Re: A tough Integral

*To*: mathgroup at smc.vnet.net*Subject*: [mg28560] Re: A tough Integral*From*: "Robert Miller" <rmiller at archimedestechnology.com>*Date*: Fri, 27 Apr 2001 03:56:14 -0400 (EDT)*References*: <9c5ndf$ir1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

The following should do it: Divide the integration region into lengths of 2 Pi and use Cos[x]==Cos[x+n*2*Pi] , n an integer to write Integrate[1/(Cos[x] + x^2), {x, 0, \[Infinity]}] = = Sum[Integrate[1/(Cos[x] + (x + n*2*Pi)^2), {x, 0, 2 \[Pi]}], {n, 0, \[Infinity]}] Interchange the order of summation and integration. Mathematica can do the sum sum=Sum[1/(Cos[x] + (x + n*2*Pi)^2), {n, 0, \[Infinity]}]//FullSimplify It gives the answer in terms of PolyGamma Then you can NIntegrate sum from 0 to 2 Pi. Note, however, that in the form in which Mathematica returns the sum, the denominator ->0 at Pi/2 and 3 Pi/2 ( the numerator ->0 as well and limit is OK). So make these endpoints of integration to avoid NIntegrate error messages: NIntegrate[sum,{x,0 ,Pi/2}]+NIntegrate[sum,{x,Pi/2, 3 Pi/2}]+NIntegrate[sum,{x,3 Pi/2, 2 Pi}] with WorkingPrecision->80, I get 1.83801769501050385269531439813992926843308653301263139425699189822880091 Robert Miller "bobbym1953" <bobbym1953 at aol.com> wrote in message news:9c5ndf$ir1 at smc.vnet.net... > Does anyone know how I can get the following Integral to at least 60 places, > using Mathematica? > > Int(1/(cos(x)+x^2)) between x=0 and x=infinity. Both Integrate and NIntegrate > seemed helpless. > > Thanks, > Angela > > bobbym1953 at aol.com >