Bug report ID: 719832 - limit(exp(x*%i)*x, x, inf) should give infinity

Hello Stavros, Hello Eric,

thank you very much for the fast response. 

I see. So, we need an additional relation to detect of what type the
'$ind is. When we are in simplimtimes we have the complete expression by
hand to check for an additional condition the expression must fullfil.

Dieter Kaiser

Am Dienstag, den 06.04.2010, 17:54 -0400 schrieb Stavros Macrakis:
> ind*inf is only infinity if the ind doesn't include 0, as shown by the
> sin(x)*x example, which should be und, not infinity.
> 
> On 2010-04-06, Dieter Kaiser <drdieterkaiser at web.de> wrote:
> > We have the open bug report ID: 719832 - limit(exp(x*%i)*x,x,inf) should
> > give infinity.
> >
> > Maxima knows the following limits:
> >
> > (%i5) limit(exp(%i*x),x,inf);
> > (%o5) ind
> > (%i6) limit(x*ind,x,inf);
> > (%o6) infinity
> >
> > But Maxima does not evaluate the reported example to infinity:
> >
> > (%i7) limit(x*exp(%i*x),x,inf);
> > (%o7) und
> >
> > In the routine simplimtimes we have code to check for an '$ind in a
> > product. For this case Maxima always returns the result '$und. We might
> > add code to check for the case '$ind*'$inf -> '$infinity:
> >
> > ...
> >   ((equal num 1)
> >    (setq sign ($csign prod))
> >    (return (cond ((and flag2
> >                        (member flag '($inf $minf)))
> >                   ;; Check in addition the case ind*inf -> infinity.
> >                   '$infinity)
> >                  (flag2 '$und)
> > ...
> >
> > With this extension we will get the desired result for the example of
> > the bug report:
> >
> > (%i1) limit(exp(%i*x)*x,x,inf);
> > (%o1) infinity
> >
> > We get the same result for the following:
> >
> > (%i2) limit(exp(-%i*x)*x,x,inf);
> > (%o2) infinity
> >
> > And as a consequence:
> >
> > (%i3) limit(sin(x)*x,x,inf);
> > (%o3) infinity
> >
> > (%i4) limit(cos(x)*x,x,inf);
> > (%o4) infinity
> >
> > Two integrals in rtestint.mac will get a problem:
> >
> > ********************** Problem 59 ***************
> > Input:
> > integrate(cos(9*x^(7/3)),x,0,inf)
> >
> >
> > Result:
> > Principal Value
> > 0
> >
> > This differed from the expected result:
> > 3*gamma(3/7)*cos(3*%pi/14)/(7*9^(3/7))
> >
> > ********************** Problem 60 ***************
> > Input:
> > integrate(sin(9*x^(7/3)),x,0,inf)
> >
> >
> > Result:
> > 'integrate(sin(9*x^(7/3)),x,0,inf)
> >
> >
> > The reason is, that Maxima no longer will get the limit '$und but
> > '$infinity in the routine limit-pole for the involved trig expressions,
> > e.g.
> >
> > (%i2) limit(x*cos(9*x^2),x,inf);
> > (%o2) infinity
> >
> > But the routine limit-pole does not handle the case '$infinity and
> > returns the default answer '$yes and not the expected result '$und. This
> > could be corrected in the routine limit-pole by adding the case
> > '$infinity with a result '$no:
> >
> > (defun limit-pole (exp var limit direction)
> >   (let ((ans (cond ((member limit '($minf $inf) :test #'eq)
> >                     (cond ((eq (special-convergent-formp exp limit)
> > '$yes)
> >                            '$no)
> >                           (t (get-limit (m* exp var) var limit
> > direction))))
> >                    (t '$no))))
> >     (cond ((eq ans '$no)   '$no)
> >           ((null ans)   nil)
> >           ((eq ans '$und) '$no)
> > --->      ((eq ans '$infinity) '$no)
> >           ((equal ans 0.)   '$no)
> >           (t '$yes))))
> >
> > I am not sure about the following two points:
> >
> > 1. Is it correct to return the result '$infinity for the limits of
> >    products with the trig functions sin and cos, e.g. for
> >    limit(x*cos(x),x,inf) -> infinity?
> >
> > 2. The routine limit-pole seems not to expect a result '$infinity.
> >    Is it correct in general to return the answer '$no for this case?
> >
> > Dieter Kaiser
> >
> >
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> >