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MathGroup Archive 2000

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Re: Demonstrate that 1==-1

  • To: mathgroup at smc.vnet.net
  • Subject: [mg23202] Re: [mg23171] Demonstrate that 1==-1
  • From: David Withoff <withoff at wolfram.com>
  • Date: Mon, 24 Apr 2000 01:12:09 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

> Compute
> Integrate[(1 + a/E^(I*u))/(-1 + a/E^(I*u)), {u, 0, 2*Pi}]
> 
> Mathematica gives -2Pi
> 
> Now multiply the numerator and the denominator by -1
> 
> Integrate[(-1 - a/E^(I*u))/(1 - a/E^(I*u)), {u, 0, 2*Pi}]
> 
> Mathematica gets 2*Pi
> 
> This is only possible if 1==-1

It is also possible if the first integral implicitly makes
the assumption Abs[a]<1 and the second integral implicitly
makes the assumption Abs[a]>1.  Implicit assumptions like
this are quite common in definite Integrate with symbolic
parameters.  Although it is often desirable for Integrate
to report such assumptions, that does not yet happen in
all examples.

> Is this another bug in Limit?

No.

Since both results are arguably correct (for different
values of a), it isn't necessarily a bug at all.  In any
case, regardless of how one sees that sort of thing, it
may be more useful to simply note that at least a few
examples involving unreported assumptions about parameters
will probably remain for the foreseeable future.  Although
enormous progress has been made in this area in recent years,
it remains a difficult and largely unsolved problem.  In any
result from definite integration with symbolic parameters,
including results computed by hand, it is usually a good idea
to verify that the result is correct for the parameter values
that are of interest.

> Alberto Verga
> irphe - Marseille

Dave Withoff
Wolfram Research


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