Timeline for Calculate asymptotic value of an integral of exponential function
Current License: CC BY-SA 4.0
8 events
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| Jul 20, 2022 at 9:24 | comment | added | Johannes Trost | Your approach works fine for $p>2$ as well. There is only one significant positive maximum, $z_{0}>0$ of the exponent, which is (asymptotically, but that is sufficient) $z_{0}\sim (2 c^{p-1}\ \zeta)^{\frac{1}{p-2}}$. Calculating the Gaussian integral around this point gives at least the leading term of the expansion. | |
| Jul 19, 2022 at 21:45 | comment | added | Carlo Beenakker | Ah, c>0, I had missed that. This p=2 answer is for c<0, otherwise one needs p>2, as you say. | |
| Jul 19, 2022 at 21:12 | comment | added | Johannes Trost | Isn't there a problem with convergence for $p\le 2$ given that $c>0$ and $\zeta >0$ ? | |
| Jul 19, 2022 at 17:33 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Jul 19, 2022 at 17:28 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Jul 19, 2022 at 17:22 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Jul 19, 2022 at 17:16 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Jul 19, 2022 at 17:11 | history | answered | Carlo Beenakker | CC BY-SA 4.0 |