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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