Nettels at unifr.ch>
To: mathgroup at smc.vnet.net
Date:		Tue, 8 Apr 2003 07:08:34 +0000 (UTC)

Hi

How can I draw arrows at both and of a line?

Daniel Nettels













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