Re: Mathematica and Maple disagree on this integral
- To: mathgroup at smc.vnet.net
- Subject: [mg33225] Re: Mathematica and Maple disagree on this integral
- From: Ben Crain <bcrain at bellatlantic.net>
- Date: Sat, 9 Mar 2002 03:20:12 -0500 (EST)
- References: <200203011152.GAA27858@smc.vnet.net> <a61vd9$gqb$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Thanks for the clarification. I have learned about the "principal value"
integral, but did not know that one must distinguish between various definitions
of the integral in Mathematica (or in Maple). I think the "default" integral
should be the most general, or most useful, definition of "integral". Clearly
(in my opinion) the most useful definition of "integral" would be (at least) the
"principal value" integral; so if you ask for Mathematica's integral of this
function, what it returns is correct (i.e., most useful). My standard for
judgment is very practical: suppose you had a model for a real-word process
(e.g., designing an engine for a rocket), which required the evaluation of this
integral. Would you reject the design because the integral is, according to
some definition, "divergent", or would you simply set the integral to zero, and
accept the model. I think almost all engineers would accept the model (assuming
that is the only possible problem), and I am certain that the engine would work
(assuming everthing else in the engine's design is correct). That means that
the "principal value" integral really is the "correct" answer. Other integrals
are simply wrong!
Philippe Dumas wrote:
> 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
> to zero in your case.
> 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: [mg33225] 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.
> >
> >
- References:
- Mathematica and Maple disagree on this integral
- From: Ben Crain <bcrain@bellatlantic.net>
- Mathematica and Maple disagree on this integral