Re: Demonstrate that 1==-1

*To*: mathgroup at smc.vnet.net*Subject*: [mg23220] Re: [mg23171] Demonstrate that 1==-1*From*: "Kevin J. McCann" <kevinmccann at home.com>*Date*: Tue, 25 Apr 2000 01:40:26 -0400 (EDT)*References*: <8e0mhl$f25@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

f1[a_] := Integrate[ (-1 - a/E^(I*u))/(1 - a/E^(I*u)), {u, 0, 2*Pi}] f2[a_] := Integrate[(1 + a/E^(I*u))/(-1 + a/E^(I*u)), {u, 0, 2*Pi}] {f1[a],f2[a]} = {2*Pi, -2*Pi} {f1[2],f2[2]} = {2*Pi, 2*Pi} {f1[-2],f2[-2]} = {2*Pi, 2*Pi} {f1[1/2],f2[1/2]} = {-2*Pi, -2*Pi} Seems that it works fine with a definite choice for a, but not when a is a variable. Not sure what Mathematica is assuming for a, or why. Ordinarily Mathematica puts out a conditional answer where appropriate. Not so here for a > 1 or a < 1. Normal behavior is that Mathematica puts out a conditional, not so here. This is a BUG. Cheers, Kevin "David Withoff" <withoff at wolfram.com> wrote in message news:8e0mhl$f25 at smc.vnet.net... > > 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. >