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