Timeline for How to solve the following $0= \int_{-\infty}^\infty e^{-\frac{(bt+\omega)^2}{2}} f(t+\omega) \frac{1}{i t} dt, \forall \omega \in \mathbb{R}$
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
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| S Mar 19, 2022 at 22:00 | history | bounty ended | CommunityBot | ||
| S Mar 19, 2022 at 22:00 | history | notice removed | CommunityBot | ||
| Mar 17, 2022 at 21:03 | vote | accept | Boby | ||
| Mar 17, 2022 at 21:04 | |||||
| Mar 15, 2022 at 20:25 | answer | added | Diger | timeline score: 0 | |
| Mar 14, 2022 at 14:40 | history | edited | Boby | CC BY-SA 4.0 |
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| Mar 13, 2022 at 19:45 | history | edited | Boby | CC BY-SA 4.0 |
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| S Mar 11, 2022 at 20:39 | history | bounty started | Boby | ||
| S Mar 11, 2022 at 20:39 | history | notice added | Boby | Authoritative reference needed | |
| Mar 8, 2022 at 15:31 | history | edited | Boby | CC BY-SA 4.0 |
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| Mar 7, 2022 at 22:59 | answer | added | Carlo Beenakker | timeline score: 5 | |
| Mar 7, 2022 at 22:03 | comment | added | Boby | @CarloBeenakker No problem. You seem to know these things well. Any idea of how to approach this? | |
| Mar 7, 2022 at 22:02 | comment | added | Carlo Beenakker | my mistake, I misread sign for step function; I'll delete this string of comments that confused you, apologies. | |
| Mar 7, 2022 at 21:44 | comment | added | Boby | @CarloBeenakker. Essentially it originate from Fourier transform of $sign(t)$ | |
| Mar 7, 2022 at 21:35 | comment | added | Boby | @CarloBeenakker Ok. I have add it. | |
| Mar 7, 2022 at 21:29 | history | edited | Boby | CC BY-SA 4.0 |
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| Mar 7, 2022 at 21:23 | comment | added | Carlo Beenakker | how will you avoid the pole at $t=0$? is this a principal value? you may want to explain where the integral comes from, as it stands it is not well-defined (and since the left-hand-side is 0, what is the meaning of the factor $i\pi$ on the right-hand-side?) | |
| Mar 7, 2022 at 21:03 | history | asked | Boby | CC BY-SA 4.0 |