Are these both the right answer for the integral?

It?s a feature, it's a feature!! (not a cookbook fortunately).  Thanks for 
your detailed analysis.

Rich




-----Original Message----- 
From: reyssat
Sent: Thursday, January 06, 2011 2:43 AM
To: Richard Hennessy
Cc: Maxima List
Subject: Re: Are these both the right answer for the integral?

Le 06/01/2011 05:41, Richard Hennessy a ?crit :
> You are right too.  I think I tried to plot realpart() of the answer which 
> is not the right way to check for correctness in all cases.  Plotting 
> realpart(%) gives the wrong sign for x < 0.
There is no right or wrong sign when the function x --> (x^2-1)^(3/2) is
involved. Both are correct (this is the multivaluedness of fractional
power functions). Each time you try to force a sign, the function will
trap you somewhere else. For instance, you may force each fractional
power (with even denominator) to be positive, but then
(-2)^(2/6) is positive and (-2)^(1/3) is negative ! Even if this is not
the most convincing example, the fact that the function is defined for
complex numbers (and continuous there, even analytic) except a finite
number of points, forces to accept ALL signs (we say all determinations)
of the function, even on R. You may choose some signs in some occasions,
but the choices CANNOT be always consistent, in particular if you look
globally (i.e. on the whole real line).

Eric
>
> Rich
>
>
>
> -----Original Message----- From: Mark H Weaver
> Sent: Wednesday, January 05, 2011 8:08 PM
> To: Richard Hennessy
> Cc: reyssat ; Maxima List
> Subject: Re: Are these both the right answer for the integral?
>
> "Richard Hennessy" <rich.hennessy at verizon.net> writes:
>> On second thought this is not okay.  You cannot add signum(x) *%i,
>> because that is not a constant.
>
> As reyssat pointed out, since the integrand is undefined at x=-1, x=0,
> and x=1, the integral is defined only up to an additive constant on
> _each_ of the intervals (-inf,-1) (-1,0) (0,1) and (1,inf).  Each of
> those intervals can have its own constant, and they needn't be the same.
>
> In other words, instead of the traditional integration constant C, in
> this case you may add f(x) to the integral, where f is defined as
> follows (for any constants C1,C2,C3,C4):
>
> f(x) := if x<-1 then C1 elseif x<0 then C2 elseif x<1 then C3 else C4;
>
> signum(x)*%i is equivalent to f(x) where C1=C2=-%i and C3=C4=%i, except
> at 0 where the integral is undefined anyway.
>
>      Mark
>