Timeline for Inequality about exponential integrals
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 18, 2020 at 10:22 | history | edited | Dr. Pi | CC BY-SA 4.0 |
edited body
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| Apr 18, 2020 at 1:57 | comment | added | Gerry Myerson | "the said authors"??? | |
| Apr 18, 2020 at 0:31 | comment | added | Mateusz Kwaśnicki | Yes, this is just the matter of continuity. The above calculation is the usual trick for the Fourier transform, re-written in terms of the Mellin transform. | |
| Apr 17, 2020 at 23:57 | comment | added | Dr. Pi | Great, it works! How could you have predicted that the terms would cancel out without looks at the specific formulas? Edit: i guess this only has to do with the fact that $f$ is continuous. | |
| Apr 17, 2020 at 23:19 | comment | added | Mateusz Kwaśnicki | OK, that was a little bit too fast; but the idea is the same: integrate by parts twice on each interval $(t_{j-1}, t_j)$ where $f$ is linear. The terms with $(i \log x)^{-1} f(t_j) x^{i t_j}$ cancel out, the terms $(i \log x)^{-2} f'(t_j) x^{i t_j}$ are $O(N^{-1} (\log x)^{-2})$, and what remains is an integral involving $f'' = 0$. | |
| Apr 17, 2020 at 23:15 | comment | added | Dr. Pi | Thanks! Can I ask what happens with $f''(t)$ when $t$ is near the points of discontinuouity? | |
| Apr 17, 2020 at 23:12 | comment | added | Mateusz Kwaśnicki | Integrate by parts twice, to get $(i \log x)^{-2} \int_{\mathbb R} f''(t) x^{it} dt$. | |
| Apr 17, 2020 at 23:06 | history | asked | Dr. Pi | CC BY-SA 4.0 |