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

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. 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?

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?

• 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? Jun 7, 2020 at 10:51
• 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? Jun 7, 2020 at 11:42
• I know the definition, but I have some trouble showing that both forms are equivalents in your first use of $\sim$. Jun 7, 2020 at 18:35
• See my edit to see where I do have some difficulties. Jun 8, 2020 at 18:05