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From: SourceForge.net <noreply@so...>  20090112 16:36:25

Bugs item #2501765, was opened at 20090112 11:36 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=2501765&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((14*x^232)/(x^4+3*x^2+1)^2,x,0,inf); Initial Comment: integrate((14*x^232)/(x^4+3*x^2+1)^2,x,0,inf); produces a division by zero error. Maxima ought to be able to do this integral. And maxima can do the indefinite integral, producing a rational expression and another integral. Maxima can compute that definite integral.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2501765&group_id=4933 
From: SourceForge.net <noreply@so...>  20090112 17:02:09

Bugs item #2501765, was opened at 20090112 11:36 Message generated for change (Comment added) made by rtoy You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2501765&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((14*x^232)/(x^4+3*x^2+1)^2,x,0,inf); Initial Comment: integrate((14*x^232)/(x^4+3*x^2+1)^2,x,0,inf); produces a division by zero error. Maxima ought to be able to do this integral. And maxima can do the indefinite integral, producing a rational expression and another integral. Maxima can compute that definite integral.  >Comment By: Raymond Toy (rtoy) Date: 20090112 12:01 Message: Maxima is using a keyhole contour and is computing the residues of the integrand. I think Maxima doesn't notice that the poles are multiple poles, and just uses the simple method of computing the residue by differentiating the denominator. This produces a division by zero.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2501765&group_id=4933 
From: SourceForge.net <noreply@so...>  20090114 13:46:58

Bugs item #2501765, was opened at 20090112 11:36 Message generated for change (Comment added) made by rtoy You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2501765&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((14*x^232)/(x^4+3*x^2+1)^2,x,0,inf); Initial Comment: integrate((14*x^232)/(x^4+3*x^2+1)^2,x,0,inf); produces a division by zero error. Maxima ought to be able to do this integral. And maxima can do the indefinite integral, producing a rational expression and another integral. Maxima can compute that definite integral.  >Comment By: Raymond Toy (rtoy) Date: 20090114 08:46 Message: I think this fails because maxima computes the multiplicities of the roots of the denominator incorrectly. The roots are double roots, but maxima thinks they're single and hence gets a division by zero error when trying to compute the residue by differentiating the denominator once instead of twice. See also https://sourceforge.net/tracker2/?func=detail&aid=2505241&group_id=4933&atid=104933  Comment By: Raymond Toy (rtoy) Date: 20090112 12:01 Message: Maxima is using a keyhole contour and is computing the residues of the integrand. I think Maxima doesn't notice that the poles are multiple poles, and just uses the simple method of computing the residue by differentiating the denominator. This produces a division by zero.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2501765&group_id=4933 
From: SourceForge.net <noreply@so...>  20090115 18:40:36

Bugs item #2501765, was opened at 20090112 11:36 Message generated for change (Settings changed) made by dgildea You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2501765&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((14*x^232)/(x^4+3*x^2+1)^2,x,0,inf); Initial Comment: integrate((14*x^232)/(x^4+3*x^2+1)^2,x,0,inf); produces a division by zero error. Maxima ought to be able to do this integral. And maxima can do the indefinite integral, producing a rational expression and another integral. Maxima can compute that definite integral.  >Comment By: Dan Gildea (dgildea) Date: 20090115 13:40 Message: Fix in solve.lisp rev 1.22 gives correct multiplicity of roots. (%i6) integrate((14*x^232)/(x^4+3*x^2+1)^2,x,0,inf); (%o6) (sqrt(3sqrt(5))*(139*sqrt(2)*sqrt(5)+345*sqrt(2))*%pi +sqrt(sqrt(5)+3)*(345*sqrt(2)139*sqrt(2)*sqrt(5))*%pi) /200 (%i7) float(%); (%o7) 14.47111834594389 (%i8) quad_qags((14*x^232)/(x^4+3*x^2+1)^2,x,0,1000); (%o8) [14.47111834594389,1.179685737489548e9,441,0]  Comment By: Raymond Toy (rtoy) Date: 20090114 08:46 Message: I think this fails because maxima computes the multiplicities of the roots of the denominator incorrectly. The roots are double roots, but maxima thinks they're single and hence gets a division by zero error when trying to compute the residue by differentiating the denominator once instead of twice. See also https://sourceforge.net/tracker2/?func=detail&aid=2505241&group_id=4933&atid=104933  Comment By: Raymond Toy (rtoy) Date: 20090112 12:01 Message: Maxima is using a keyhole contour and is computing the residues of the integrand. I think Maxima doesn't notice that the poles are multiple poles, and just uses the simple method of computing the residue by differentiating the denominator. This produces a division by zero.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2501765&group_id=4933 