Timeline for Holomorphic extension of the Fourier transform of a measure
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Aug 13, 2023 at 3:38 | history | suggested | user509119 | CC BY-SA 4.0 |
fixed grammar
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| Aug 13, 2023 at 0:30 | review | Suggested edits | |||
| S Aug 13, 2023 at 3:38 | |||||
| Aug 12, 2023 at 20:18 | answer | added | Christian Remling | timeline score: 2 | |
| Aug 12, 2023 at 8:10 | review | Suggested edits | |||
| Aug 12, 2023 at 17:59 | |||||
| Aug 12, 2023 at 0:45 | comment | added | user509119 | The expression does not make sense if $\int_\mathbb{R}e^{y\xi} e^{-|\xi|}\,d\xi=\infty$. For instance, $\int_\mathbb{R} e^{-i\xi x}e^{-|\xi|}\,d\xi$ extends to $\frac{1}{1+z^2}$. This is not an entire function, but $\int_\mathbb{R}e^{y\xi} e^{-|\xi|}\,d\xi=\infty$ is remarkable. | |
| Aug 11, 2023 at 17:16 | comment | added | Gerald Edgar | Is it true that the analytic continuation (if it exists) must be $f(z)=\int_\mathbb{R}e^{-i z\xi}\,d\mu(\xi), z \in \mathbb C$ ? | |
| Aug 11, 2023 at 14:48 | comment | added | user509119 | Under the assumption, for any $y\in\mathbb{R}$, $\int e^{y\xi}\,d\mu(\xi)<\infty$ is ture? | |
| Aug 11, 2023 at 14:09 | history | asked | user509119 | CC BY-SA 4.0 |