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maxima-bugs

 [Maxima-bugs] [ maxima-Bugs-3585415 ] integrate loops forever with simple expression From: SourceForge.net - 2012-11-08 15:45:51 ```Bugs item #3585415, was opened at 2012-11-08 07:45 Message generated for change (Tracker Item Submitted) made by tomasriker You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=3585415&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: David Scherfgen (tomasriker) Assigned to: Nobody/Anonymous (nobody) Summary: integrate loops forever with simple expression Initial Comment: if you calculate this: integrate(x^(1/3)/(x^(2/3)+1), x, 0, 8); Maxima will compute forever and never return. However, the indefinite integral works: integrate(x^(1/3)/(x^(2/3)+1), x); gives you the antiderivative: F(x) := (3*(x^(2/3)+1))/2-(3*log(x^(2/3)+1))/2; Now the original definite integral can be computed as F(8) - F(0). This is OK, because F(x) doesn't change sign or shows any other notable behavior between 0 and 8. The question is: why doesn't Maxima do this? Instead it loops forever ... ---------------------------------------------------------------------- You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=3585415&group_id=4933 ```
 [Maxima-bugs] [ maxima-Bugs-3585415 ] integrate loops forever with simple expression From: SourceForge.net - 2012-11-08 15:50:36 ```Bugs item #3585415, was opened at 2012-11-08 07:45 Message generated for change (Settings changed) made by tomasriker You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=3585415&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: 7 Private: No Submitted By: David Scherfgen (tomasriker) Assigned to: Nobody/Anonymous (nobody) Summary: integrate loops forever with simple expression Initial Comment: if you calculate this: integrate(x^(1/3)/(x^(2/3)+1), x, 0, 8); Maxima will compute forever and never return. However, the indefinite integral works: integrate(x^(1/3)/(x^(2/3)+1), x); gives you the antiderivative: F(x) := (3*(x^(2/3)+1))/2-(3*log(x^(2/3)+1))/2; Now the original definite integral can be computed as F(8) - F(0). This is OK, because F(x) doesn't change sign or shows any other notable behavior between 0 and 8. The question is: why doesn't Maxima do this? Instead it loops forever ... ---------------------------------------------------------------------- You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=3585415&group_id=4933 ```
 [Maxima-bugs] [ maxima-Bugs-3585415 ] integrate loops forever with simple expression From: SourceForge.net - 2012-11-08 16:04:25 ```Bugs item #3585415, was opened at 2012-11-08 07:45 Message generated for change (Comment added) made by rtoy You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=3585415&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: 7 Private: No Submitted By: David Scherfgen (tomasriker) Assigned to: Nobody/Anonymous (nobody) Summary: integrate loops forever with simple expression Initial Comment: if you calculate this: integrate(x^(1/3)/(x^(2/3)+1), x, 0, 8); Maxima will compute forever and never return. However, the indefinite integral works: integrate(x^(1/3)/(x^(2/3)+1), x); gives you the antiderivative: F(x) := (3*(x^(2/3)+1))/2-(3*log(x^(2/3)+1))/2; Now the original definite integral can be computed as F(8) - F(0). This is OK, because F(x) doesn't change sign or shows any other notable behavior between 0 and 8. The question is: why doesn't Maxima do this? Instead it loops forever ... ---------------------------------------------------------------------- >Comment By: Raymond Toy (rtoy) Date: 2012-11-08 08:04 Message: Definite integration in maxima often does not compute the indefinite integral first, but in this case it does compute the indefinite integral. Maxima appears to be stuck computing the limits. Don't know why that should be. ---------------------------------------------------------------------- You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=3585415&group_id=4933 ```