For an integer $n \geq 3$, define

$$A_n = \int\limits_{-\pi}^\pi\frac{|\sin(x)|}{|\cos(nx)|^{2/n}}dx.$$

It is a fact that $A_n$ is finite for all such $n$. I am interested in the behaviour of a sequence $A_n$. From the integral representation of the Beta function

$$B(x, y) = 2\int_0^{\frac{\pi}{2}}(\sin \theta)^{2x - 1}(\cos \theta)^{2y - 1}d\theta,$$

it is rather easy to derive an upper bound

$$A_n \leq 2B\left(\frac{1}{2}, \frac{1}{2} - \frac{1}{n}\right).$$

However, I think that more accurate estimates can be made. In particular, note that $2B\left(\frac{1}{2},\frac{1}{2} - \frac{1}{n}\right)$ approaches $2\pi$ as $n$ tends to infinity, whereas I believe that $A_n$ should approach a value strictly smaller. What is the limit? Can we say that $A_n$ is monotonously decreasing? I would value any suggestions on how to approach this problem.

Below is the graph of the function $\left|\sin(x)\cos(nx)^{-2/n}\right|$ for $x \in [-\pi, \pi]$ for $n = 17$.

enter image description here

  • $\begingroup$ But $A_n\geqslant \int_{-\pi}^\pi |\sin x|dx=4$. $\endgroup$ – Fedor Petrov Nov 12 '19 at 8:11
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    $\begingroup$ I guess $B(1/2,1/2)=\pi$, not $\sqrt{\pi}$. $\endgroup$ – Fedor Petrov Nov 12 '19 at 8:13
  • $\begingroup$ @Fedor Petrov thanks for spotting the typo! $\endgroup$ – Anton Nov 12 '19 at 13:03

Consider the integral over $\frac\pi{n}[k,k+1]$, where $k=-n,\ldots,n-1$. Denote $x=\frac{k\pi+y}n$, you get $$I_k:=\frac1n\int_{0}^{\pi}\frac{|\sin \frac{k\pi+y}n|}{|\cos y|^{2/n}}dy.$$ Denote $$ \varepsilon_k:=I_k-\frac1n\int_0^\pi \left|\sin \frac{k\pi+y}n\right|dy= \frac1n \int_0^\pi \left|\sin \frac{k\pi+y}n\right|\left(|\cos y|^{-2/n}-1\right)dy. $$ We see that $$\frac1n\int_0^{\pi} \left(|\cos y|^{-2/n}-1 \right)dy\geqslant \varepsilon_k\geqslant 0.$$ By Monotone Convergence theorem, we have $$\lim_n \int_0^{\pi} \left(|\cos y|^{-2/n}-1 \right)dy=0.$$ Thus $\varepsilon_k=o(1/n)$ uniformly by $k$ and we get $$ \sum_{k=-n}^{n-1} I_k=\frac1n\sum_{k=-n}^{n-1} \int_0^\pi \left|\sin \frac{k\pi+y}n\right|dy+o(1)=\int_{-\pi}^\pi |\sin x|dx+o(1)=4+o(1). $$

  • $\begingroup$ thank you for the solution. Also there was a typo in the original question: the upper bound is 2B(1/2, 1/2 - 1/n), not B(1/2, 1/2 - 1/n), so the limiting value for my upper bound is 2\pi. $\endgroup$ – Anton Nov 12 '19 at 13:47

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