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 ("Â²" 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