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From: SourceForge.net <noreply@so...>  20100501 07:37:42

Bugs item #2995089, was opened at 20100501 03:37 Message generated for change (Tracker Item Submitted) made by neanderslob You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2995089&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: Sam Hagen (neanderslob) Assigned to: Nobody/Anonymous (nobody) Summary: arbitrary constant exponent seems to confound integration Initial Comment: It can best be explained by a look at the HTML file attached, I have comments in there as well to point out what to look at. It appears that the power of an arbitrary constant will affect the outcome of an integration.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2995089&group_id=4933 
From: SourceForge.net <noreply@so...>  20100501 12:33:49

Bugs item #2995089, was opened at 20100501 09:37 Message generated for change (Comment added) made by crategus You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2995089&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: Sam Hagen (neanderslob) Assigned to: Nobody/Anonymous (nobody) Summary: arbitrary constant exponent seems to confound integration Initial Comment: It can best be explained by a look at the HTML file attached, I have comments in there as well to point out what to look at. It appears that the power of an arbitrary constant will affect the outcome of an integration.  >Comment By: Dieter Kaiser (crategus) Date: 20100501 14:33 Message: Thank you very much for the report. The underlying problem of this bug report is the following definite integral. We have a constant a in the exponent of the exp function: (%i1) integrate(cos(r)*exp(r^2/a),r,0,inf); Is a positive or negative? p; (%o1) sqrt(%pi)*sqrt(a)*%e^(a/4)/2 When we take the square of the constant, Maxima no longer gets a result: (%i2) integrate(cos(r)*exp(r^2/a^2),r,0,inf); (%o2) und It works again if we assume a to be positive. This is a bit strange, because a^2 is known to be positive too: (%i3) assume(a>0)$ (%i4) integrate(cos(r)*exp(r^2/a^2),r,0,inf); (%o4) sqrt(%pi)*a*%e^(a^2/4)/2 There is a second problem. The derivative of 'und gives 0. 'und is the intermediate result of the integration of the example of this bug report: (%i2) diff(und,q); (%o2) 0 Dieter Kaiser  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2995089&group_id=4933 
From: SourceForge.net <noreply@so...>  20100709 21:32:58

Bugs item #2995089, was opened at 20100501 09:37 Message generated for change (Comment added) made by crategus You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2995089&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: Sam Hagen (neanderslob) Assigned to: Nobody/Anonymous (nobody) Summary: arbitrary constant exponent seems to confound integration Initial Comment: It can best be explained by a look at the HTML file attached, I have comments in there as well to point out what to look at. It appears that the power of an arbitrary constant will affect the outcome of an integration.  >Comment By: Dieter Kaiser (crategus) Date: 20100709 23:32 Message: The problem of this bug report is caused by the behavior of Maxima to return INFINITY for the following limit: limit(r/a, r, inf) > infinity If the sign of the parameter a is known to be positive Maxima gives limit(r/a, r, inf) > inf This is related to the problem of this bug report the following way. The indefinite integral is (%i1) integrate(cos(r)*exp(r^2/a^2),r),factor; (%o1) sqrt(%pi)*a*%e^(a^2/4) *(erf((2*r+%i*a^2)/(2*a))+erf((2*r%i*a^2)/(2*a))) /4 This answer is correct. It contains two erf terms. The problem is the limit of the erf function for r > inf. We take one of the terms: (%i2) limit(erf((2*r+%i*a^2)/(2*a)), r, inf); (%o2) und The realpart of the argument is r/a. Maxima returns INFINITY for the limit of the argument