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From: SourceForge.net <noreply@so...>  20070622 16:27:36

Bugs item #1552789, was opened at 20060905 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 followon 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: 20070622 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: 20060906 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 samesheetsubs 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: 20060906 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/%pi1/2)/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 
From: SourceForge.net <noreply@so...>  20070622 15:59:28

Bugs item #1741705, was opened at 20070622 11:59 Message generated for change (Tracker Item Submitted) made by Item Submitter You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1741705&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: Open Resolution: None Priority: 5 Private: No Submitted By: Raymond Toy (rtoy) Assigned to: Nobody/Anonymous (nobody) Summary: integrate(1/(sin(x)^2+1),x,0,8) wrong Initial Comment: integrate(1/(sin(x)^2+1),x,0,8) returns sqrt(2)/2*atan(sqrt(2)*tan(8)) + %pi/sqrt(2) This is not right. But integrate(1/(sin(x)^2+1),x,0,5*%pi/2) returns 5*sqrt(2)*%pi/4, which is probably correct according to quad_qags. This latter integral works because intsc1 notices that the interval length is a rational multiple of %pi and breaks up the integral. However, for the former integral, intsc1 gives up because the interval length is not a multiple of %pi. Since we now have a floor function that works well, we should try to extend intsc1 to accept all numeric limits. This issue affects all integrals of trig functions that are handled by intsc1. See also the related bug 1552789.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1741705&group_id=4933 