Re: Mathematica and Maple disagree on this integral

• To: mathgroup at smc.vnet.net
• Subject: [mg33144] Re: [mg33091] Mathematica and Maple disagree on this integral
• From: "Philippe Dumas" <dumasphi at noos.fr>
• Date: Tue, 5 Mar 2002 03:08:51 -0500 (EST)
• References: <200203011152.GAA27858@smc.vnet.net>
• Reply-to: "Philippe Dumas" <dumasphi at noos.fr>
• Sender: owner-wri-mathgroup at wolfram.com

```Your question deals with the problem of defining correctly the value of such
integral of a function taking on infinite value. The proper way is to use
the so called "Cauchy principal value" being defined (in your case) as:
limit of  Integral[Sec[x],{x,eps,Pi-eps}] when eps-->0
Such limit is called the Cauchy principal value (noted vp) and does converge
Have a look to "PrincipalValue" in the on-line help.

Regards

Philippe Dumas
99, route du polygone
03 88 84 67 80
67100 Strasbourg

----- Original Message -----
From: "Ben Crain" <bcrain at bellatlantic.net>
To: mathgroup at smc.vnet.net
Subject: [mg33144] [mg33091] Mathematica and Maple disagree on this integral

> What is the definite integral of Sec(x), from 0 to pi?  A textbook
> answer (Stewart, Calculus) is that it diverges.
> And that is the answer Maple gives (calling it "undefined").  But
> Mathematica returns 0.
>
> The integral is split into two improper integrals, from 0 to pi/2 and
> from pi/2 to pi.  Each, by itself, diverges.  The textbook definition
> requires both improper integrals to separately converge for the total
> integral to converge.  By that definition, Maple is right.  But does
> that make sense?  The second improper integral is just the negative of
> the first, and they exactly cancel out for the antiderivative
> ln(abs(sec(t) + tan(t)) at any t close to pi/2.  Why don't they exactly
> offset each other in the limit, as t goes to pi/2, and yield 0?  Why
> shouldn't the integral be so defined, instead of the textbook
> requirement that the improper integrals must separately converge.
>
>

```

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