Timeline for Which method should I use for NIntegrate near a singularity?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 28, 2020 at 12:10 | vote | accept | Pxx | ||
| Mar 28, 2020 at 12:10 | comment | added | Pxx | The integral converges, I have understood my mistake in the analysis of the original integral. Therefore I accept your answer, since it was the first one chronologically to show the convergence. | |
| Mar 26, 2020 at 22:29 | comment | added | Pxx | I still could not determine for sure whether the integral diverges or not, but at least I am now sure I have not made a mistake. I opened a new post in the math stackexchange to try to resolve that question before working further on the numerical side. | |
| Mar 26, 2020 at 12:59 | comment | added | Anton Antonov | @AlexeiBoulbitch Try to put "SingularityHandler" -> None. -- That one is my second favorite singularity handler! :) | |
| Mar 26, 2020 at 12:46 | comment | added | Pxx | It changes nothing unfortunately. When I do the naive NIntegrate[ Integrand[[Tau]3, [Tau]4], {[Tau]3, -[Infinity], [Infinity]}, {\ [Tau]4, -[Infinity], [Infinity]}, WorkingPrecision -> 10], it seems to diverge though. But how can I be sure? | |
| Mar 26, 2020 at 12:42 | comment | added | Alexei Boulbitch | Try to put "SingularityHandler" -> None. | |
| Mar 26, 2020 at 12:29 | comment | added | Pxx | Hmm well, the idea was to reproduce the $-\log \epsilon^2$ divergence in the plot that you produced. What about disabling the SingularityHandler? | |
| Mar 26, 2020 at 12:19 | comment | added | Alexei Boulbitch | The numeric estimate does not show it. If you are right and it, indeed, diverges, this means that you cannot calculate it. No way. The SingularityHandler supresses the singularity during the numeric estimate of the integral, and I guess the weak singularity, such as the logarithmic one, can be erroneously removed. In this case you may obtain the value of the integral when in reality it does not exist. So, be careful. | |
| Mar 26, 2020 at 12:15 | comment | added | Pxx | Hi, and thanks for your answer! Shouldn't the integral diverge as $\epsilon \to 0$? That is the result that I found in $(2)$ and would like to reproduce. Why does it appear finite numerically? | |
| Mar 26, 2020 at 12:11 | history | answered | Alexei Boulbitch | CC BY-SA 4.0 |