Timeline for Factoring a separable integral with a product of independent integrands
Current License: CC BY-SA 3.0
9 events
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| May 16, 2015 at 7:34 | comment | added | Dmitri | I would like only to mention, that @Vitaliy Kaurov method fails in the case then the integral is not multiple (i.e. just one integral). More difficult case is then we have to factorize if only it is possible; i.e. inTfaC[Integrate[p[x,y], {x, -1, 1}, {y, -1, 1}]] will return the same expressions, whereas inTfaC[Integrate[p[x]p[y], {x, -1, 1}, {y, -1, 1}]] will be factorised. | |
| May 15, 2015 at 21:32 | comment | added | evanb | Ah, you're right. Nevertheless, it's important to emphasize! | |
| May 15, 2015 at 21:22 | comment | added | Vitaliy Kaurov | @evanb Your integral does not satisfy the question title: "Factoring a separable integral with a product of independent integrands " - and I assume that OP deals with those only. | |
| May 15, 2015 at 21:17 | comment | added | evanb |
Be very careful! This will even factor integrals where the variables only "talk to each other" through the integrals' limits! For example, try inTfaC[Integrate[p[x] p[y], {x, -1, y}, {y, -1, 1}]].
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| May 13, 2015 at 22:52 | comment | added | Dmitri | Great! Thank you very much!!! | |
| May 13, 2015 at 22:34 | comment | added | Vitaliy Kaurov | @Dmitri updated | |
| May 13, 2015 at 22:34 | history | edited | Vitaliy Kaurov | CC BY-SA 3.0 |
added 352 characters in body
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| May 13, 2015 at 22:18 | comment | added | Dmitri | Probably, I was not exact. I do not want to teach Mathematica how to factorize the answer. I would like to ask it to do this. The same should be done for a triple integral and so on, | |
| May 13, 2015 at 21:35 | history | answered | Vitaliy Kaurov | CC BY-SA 3.0 |