Timeline for Asymptotic behavior of an integral depending on an integer
Current License: CC BY-SA 4.0
20 events
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| Jun 23, 2018 at 23:08 | comment | added | GH from MO | My response now contains a proof that $4\pi n-6\pi<f(n)<4\pi n-2\pi$. | |
| Jun 23, 2018 at 22:16 | vote | accept | Mahdi - Free Palestine | ||
| Jun 23, 2018 at 21:48 | history | edited | GH from MO |
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| Jun 23, 2018 at 21:47 | answer | added | GH from MO | timeline score: 7 | |
| Jun 23, 2018 at 18:18 | comment | added | Henri Cohen | numerically it seems that $f(n)=4\pi n-6\pi+O(1/\sqrt{n})$ | |
| Jun 23, 2018 at 16:59 | comment | added | Mahdi - Free Palestine | @fedja: Note that in the comment of Greg, there is a typo. Actually, we have $\int_0^{\infty} \log((1+t)^n)t^{-3/2}dt = 2n\pi$ | |
| Jun 23, 2018 at 15:05 | comment | added | fedja | Something is fishy. For large even $n$, if I divide the expression under the logarithm by $(1+t)^n$, I get $1+\frac12[(\frac{1-t}{1+t})^n-1]+n(n-1)t\frac{1}{(1+t)^2}\ge 1+\frac 12(-2tn)+\frac 14n(n-1)t$ for $t\le 1$, which is never below $1$ and tends to $+\infty$ as $n\to\infty$. The integrals from $1$ to $\infty$ are uniformly bounded from below, so you get more than $4\pi n$, which is impossible since you have an example of a matrix giving $2n-2$. What am I missing? | |
| Jun 23, 2018 at 9:19 | history | undeleted | Mahdi - Free Palestine | ||
| Jun 23, 2018 at 9:18 | history | deleted | Mahdi - Free Palestine | via Vote | |
| Jun 23, 2018 at 7:56 | comment | added | Mostafa - Free Palestine | @GregMartin In fact using this idea and and a change of variable $nt=u$ gives $f(n)=4\pi n +O(1)$. | |
| Jun 23, 2018 at 7:47 | comment | added | Mahdi - Free Palestine | @GregMartin: Thanks. It's a good idea. | |
| Jun 23, 2018 at 6:54 | comment | added | Greg Martin | Can't you just consider the difference $f(n) - 4n\pi = f(n) - \int_0^\infty \log((1+t)^n)t^{-3/2}\,dt$, combined into a single integral that looks like $\int_0^\infty \log(g(n,t))t^{-3/2}\,dt$, where $g(n,t)$ is bounded below away from $0$ and bounded above by say $n^2$? That would show that $f(n) = 4n\pi + O(\log n)$. | |
| Jun 23, 2018 at 6:25 | history | edited | Mahdi - Free Palestine | CC BY-SA 4.0 |
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| Jun 22, 2018 at 17:06 | comment | added | Suvrit | For $n=2,3,4,5$ this is $4\pi, 4\sqrt{3}\pi, 4\cdot (2.782...)\pi,4\cdot 3.65\pi$, interesting.. | |
| Jun 22, 2018 at 16:57 | history | edited | Mahdi - Free Palestine | CC BY-SA 4.0 |
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| Jun 22, 2018 at 16:24 | history | edited | Mahdi - Free Palestine | CC BY-SA 4.0 |
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| Jun 22, 2018 at 16:13 | comment | added | Mahdi - Free Palestine | @NateEldredge The function under integral divided by $n$ converges pointwise to $log(1+t)$ but it seems that there is no dominating function to use convergence theorems for integrals. | |
| Jun 22, 2018 at 15:25 | comment | added | Nate Eldredge | Does anything useful happen if you apply L'Hospital's rule and differentiate under the integral sign? | |
| Jun 22, 2018 at 15:12 | history | edited | Mahdi - Free Palestine | CC BY-SA 4.0 |
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| Jun 22, 2018 at 14:57 | history | asked | Mahdi - Free Palestine | CC BY-SA 4.0 |