Re: Definite Integration in Mathematica

*To*: mathgroup at smc.vnet.net*Subject*: [mg74606] Re: Definite Integration in Mathematica*From*: "David W.Cantrell" <DWCantrell at sigmaxi.net>*Date*: Wed, 28 Mar 2007 01:40:22 -0500 (EST)*References*: <etqo3f$10i$1@smc.vnet.net> <eu7rtg$bgr$1@smc.vnet.net> <euan9u$dpo$1@smc.vnet.net>

"dimitris" <dimmechan at yahoo.com> wrote: [snip] > It's just happen in this case the antiderivative returned by the other > CAS to be continuous in the real axis. For other integrals usually > discontinuous antiderivatives are returned. For example > > (*other CAS*) > int(1/(5+cos(x)),x); > 1/6*6^(1/2)*arctan(1/3*tan(1/2*x)*6^(1/2)) > > (*mathematica*) > Integrate[1/(5 + Cos[x]), x] > ArcTan[Sqrt[2/3]*Tan[x/2]]/Sqrt[6] > > both have jump discontinuity at x=+/- n*Pi, n=odd. > > See this thread for a clever method by Peter Pein in order > to get a continuous antiderivative with Mathematica. You seem to be implying that Peter's method can be used to produce an antiderivative, continuous on R, for 1/(5 + Cos[x]). I don't see how. Could you please show us? > But in case of recognizing the jump discontinuity and its position, it > is very easy to obtain a continous antiderivative adding the piecewise > constant (see the relevant example from Trott's book!) Unfortunately, I don't have Trott's book. Since it's very easy to obtain, could you also please show us the continuous antiderivative for 1/(5 + Cos[x]) obtained by Trott's method? David W. Cantrell