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Let $f(x) = \frac{1}{\sqrt{|\sin 2 \pi x|}}$, and take $$F(x) = \sum_{k=1}^\infty \frac{f(kx)}{k^2}.$$ Show that $F$ is finite a.e.

It is enough to show that $\int_n^{n+1} F(x) \, dx < \infty$ for all $n \in \mathbb{Z}$. Note that $$\int_{n}^{n+1} F(x) \, dx = \sum_{k=1}^\infty \frac{1}{k^2} \int_n^{n+1} f(kx) \, dx.$$ If we can show that this integral is finite, then we are done. But we calculate $$\int_n^{n+1} f(kx) \, dx = \int_n^{n+1} \frac{1}{\sqrt{|\sin 2 \pi kx|}} \, dx = \int_0^1 \frac{1}{\sqrt{|\sin 2 \pi kx|}} \, dx \\ = \frac1k \int_0^k \frac{1}{\sqrt{|\sin 2 \pi x|}} \, dx = \int_0^1 \frac{1}{\sqrt{|\sin 2 \pi x|}} \, dx,$$ where we used the periodicity of $f$. But the problem is I do not think this integral is finite.

BingerBiter
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  • https://math.stackexchange.com/questions/2510876/show-that-sum-k-1-infty-frac1k2-frac1-sqrt-sinkx-converge?rq=1 – Reza Rajaei Nov 23 '22 at 05:35

1 Answers1

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\begin{aligned} 0\leq \int_0^1 \frac{1}{\sqrt{|\sin 2 \pi x|}} \, dx&=4\int_0^\frac{1}{4} \frac{1}{\sqrt{\sin 2 \pi x}} \, dx \\&=\frac{2}{\pi}\int_0^\frac{\pi}{2}\frac{1}{\sqrt{\sin x}} \, dx \\&\leq \frac{2}{\pi}\int_0^\frac{\pi}{2}\sqrt{\frac{\pi}{2x}} \, dx \quad\quad(*) \\&=2 \end{aligned}

$(*)$ : $\sin(x) > \frac2\pi x$ for $0<x<\frac\pi2 $

M A Pelto
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