[Maxima-bugs] [ maxima-Bugs-2875784 ] fourier integral incorrect

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Bugs item #2875784, was opened at 2009-10-09 11:38
Message generated for change (Comment added) made by dgildea
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Category: Lisp Core - Integration
Group: None
>Status: Closed
>Resolution: Fixed
Priority: 5
Private: No
Submitted By: Leo Butler (l_butler)
Assigned to: Nobody/Anonymous (nobody)
Summary: fourier integral incorrect

Initial Comment:
The following is incorrect:
(%i2) I : 'integrate(%e^(%i*x)*sin(x)/x,x,minf,inf);
(%o2) 'integrate(%e^(%i*x)*sin(x)/x,x,minf,inf)
(%i3) J : changevar(I,y+x,y,x);
(%o3) 'integrate(%e^-(%i*y)*sin(y)/y,y,minf,inf)
(%i4) ev(I,nouns);
(%o4) %pi
(%i5) ev(J,nouns);
(%o5) 0


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>Comment By: Dan Gildea (dgildea)
Date: 2009-12-13 19:59

Message:
Fixed in defint.lisp rev 1.68.  
(%i15) integrate(%e^(%i*x)*sin(x)/x,x,minf,inf);
Principal Value
(%o15) %pi/2
(%i16) integrate(%e^(-%i*x)*sin(x)/x,x,minf,inf);
Principal Value
(%o16) %pi/2
(%i17) integrate(cos(x)*sin(x)/x,x,minf,inf);
(%o17) %pi/2


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Comment By: Dan Gildea (dgildea)
Date: 2009-12-07 08:48

Message:
The problem seems to be in mtosc, which corresponds to page 77 of Paul
Wang's thesis:
http://publications.csail.mit.edu/lcs/pubs/pdf/MIT-LCS-TR-092.pdf

This routine handles integrals of the form exp(%i*m*x)*r(x) where r(x) is
rational.
The thesis indicates that r(x) should have no real poles.  Most integrands
with real
poles are caught by poles-in-interval before being dispatched to mtosc. 
However,
poles-in-interval does not consider sin(x)/x to have any poles.  I think
mtosc is finding
the wrong answer because 1/x does have a pole at zero.

I tried adding a test for this case to mtosc:
	  ((and (or (setq plf (polfactors n))  t)
+		(eq (real-roots d var) '$no)  ; fail if rational part has poles

Now we get noun forms for the following integrals:
(%i2) integrate(%e^(%i*x)*sin(x)/x,x,minf,inf);
(%o2) 'integrate(%e^(%i*x)*sin(x)/x,x,minf,inf)
(%i3) integrate(%e^(-%i*x)*sin(x)/x,x,minf,inf);
(%o3) 'integrate(%e^-(%i*x)*sin(x)/x,x,minf,inf)

%pi/2 would be better, but at least these aren't wrong.  However, now
we get the wrong answer for:
(%i4) integrate(cos(x)*sin(x)/x,x,minf,inf);
(%o4) 0

Where previously mtosc gave the correct answer, now mtosc gives up,
and intsubs seems to give a wrong answer because it misses
discontinuities
in gamma_incomplete.


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Comment By: Leo Butler (l_butler)
Date: 2009-10-09 12:13

Message:
I should add that there are two problems above: first, I and J should be
equal; second, the correct answer should be
(%i19) integrate(cos(x)*sin(x)/x,x,minf,inf);
(%o19) %pi/2


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