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MathGroup Archive 2006

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Re: Solving integrals

  • To: mathgroup at smc.vnet.net
  • Subject: [mg68685] Re: Solving integrals
  • From: Jens Benecke <jens-news at spamfreemail.de>
  • Date: Wed, 16 Aug 2006 03:36:16 -0400 (EDT)
  • References: <ebhi41$11p$1@smc.vnet.net> <ebpm4o$p9i$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Jens Benecke wrote:

> sorry, this post apparently got mangled ("²" aka "^2" wasn't recognized).
> Here is the integral again:
> 
> f := int ( cos(x) / sqrt[ A*A*sin(x/2)*sin(x/2) + (Bx)^2] dx, x=0..N*2*pi
>
> where N==(int)0..20, and A and B are small fixed values (around
> 0.0001..0.05).
 
Hello everybody,

since I've been told by email a number of times:
yes, I know this integral has a singularity. However, it represents a
physical measure (so it must exist), and a simpler form (without the (Bx)^2
in the denominator) can be and has been solved via conversion to Elliptic
Integrals.

I'm just at a los as how to split this one up.

Please help me a bit :)


-- 
Jens Benecke


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