Seems like a bug in integrate



Hi,

You mentioned that "the integrand is not Lebesgue integrable on R" which I would call just not in L^2(R).  Yes I noticed that but I 
didn't really expect an error message like the one I received.  It should have said that the integral does not converge or something 
like that.  Also you said that "%o2 shows that one needs to be careful in using the output of maxima's computations--maxima told you 
that its answer is valid on t>0 and made no promises about t<=0.".  I would like to say about this that the answer is not clear 
about the domain of t because sometimes when Maxima asks a question about the sign of a variable the answer is the same no matter 
what you say and so it is true for all t, but I would prefer an answer that is valid for all t even if you have to use piecewise 
functions in the answer.  The fact that piecewise solutions are (or seem to be) banned from Maxima is a strange weakness in Maxima 
(and I suspect other CAS also).  abs(-t) = pw([minf,t,0,-t,inf],t) so even abs(t) seems to be or is treated as if it is not a good 
answer.

Rich


On Sun, 24 May 2009, Dieter Kaiser wrote:

< Am Samstag, den 23.05.2009, 23:21 -0600 schrieb Robert Dodier:
< > On 5/22/09, Richard Hennessy <rich.hennessy at verizon.net> wrote:
< >
< > > (%i1) integrate(1/(x^2+1)*exp(-%i*%pi*x*t),x,minf,inf);
< > > Is  t  positive, negative, or zero?
< > > p;
< > > (%o1) %pi*%e^(-%pi*t)
< > > (%i2) integrate(1/(x^2+1)*exp(-2*%i*%pi*x*t),x,minf,inf);
< > > Is  t  positive, negative, or zero?
< > > p;
< > > (%o2) %pi*%e^(- 2*%pi*t)

A mathematical comment for RH:

Note that %o2 is correct only for t>=0. The correct answer for all t
requires abs(t) in place of t. Indeed, the modulus of %o2 should be bounded by 0 and
%pi for all t and be in L^2(R). %o2 shows that one needs to be careful
in using the output of maxima's computations--maxima told you that its
answer is valid on t>0 and made no promises about t<=0.

Also, %i3 will produce rubbish (the integrand is not Lebesgue
integrable on R).

< > > (%i3) integrate(%*exp(2*%i*%pi*x*t),t,minf,inf);

I am sure that you know this, but since you did not mention it in your
post, I thought I should point it out.

Leo

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