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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| Mar 17, 2022 at 21:03 | vote | accept | Boby | ||
| Mar 17, 2022 at 21:04 | |||||
| Mar 15, 2022 at 20:09 | comment | added | Iosif Pinelis | My apologies for the hastiness again. I mistook $f(t+\omega)$ for $f(t)$. However, as Boby said, $f(t)$ should not depend on $\omega$. | |
| Mar 15, 2022 at 7:23 | comment | added | Carlo Beenakker | @Diger --- I added the explanation to the text. | |
| Mar 15, 2022 at 7:19 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Mar 15, 2022 at 7:17 | comment | added | Carlo Beenakker | @IosifPinelis -- perhaps I am just misunderstanding you, I added a line to show how the regularized integral vanishes. | |
| Mar 15, 2022 at 7:07 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Mar 15, 2022 at 0:13 | comment | added | Iosif Pinelis | Sorry for the typo. What I meant to say is this: If $\text{constant}\ne0$ and $b\ne0$, the integral equals $\infty$ (not $0$) for your general solution. | |
| Mar 15, 2022 at 0:11 | comment | added | Diger | Why is $G_0(x) \propto \delta(x)$ the only solution? | |
| Mar 14, 2022 at 16:51 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Mar 14, 2022 at 16:47 | comment | added | Carlo Beenakker | yes, it is also a function of $\omega$, I don't think there is an $\omega$-independent solution for $b\neq 0,1$, but you're right, this is not a useful solution. | |
| Mar 14, 2022 at 16:44 | comment | added | Boby | Thanks. Quick question. Shouldn't $f(t)$ be only a function of $t$? Here it is also a function of $\omega$. | |
| Mar 14, 2022 at 16:33 | comment | added | Carlo Beenakker | I don't think so: when $b=0$ and $f(t)=\text{constant}$ we have the principal value integral $\int_{-\infty}^\infty dt/t$ which vanishes --- when interpreted as $\lim_{b\rightarrow\infty}\lim_{a\rightarrow 0}\left(\int_{-b}^{-a}dt/t+\int_{a}^{b}dt/t\right)$ --- my understanding is that this is how the OP wishes to interpret the singular integral | |
| Mar 14, 2022 at 16:30 | comment | added | Iosif Pinelis | But, if $constant\ne0$ and $b=0$, the integral equals $\infty$ (not $0$) for your general solution. | |
| Mar 14, 2022 at 16:28 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Mar 14, 2022 at 16:10 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
updated with the solution for general $b$
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| Mar 8, 2022 at 2:57 | comment | added | Boby | Thanks. Any other ideas or places I can look would be greatly appreciated. | |
| Mar 7, 2022 at 22:59 | history | answered | Carlo Beenakker | CC BY-SA 4.0 |