From: SourceForge.net <no...@so...> - 2007-06-22 16:27:36
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Bugs item #1552789, was opened at 2006-09-05 12:08 Message generated for change (Comment added) made by rtoy You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1552789&group_id=4933 Please note that this message will contain a full copy of the comment thread, including the initial issue submission, for this request, not just the latest update. Category: Lisp Core - Integration Group: None >Status: Closed >Resolution: Fixed Priority: 5 Private: No Submitted By: Raymond Toy (rtoy) Assigned to: Nobody/Anonymous (nobody) Summary: integrate(1/(sin(x)^2+1),x,1,1+%pi) takes forever Initial Comment: This is a follow-on to Bug [ 1044318 ] defint(1/(sin(x)^2+1),x,0,3*%pi) wrong. integrate(1/(sin(x)^2+1),x,1,1+%pi) seems to take forever. But if the limits are 0 and %pi, the integral is evaluated instantly with the correct value. ---------------------------------------------------------------------- >Comment By: Raymond Toy (rtoy) Date: 2007-06-22 12:27 Message: Logged In: YES user_id=28849 Originator: YES This integral no longer takes forever. Don't know how it got fixed. The result is wrong, though. See bug 1741705 Closing this since it no longer directly applies. ---------------------------------------------------------------------- Comment By: Raymond Toy (rtoy) Date: 2006-09-06 09:12 Message: Logged In: YES user_id=28849 Looking at the code for ldefint shows why it returns 0. As the docs say, it evaluates the antiderivative and plugs in the limits. I believe this is why defint uses the routine intsubs and same-sheet-subs to handle these issues. Perhaps they're not doing what they're supposed to do, but in this particular case, maxima is stuck computing the antiderivative. I don't understand why. ---------------------------------------------------------------------- Comment By: Barton Willis (willisbl) Date: 2006-09-06 07:38 Message: Logged In: YES user_id=895922 ldefint gives a wrong answer (basically bug 1044318) (%i10) ldefint(1/(sin(x)^2+1),x,1,1+%pi); (%o10) 0 integrate gives an antiderivative that isn't continuous at odd integer multiples of %pi / 2 (%i11) integrate(1/(sin(x)^2+1),x); (%o11) atan((2*tan(x))/sqrt(2))/sqrt(2) The expression atan(2*tan(x)/sqrt(2))/sqrt(2)+sqrt(2)*%pi*floor(x/%pi-1/2)/2 might be a valid antiderivative on all of R provided the first term is assumed left continuous at each odd integer multiple of %pi/2, I think. Barton ---------------------------------------------------------------------- You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1552789&group_id=4933 |