The integral of a periodic function $\int_{1}^{\infty}\frac{f(x)}{x}\,dx$

Question

Let $f$ be a periodic function that is positive and not equal to zero. Prove that

$$\int_{1}^{\infty}\frac{f(x)}{x}\,dx$$

doesn't diverge. Please help me find the answer.

Answer 1

Through a suitable substitution we may assume, without loss of generality, that the given function is $1$-periodic. If $f(x)\geq K >0$ and $a\in\mathbb{N}$ we have

$$ \int_{2^a}^{2^{a+1}}\frac{f(x)}{x}\,dx \geq \int_{2^a}^{2^{a+1}}\frac{K}{x}\,dx = K\log(2) $$ hence $$ \int_{1}^{+\infty}\frac{f(x)}{x}\,dx = \int_{2^0}^{2^1}\frac{f(x)}{x}\,dx+\int_{2^1}^{2^2}\frac{f(x)}{x}\,dx+\int_{2^2}^{2^3}\frac{f(x)}{x}\,dx+\ldots$$ cannot be convergent.