Timeline for Simple-looking problem with integrals
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| May 4, 2021 at 17:20 | comment | added | Mateusz Kwaśnicki | Yes, that is the idea of the argument. And sorry for the typo. | |
| May 4, 2021 at 17:19 | history | edited | Mateusz Kwaśnicki | CC BY-SA 4.0 |
added 127 characters in body
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| May 4, 2021 at 13:02 | comment | added | alvarezpaiva | BTW, I think there is no $u$ in the numerator in the next to last integral. | |
| May 4, 2021 at 12:50 | comment | added | alvarezpaiva | Let me see if I'm getting this from the conceptual viewpoint: forgetting some multiplicative constants, Fedor Petrov's change of variables makes the integral into the Riemann-Liouville integral $I^{1/2} g$, then you use the $I^{1/2} I^{1/2} = I^1$ relation to show that the antiderivative of $g$ must be zero. Is that an accurate summary? | |
| May 4, 2021 at 11:53 | comment | added | alvarezpaiva | @MateuszKwaśnicki, that's lovely !! | |
| May 4, 2021 at 11:52 | vote | accept | alvarezpaiva | ||
| May 4, 2021 at 8:23 | comment | added | Mateusz Kwaśnicki | @FedorPetrov: Right, of course. And so my contribution to this answer decreased to 1/3. :-) | |
| May 4, 2021 at 8:03 | comment | added | Fedor Petrov | ...and since $g$ is continuous, almost everywhere yields everywhere | |
| May 4, 2021 at 7:26 | history | answered | Mateusz Kwaśnicki | CC BY-SA 4.0 |