because the sign of r/a is not known. This gives UND as the result for the erf function and as a consequence UND as the result for the definite integral. So the problem of this bug report is, that Maxima returns immediately INFINITY for the limit of r/a and does not ask for the sign of the parameter a. Of course, Maxima gets the correct answer, if we assume the parameter a to be positive, but the user might not recognize that this information is necessary, because a limit has to be evaluated for an expression r/a. (%i6) assume(a>0)$ (%i7) integrate(cos(r)*exp(r^2/a^2),r,0,inf),factor; (%o7) sqrt(%pi)*a*%e^(a^2/4)/2 Dieter Kaiser  Comment By: Dieter Kaiser (crategus) Date: 20100501 14:33 Message: Thank you very much for the report. The underlying problem of this bug report is the following definite integral. We have a constant a in the exponent of the exp function: (%i1) integrate(cos(r)*exp(r^2/a),r,0,inf); Is a positive or negative? p; (%o1) sqrt(%pi)*sqrt(a)*%e^(a/4)/2 When we take the square of the constant, Maxima no longer gets a result: (%i2) integrate(cos(r)*exp(r^2/a^2),r,0,inf); (%o2) und It works again if we assume a to be positive. This is a bit strange, because a^2 is known to be positive too: (%i3) assume(a>0)$ (%i4) integrate(cos(r)*exp(r^2/a^2),r,0,inf); (%o4) sqrt(%pi)*a*%e^(a^2/4)/2 There is a second problem. The derivative of 'und gives 0. 'und is the intermediate result of the integration of the example of this bug report: (%i2) diff(und,q); (%o2) 0 Dieter Kaiser  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2995089&group_id=4933 
From: SourceForge.net <noreply@so...>  20100821 12:14:14

Bugs item #2995089, was opened at 20100501 09:37 Message generated for change (Comment added) made by crategus You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2995089&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: Sam Hagen (neanderslob) Assigned to: Nobody/Anonymous (nobody) Summary: arbitrary constant exponent seems to confound integration Initial Comment: It can best be explained by a look at the HTML file attached, I have comments in there as well to point out what to look at. It appears that the power of an arbitrary constant will affect the outcome of an integration.  >Comment By: Dieter Kaiser (crategus) Date: 20100821 14:14 Message: We had a bug fix for the bug ID: 3034140 "incorrect integration of %e^((x^2*a^2))*cos(b*x)". We get the example of this bug report when we set b=1. At first the correct results. We start with the assumption a>0: (%i1) assume(a>0)$ (%i2) integrate(cos(b*x)*exp(x^2/a^2),x,minf,inf); (%o2) sqrt(%pi)*a*%e^(a^2*b^2/4) (%i3) integrate(cos(b*x)*exp(x^2/a^2),x,0,inf); (%o3) sqrt(%pi)*a*%e^(a^2*b^2/4)/2 Both integrals are correct. There is no problem. Now again with no assumption. The first integral again is correct: (%i4) forget(a>0)$ (%i5) integrate(cos(b*x)*exp(x^2/a^2),x,minf,inf); Is a positive or negative? p; (%o5) sqrt(%pi)*a*%e^(a^2*b^2/4) But the following integral gives an extra factor erf(%i*a*b/2). This is wrong: (%i6) integrate(cos(b*x)*exp(x^2/a^2),x,0,inf); Is a positive or negative? p; Is b positive or negative? p; (%o6) sqrt(%pi)*a*%e^(a^2*b^2/4)*erf(%i*a*b/2)/2 By the way: The behavior of Maxima for this integral differs for Maxima 5.21, 5.22 and current CVS. In all versions of Maxima the indefinite integral is the same: (%i7) integrate(cos(b*x)*exp(x^2/a^2),x); (%o7) sqrt(%pi)*%e^(a^2*b^2/4) *(a*erf((2*x+%i*a^2*b)/(2*a))+a*erf((2*x%i*a^2*b)/(2*a))) /4 Dieter Kaiser  Comment By: Dieter Kaiser (crategus) Date: 20100709 23:32 Message: The problem of this bug report is caused by the behavior of Maxima to return INFINITY for the following limit: limit(r/a, r, inf) > infinity If the sign of the parameter a is known to be positive Maxima gives limit(r/a, r, inf) > inf This is related to the problem of this bug report the following