Re: How to simplify to a result that is real

• To: mathgroup at smc.vnet.net
• Subject: [mg50837] Re: How to simplify to a result that is real
• From: "David W. Cantrell" <DWCantrell at sigmaxi.org>
• Date: Thu, 23 Sep 2004 05:27:18 -0400 (EDT)
• References: <20040921115025.QQQW18891.lakermmtao10.cox.net@smtp.east.cox.net> <ciqv1p\$ihc\$1@smc.vnet.net> <cirejr\$mmc\$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```Paul Abbott <paul at physics.uwa.edu.au> wrote:
>
>   SetOptions[Integrate, GenerateConditions -> False];
>
>   Simplify[Integrate[1/(a + b Cos[t]), {t, 0, c}], a > b > 0]

But of course, since you asked that no conditions be generated, that result
could then be deceptive. Why? The integrand is continuous on R (since we
know that a > b > 0) and thus there is a result which is valid for all
real c. However, Mathematica does not give us such a result. If one naively
were to replace c by 2*Pi in Mathematica's result

(2*ArcTan[((a - b)*Tan[c/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2]

one would get 0 for the value of the integral, and that is absurd.

A result which is valid for all real c would have been

1/Sqrt[a^2-b^2]*(c - 2*ArcTan[b*Sin[c]/(b*Cos[c] + a + Sqrt[a^2-b^2])])

but I don't think that Mathematica can be pursuaded to give that. Of
course it would be nice if it could because then there would be no need to
generate conditions (since a > b > 0 had already been specified).

David Cantrell

```

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