# An integral of $|\sin(x)\cos(nx)^{-2/n}|$ from $-\pi$ to $\pi$

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 $$n = 17$$.

• But $A_n\geqslant \int_{-\pi}^\pi |\sin x|dx=4$. – Fedor Petrov Nov 12 '19 at 8:11
• I guess $B(1/2,1/2)=\pi$, not $\sqrt{\pi}$. – Fedor Petrov Nov 12 '19 at 8:13
• @Fedor Petrov thanks for spotting the typo! – 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 $$\sum_{k=-n}^{n-1} I_k=\frac1n\sum_{k=-n}^{n-1} \int_0^\pi \left|\sin \frac{k\pi+y}n\right|dy+o(1)=\int_{-\pi}^\pi |\sin x|dx+o(1)=4+o(1).$$