way. The indefinite integral is (%i1) integrate(cos(r)*exp(r^2/a^2),r),factor; (%o1) sqrt(%pi)*a*%e^(a^2/4) *(erf((2*r+%i*a^2)/(2*a))+erf((2*r%i*a^2)/(2*a))) /4 This answer is correct. It contains two erf terms. The problem is the limit of the erf function for r > inf. We take one of the terms: (%i2) limit(erf((2*r+%i*a^2)/(2*a)), r, inf); (%o2) und The realpart of the argument is r/a. Maxima returns INFINITY for the limit of the argument because the sign of r/a is not known. This gives UND as the result for the erf function and as a consequence UND as the result for the definite integral. So the problem of this bug report is, that Maxima returns immediately INFINITY for the limit of r/a and does not ask for the sign of the parameter a. Of course, Maxima gets the correct answer, if we assume the parameter a to be positive, but the user might not recognize that this information is necessary, because a limit has to be evaluated for an expression r/a. (%i6) assume(a>0)$ (%i7) integrate(cos(r)*exp(r^2/a^2),r,0,inf),factor; (%o7) sqrt(%pi)*a*%e^(a^2/4)/2 Dieter Kaiser  Comment By: Dieter Kaiser (crategus) Date: 20100501 14:33 Message: Thank you very much for the report. The underlying problem of this bug report is the following definite integral. We have a constant a in the exponent of the exp function: (%i1) integrate(cos(r)*exp(r^2/a),r,0,inf); Is a positive or negative? p; (%o1) sqrt(%pi)*sqrt(a)*%e^(a/4)/2 When we take the square of the constant, Maxima no longer gets a result: (%i2) integrate(cos(r)*exp(r^2/a^2),r,0,inf); (%o2) und It works again if we assume a to be positive. This is a bit strange, because a^2 is known to be positive too: (%i3) assume(a>0)$ (%i4) integrate(cos(r)*exp(r^2/a^2),r,0,inf); (%o4) sqrt(%pi)*a*%e^(a^2/4)/2 There is a second problem. The derivative of 'und gives 0. 'und is the intermediate result of the integration of the example of this bug report: (%i2) diff(und,q); (%o2) 0 Dieter Kaiser  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2995089&group_id=4933 
From: SourceForge.net <noreply@so...>  20110520 21:06:34

Bugs item #2995089, was opened at 20100501 03:37 Message generated for change (Comment added) made by dgildea You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2995089&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: Pending >Resolution: Works For Me Priority: 5 Private: No Submitted By: Sam Hagen (neanderslob) Assigned to: Nobody/Anonymous (nobody) Summary: arbitrary constant exponent seems to confound integration Initial Comment: It can best be explained by a look at the HTML file attached, I have comments in there as well to point out what to look at. It appears that the power of an arbitrary constant will affect the outcome of an integration.  >Comment By: Dan Gildea (dgildea) Date: 20110520 17:06 Message: Seems OK in current git: (%i2) integrate(cos(b*x)*exp(x^2/a^2),x,0,inf); Is a positive or negative? p; Is b positive or negative? p; (%o2) sqrt(%pi)*a*%e^(a^2*b^2/4)/2  Comment By: Dieter Kaiser (crategus) Date: 20100821 08:14 Message: We had a bug fix for the bug ID: 3034140 "incorrect integration of %e^((x^2*a^2))*cos(b*x)". We get the example of this bug report when we set b=1. At first the correct results. We start with the assumption a>0: (%i1) assume(a>0)$ (%i2) integrate(cos(b*x)*exp(x^2/a^2),x,minf,inf); (%o2) sqrt(%pi)*a*%e^(a^2*b^2/4) (%i3) integrate(cos(b*x)*exp(x^2/a^2),x,0,inf); (%o3) sqrt(%pi)*a*%e^(a^2*b^2/4)/2 Both integrals are correct. There is no problem. Now again with no assumption. The first integral again is correct: (%i4) forget(a>0)$ (%i5) integrate(cos(b*x)*exp(x^2/a^2),x,minf,inf); Is a positive or negative? p; (%o5) sqrt(%pi)*a*%e^(a^2*b^2/4) But the following integral gives an extra factor erf(%i*a*b/2). This is wrong: (%i6) integrate(cos(b*x)*exp(x^2/a^2),x,0,inf); Is a positive or negative? p; Is