Re: Equality question?
- To: mathgroup at smc.vnet.net
- Subject: [mg59476] Re: Equality question?
- From: Paul Abbott <paul at physics.uwa.edu.au>
- Date: Wed, 10 Aug 2005 02:57:56 -0400 (EDT)
- Organization: The University of Western Australia
- References: <dd9mq3$is1$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
In article <dd9mq3$is1$1 at smc.vnet.net>, "ab at sd.com" <at2 at ads.com>
wrote:
> EQ1 = ((v*a*(x^2 - a^2))/(4*Pi))*Integrate[1, {\[Phi]p, 0, 2*Pi}]*
> Integrate[(-Sin[\[Theta]p])*((a^2 + x^2 - 2*a*x*Cos[\[Theta]p])^(-3/2) -
> (a^2 + x^2 + 2*a*x*Cos[\[Theta]p])^(-3/2)), {\[Theta]p, Pi/2, 0}]
>
> EQ2 = v*(1 - (z^2 - a^2)/(z*Sqrt[z^2 + a^2]))
>
> Is EQ1 equal to EQ2 ? or not?
Clearly not. EQ1 involves x and EQ2 does not. Also, the result you are
after requires the assumption x > a > 0.
Note that computing the indefinite integral and substituting in the
integration limits is much faster than computing the definite integral
directly.
Cheers,
Paul
--
Paul Abbott Phone: +61 8 6488 2734
School of Physics, M013 Fax: +61 8 6488 1014
The University of Western Australia (CRICOS Provider No 00126G)
AUSTRALIA http://physics.uwa.edu.au/~paul
http://InternationalMathematicaSymposium.org/IMS2005/