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seen Nov 22 at 10:13

May
5
comment How prove this linear algebra $AB=BA$?
It is also related, Motzkin & Taussky's original proof.
May
5
comment How prove this linear algebra $AB=BA$?
I hope if there's some pure elementary approach.
May
5
comment How prove this linear algebra $AB=BA$?
The second link seems unavailable.
May
5
revised Tricky elementary integral
added 147 characters in body
May
5
answered Tricky elementary integral
May
5
comment Snags when discovering the asymptotic behavior of an integral
Although it's not a complete answer, it's well-informed and informative.
May
5
accepted Snags when discovering the asymptotic behavior of an integral
May
4
awarded  Announcer
May
4
comment Laplace integration after the first term
Bibliography: de Bruijn's Asymptotic methods in analysis
May
4
revised Prove an integral limit
edited body
May
4
comment Laplace integration after the first term
The substitution $t=h(s)$ does work; however, the integral become an improper integral. Note that the improper integral $\Gamma(1+\alpha)=\int_0^\infty e^{-q}q^\alpha dq$ converges even when $-1<\alpha<0$.
May
4
answered Prove an integral limit
May
4
revised Asymptotic related to the infinite product of sine
added 273 characters in body
May
3
revised Asymptotic related to the infinite product of sine
added 356 characters in body
May
3
comment Asymptotic related to the infinite product of sine
@AntonioVargas $x$ is a constant, as I've said. What I really need, is methods, discipline to deal with such a summand. Comparing is just yielding by-products.
May
2
comment Snags when discovering the asymptotic behavior of an integral
Huh, since these days I read part of de Bruijn's, I picked some stuff out from my mathematical analysis textbook and tried to determine the asymptotic behavior. Unfortunately, I found the road is extremely distorted. This one is just one of these problems. I've posted a new question related to the proof of the infinite product of sine. I will be pleased if there's some asymptotic analyst (professor) around me.
May
2
comment Snags when discovering the asymptotic behavior of an integral
I doubt that it could be expanded. I guess the term after $Cn^{-2}$ is $O(n^{-3}\log n)$, not $O(n^{-3})$.
May
2
comment Snags when discovering the asymptotic behavior of an integral
Typo, the remainder is $\displaystyle\frac1{n^2\pi^2}\int_0^{\pi/2}\frac{x\sin xdx}{(\sin x+1/n\pi)(\sin x+(1+x\sin x)/n\pi)}$.
May
2
comment Snags when discovering the asymptotic behavior of an integral
The remainder is $\displaystyle\frac1{n^2\pi^2}\int_0^{\pi/2}\frac{x\sin xdx}{\sin x+1/n\pi}{\sin x+(1+x\sin x)/n\pi}\asymp\frac1{n^2\pi^2}\int_0^{\pi/2}\frac{xdx}{\sin x}$, where the absolute error is $\displaystyle-\frac1{n^3\pi^3}\int_0^{\pi/2}\frac{(2\sin x+x\sin^2x+(1+x\sin x)/n\pi)xdx}{(\sin x+1/n\pi)(\sin x+(1+x\sin x)/n\pi)\sin x}$, which is not easy to determine the major part (it seems that Lebesgue's dominated convergence theorem doesn't work well).
May
2
comment Snags when discovering the asymptotic behavior of an integral
Well, I'll compute it soon. Incidentally, does some method called saddle point method work for this integral? I have no idea about complex analysis and Cauchy integral theorem. It is said that such a method is very powerful in de Bruijn's Asymptotic methods in analysis.