b positive or negative? p; (%o6) sqrt(%pi)*a*%e^(a^2*b^2/4)*erf(%i*a*b/2)/2 By the way: The behavior of Maxima for this integral differs for Maxima 5.21, 5.22 and current CVS. In all versions of Maxima the indefinite integral is the same: (%i7) integrate(cos(b*x)*exp(x^2/a^2),x); (%o7) sqrt(%pi)*%e^(a^2*b^2/4) *(a*erf((2*x+%i*a^2*b)/(2*a))+a*erf((2*x%i*a^2*b)/(2*a))) /4 Dieter Kaiser  Comment By: Dieter Kaiser (crategus) Date: 20100709 17:32 Message: The problem of this bug report is caused by the behavior of Maxima to return INFINITY for the following limit: limit(r/a, r, inf) > infinity If the sign of the parameter a is known to be positive Maxima gives limit(r/a, r, inf) > inf This is related to the problem of this bug report the following way. The indefinite integral is (%i1) integrate(cos(r)*exp(r^2/a^2),r),factor; (%o1) sqrt(%pi)*a*%e^(a^2/4) *(erf((2*r+%i*a^2)/(2*a))+erf((2*r%i*a^2)/(2*a))) /4 This answer is correct. It contains two erf terms. The problem is the limit of the erf function for r > inf. We take one of the terms: (%i2) limit(erf((2*r+%i*a^2)/(2*a)), r, inf); (%o2) und The realpart of the argument is r/a. Maxima returns INFINITY for the limit of the argument because the sign of r/a is not known. This gives UND as the result for the erf function and as a consequence UND as the result for the definite integral. So the problem of this bug report is, that Maxima returns immediately INFINITY for the limit of r/a and does not ask for the sign of the parameter a. Of course, Maxima gets the correct answer, if we assume the parameter a to be positive, but the user might not recognize that this information is necessary, because a limit has to be evaluated for an expression r/a. (%i6) assume(a>0)$ (%i7) integrate(cos(r)*exp(r^2/a^2),r,0,inf),factor; (%o7) sqrt(%pi)*a*%e^(a^2/4)/2 Dieter Kaiser  Comment By: Dieter Kaiser (crategus) Date: 20100501 08:33 Message: Thank you very much for the report. The underlying problem of this bug report is the following definite integral. We have a constant a in the exponent of the exp function: (%i1) integrate(cos(r)*exp(r^2/a),r,0,inf); Is a positive or negative? p; (%o1) sqrt(%pi)*sqrt(a)*%e^(a/4)/2 When we take the square of the constant, Maxima no longer gets a result: (%i2) integrate(cos(r)*exp(r^2/a^2),r,0,inf); (%o2) und It works again if we assume a to be positive. This is a bit strange, because a^2 is known to be positive too: (%i3) assume(a>0)$ (%i4) integrate(cos(r)*exp(r^2/a^2),r,0,inf); (%o4) sqrt(%pi)*a*%e^(a^2/4)/2 There is a second problem. The derivative of 'und gives 0. 'und is the intermediate result of the integration of the example of this bug report: (%i2) diff(und,q); (%o2) 0 Dieter Kaiser  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2995089&group_id=4933 
From: SourceForge.net <noreply@so...>  20110603 21:20:06

Bugs item #2995089, was opened at 20100501 07:37 Message generated for change (Settings changed) made by sfrobot You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2995089&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: Works For Me Priority: 5 Private: No Submitted By: Sam Hagen (neanderslob) Assigned to: Nobody/Anonymous (nobody) Summary: arbitrary constant exponent seems to confound integration Initial Comment: It can best be explained by a look at the HTML file attached, I have comments in there as well to point out what to look at. It appears that the power of an arbitrary constant will affect the outcome of an integration.  >Comment By: SourceForge Robot (sfrobot) Date: 20110603 21:20 Message: This Tracker item was closed automatically by the system. It was previously set to a Pending status, and the original submitter did not respond within 14 days (the time period specified by the administrator of this Tracker).  