Re: Solving integrals

*To*: mathgroup at smc.vnet.net*Subject*: [mg68672] Re: Solving integrals*From*: Peter Pein <petsie at dordos.net>*Date*: Wed, 16 Aug 2006 03:35:58 -0400 (EDT)*References*: <ebhi41$11p$1@smc.vnet.net> <ebpm4o$p9i$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Jens Benecke schrieb: > Jens Benecke wrote: > >> Hello everybody, >> >> I am trying to solve this integral: >> >> f :== int ( cos(x) / sqrt[A=C2=B2=C2=B7sin=C2=B2(x/2) + (Bx)=C2=B2 = >> ] dx, x==0..N*2*pi > > Hello, > > 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). ... >> Thank you! :) > Hi Jens, sorry, your integral does not converge on the interval [0, something], because your integrand is c_(-1)*x^(-1) + c_(1)*x + ... near x=0: expr = Cos[x]/Sqrt[(a*Sin[x/2])^2 + (b*x)^2]; Normal[Simplify[Series[expr, {x, 0, 1}], a > 0 && b > 0]]//InputForm Out[2]//InputForm= 2/(Sqrt[a^2 + 4*b^2]*x) + ((-11*a^2 - 48*b^2)*x)/(12*(a^2 + 4*b^2)^(3/2)) Peter