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For which $\alpha \in (0,+\infty)$ does $\sum_n a_n$ converges where $a_n = \int_0^{\pi/2} (\sin(t))^{n^\beta(\ln n)^\alpha}dt$?

I feel that it converges iff ($\alpha>1$ and $\beta = 1$) or $\beta >1$.

But I have some trouble showing that it diverges for $\alpha<1$. Every lower bound I take for $\sin x$ lead me to something of the form $a_n \geq C\times a^{n^\beta\ln^\alpha n} \frac{1}{n^\beta \ln^\alpha n}$ where $C>0$ and $0<a<1$. Because of the geometric factor, it converges, so the lower bounds I used are not precise enough.

EDIT: actually, I'm not sure when it does converge... I edited the question to its correct form.


As @Zachary answer's suggested, I considered the following steps:

  • Considering $\cos t$ instead of $\sin t$ (since it's the same integral). I'm OK with that.
  • $\cos t = 1 - \frac{t^2}{2} + O(t^4)$ and then $\ln(\cos t) = - \frac{t^2}{2} + O(t^4)$ around $0$. Still fine here.
  • Thus,

$$ \begin{align*} \cos(t)^{n^\beta (\ln n)^\alpha} &= \exp\left((- \frac{t^2}{2} + O(t^4))n^\beta \log^\alpha(n)\right) \\ &= \exp\left(-\frac{t^2}{2}n^\beta \log^\alpha(n)\right) \times \exp\left(O(t^4) n^\beta \log^\alpha(n)\right) \end{align*} $$

How do you deal with the big O?

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Rewrite the integrand as $\exp\left[n^\beta \log^\alpha (n)\log \cos t\right]$ (I've used cosine to perform the integral, but it's the same one by symmetry). The dominant contribution in the limit of the integral as $n \to \infty$ lies in the region near $t=0$. As such, we taylor expand the $\log\cos t$ around there: $$\log \cos t = -\frac{1}{2}t^2 + \mathcal{O}(t^4).$$ Your integral is now $$a_n \sim \frac{1}{2}\int_{-\pi/2}^{\pi/2}\exp\left(-\frac{1}{2}t^2n^\beta \log^\alpha(n)\right)\,dt.$$ We may extend the integration interval to $(-\infty, \infty)$ in this limit since the integrand will be negligible for any $t$ that is greater in magnitude than about $ n^{-\beta/2}\log^{-\alpha/2}(n)$ as $n\to\infty$. The integral is now $$a_n \sim \frac{1}{2} \int_{-\infty}^{\infty}\exp\left(-\frac{1}{2}t^2n^\beta \log^\alpha(n)\right)\,dt = \sqrt{\frac{\pi}{2}}\frac{1}{n^{\beta/2}\log^{\alpha/2}(n)}.$$ Can you take it from there?

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  • $\begingroup$ Thanks for your answer. I understand what you did, however it seems too much "qualitative". I need more details when you write the "$\sim$". How to show it rigorously? $\endgroup$
    – MiKiDe
    Jun 7, 2020 at 10:51
  • $\begingroup$ The $\sim$ notation is a rigorous statement, though admittedly I wasn't very clear about this and left out some details. We say that $b_n \sim c_n$ if and only if $\lim_{n\to\infty} b_n/c_n = 1$; it is often used when and useful in studying the asymptotics of integrals, sums, sequences, etc. What specifically are you interested in? $\endgroup$
    – Diffusion
    Jun 7, 2020 at 11:42
  • $\begingroup$ I know the definition, but I have some trouble showing that both forms are equivalents in your first use of $\sim$. $\endgroup$
    – MiKiDe
    Jun 7, 2020 at 18:35
  • $\begingroup$ See my edit to see where I do have some difficulties. $\endgroup$
    – MiKiDe
    Jun 8, 2020 at 18:05

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