Timeline for Integral of the integral using NIntegrate
Current License: CC BY-SA 4.0
21 events
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| Nov 6, 2020 at 2:27 | history | edited | bbgodfrey |
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| Aug 1, 2019 at 11:29 | answer | added | Anton Antonov | timeline score: 3 | |
| Jul 27, 2019 at 20:13 | comment | added | Michael E2 |
Possible duplicate: mathematica.stackexchange.com/questions/18393/…, esp. ref. 6) under NIntegrate.
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| Jul 27, 2019 at 19:43 | answer | added | Carl Woll | timeline score: 5 | |
| Jul 27, 2019 at 18:32 | comment | added | bbgodfrey |
Using Method -> {Automatic, "SymbolicProcessing" -> False} indeed does not eliminate the error messages, but does reduce the number of them to two when applied to the outer integration. It seems to have no effect on the inner integration.
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| Jul 27, 2019 at 18:25 | answer | added | bbgodfrey | timeline score: 4 | |
| Jul 27, 2019 at 15:00 | history | tweeted | twitter.com/StackMma/status/1155130854609489920 | ||
| Jul 27, 2019 at 13:57 | comment | added | Henrik Schumacher |
@Turgo Ah, you mean Method -> {Automatic, "SymbolicProcessing" -> False}. Yes. Unfortunately, that does not remove the error messages.
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| Jul 27, 2019 at 13:41 | comment | added | Turgon |
@HenrikSchumacher actually it is an option of NIntegrate. Check the help page, it's under Options->Method->SymbolicProcessing.
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| Jul 27, 2019 at 12:59 | comment | added | Henrik Schumacher |
@Turgon How would you do that? SymbolicProcessing is not an option of NIntegrate.
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| Jul 27, 2019 at 8:34 | comment | added | Turgon |
@HenrikSchumacher I'm wondering if I turn off SymbolicProcessing, would this error still present?
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| Jul 27, 2019 at 6:42 | history | became hot network question | |||
| Jul 26, 2019 at 23:02 | vote | accept | John Taylor | ||
| Jul 26, 2019 at 22:46 | comment | added | Henrik Schumacher | Mathematica does not have to evaluate the integrand symbolically, in order to apply numerical quadrature. However, it does so usually in order to analyze the integrand. If it fails, it throws an error and tries purely numerical methods. This why it works in the end. | |
| Jul 26, 2019 at 22:44 | answer | added | Wen Chern | timeline score: 8 | |
| Jul 26, 2019 at 22:43 | history | edited | Henrik Schumacher | CC BY-SA 4.0 |
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| Jul 26, 2019 at 22:43 | comment | added | David G. Stork | Hmmm.... a tricky question about the internal methods of Mathematica. Frankly, I don't know why the correct answer is given (despite the error messages). | |
| Jul 26, 2019 at 22:41 | answer | added | David G. Stork | timeline score: 3 | |
| Jul 26, 2019 at 22:40 | comment | added | John Taylor | @DavidG.Stork : however, it turned out that Mathematica gives the correct answer. Why this is possible? | |
| Jul 26, 2019 at 22:39 | comment | added | David G. Stork | Of course you cannot perform the inner numerical integral with an un-specified (free) variable. | |
| Jul 26, 2019 at 22:34 | history | asked | John Taylor | CC BY-SA 4.0 |