3 votes
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Can we bound the lower and upper asymptotic density of these subsets of $\mathbb N$?

We can prove the conjecture true by first considering the logarithmic density of a set $T$ of positive integers, which is $$ \lim_{x\to\infty} \frac1{\log x} \sum_{m\in T\cap[1,x]} \frac1m. $$ ...
Greg Martin's user avatar
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2 votes
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Prove function is in $\mathcal{O}(x\log(x))$

You don't need to prove that the limit is finite the limit can even not exist. However, you can easily prove that the function is bounded. Indeed, for $x\in \mathbb R_{>0}$, $n = \left\lceil\log_2(...
Kroki's user avatar
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2 votes

Question regarding asymptotic distribution and the second order delta method

The technique behind the delta method is the Taylor's approximation. You have: $$f(\bar X_n) =-f(\mu)+(\bar X_n-\mu)f'(\mu) + \frac{1}{2}(\bar X_n-\mu)^2 f''(\mu) + \mathcal{o}((\bar X_n-\mu)^2)$$ We ...
NN2's user avatar
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2 votes

Calculating asymptotics of integral $B(r) = \int_0^\infty \frac{2}{\pi} \frac{k^2 \sin(k r )}{k^3 + a} dk$

Let $$ h(t) = \sin t,\quad f(t) = \frac{2}{\pi }\frac{{t^2 }}{{t^3 + a}}. $$ Then $$ \mathscr{M}B(z) = \mathscr{M}h(z)\mathscr{M}f(1 - z), $$ where $\mathscr{M}$ denotes the Mellin transform. Now $$ \...
Gary's user avatar
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2 votes
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Calculating asymptotics of integral $B(r) = \int_0^\infty \frac{2}{\pi} \frac{k^2 \sin(k r )}{k^3 + a} dk$

$$B(r)=\frac{2}{\pi}\int_0^\infty \frac{k^2 \sin(k r )}{k^3 + a} dk\overset{k=a^{\frac13}t}{=}\frac{2}{\pi}\int_0^\infty \frac{t^2 \sin(a^{\frac13}rt)}{t^3 + 1} dt$$ Denoting for a while $\,a^{\...
Svyatoslav's user avatar
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1 vote

Calculating asymptotics of integral $B(r) = \int_0^\infty \frac{2}{\pi} \frac{k^2 \sin(k r )}{k^3 + a} dk$

This is probably an unsatisfying answer, but at least it should produce the correct answer. Computational software (I used Mathematica) can evaluate the integral: $$ B(r) = \frac1{\pi^{3/2}\sqrt3} G_{...
Greg Martin's user avatar
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