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.
>

```

• Prev by Date: Re: Q: open palletes in a specific on screen position.
• Next by Date: Re: how to rank a list of elements?
• Previous by thread: Re: Re: Demonstrate that 1==-1
• Next by thread: Re: Re: Demonstrate that 1==-1