Comment By: Dan Gildea (dgildea) Date: 20110520 21:06 Message: Seems OK in current git: (%i2) integrate(cos(b*x)*exp(x^2/a^2),x,0,inf); Is a positive or negative? p; Is b positive or negative? p; (%o2) sqrt(%pi)*a*%e^(a^2*b^2/4)/2  Comment By: Dieter Kaiser (crategus) Date: 20100821 12:14 Message: We had a bug fix for the bug ID: 3034140 "incorrect integration of %e^((x^2*a^2))*cos(b*x)". We get the example of this bug report when we set b=1. At first the correct results. We start with the assumption a>0: (%i1) assume(a>0)$ (%i2) integrate(cos(b*x)*exp(x^2/a^2),x,minf,inf); (%o2) sqrt(%pi)*a*%e^(a^2*b^2/4) (%i3) integrate(cos(b*x)*exp(x^2/a^2),x,0,inf); (%o3) sqrt(%pi)*a*%e^(a^2*b^2/4)/2 Both integrals are correct. There is no problem. Now again with no assumption. The first integral again is correct: (%i4) forget(a>0)$ (%i5) integrate(cos(b*x)*exp(x^2/a^2),x,minf,inf); Is a positive or negative? p; (%o5) sqrt(%pi)*a*%e^(a^2*b^2/4) But the following integral gives an extra factor erf(%i*a*b/2). This is wrong: (%i6) integrate(cos(b*x)*exp(x^2/a^2),x,0,inf); Is a positive or negative? p; Is b positive or negative? p; (%o6) sqrt(%pi)*a*%e^(a^2*b^2/4)*erf(%i*a*b/2)/2 By the way: The behavior of Maxima for this integral differs for Maxima 5.21, 5.22 and current CVS. In all versions of Maxima the indefinite integral is the same: (%i7) integrate(cos(b*x)*exp(x^2/a^2),x); (%o7) sqrt(%pi)*%e^(a^2*b^2/4) *(a*erf((2*x+%i*a^2*b)/(2*a))+a*erf((2*x%i*a^2*b)/(2*a))) /4 Dieter Kaiser  Comment By: Dieter Kaiser (crategus) Date: 20100709 21:32 Message: The problem of this bug report is caused by the behavior of Maxima to return INFINITY for the following limit: limit(r/a, r, inf) > infinity If the sign of the parameter a is known to be positive Maxima gives limit(r/a, r, inf) > inf This is related to the problem of this bug report the following way. The indefinite integral is (%i1) integrate(cos(r)*exp(r^2/a^2),r),factor; (%o1) sqrt(%pi)*a*%e^(a^2/4) *(erf((2*r+%i*a^2)/(2*a))+erf((2*r%i*a^2)/(2*a))) /4 This answer is correct. It contains two erf terms. The problem is the limit of the erf function for r > inf. We take one of the terms: (%i2) limit(erf((2*r+%i*a^2)/(2*a)), r, inf); (%o2) und The realpart of the argument is r/a. Maxima returns INFINITY for the limit of the argument because the sign of r/a is not known. This gives UND as the result for the erf function and as a consequence UND as the result for the definite integral. So the problem of this bug report is, that Maxima returns immediately INFINITY for the limit of r/a and does not ask for the sign of the parameter a. Of course, Maxima gets the correct answer, if we assume the parameter a to be positive, but the user might not recognize that this information is necessary, because a limit has to be evaluated for an expression r/a. (%i6) assume(a>0)$ (%i7) integrate(cos(r)*exp(r^2/a^2),r,0,inf),factor; (%o7) sqrt(%pi)*a*%e^(a^2/4)/2 Dieter Kaiser  Comment By: Dieter Kaiser (crategus) Date: 20100501 12:33 Message: Thank you very much for the report. The underlying problem of this bug report is the following definite integral. We have a constant a in the exponent of the exp function: (%i1) integrate(cos(r)*exp(r^2/a),r,0,inf); Is a positive or negative? p; (%o1) sqrt(%pi)*sqrt(a)*%e^(a/4)/2 When we take the square of the constant, Maxima no longer gets a result: (%i2) integrate(cos(r)*exp(r^2/a^2),r,0,inf); (%o2) und It works again if we assume a to be positive. This is a bit strange, because a^2 is known to be positive too: (%i3) assume(a>0)$ (%i4) integrate(cos(r)*exp(r^2/a^2),r,0,inf); (%o4) sqrt(%pi)*a*%e^(a^2/4)/2 There is a second problem. The derivative of 'und gives 0. 'und is the intermediate result of the integration of the example of this bug report: (%i2) diff(und,q); (%o2) 0 Dieter Kaiser  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2995089&group_id=4933 
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