Say we have a function $f:\mathbb{R}\to\mathbb{R}$ that is periodic, non-negative, continuous, and not identically zero. I have to prove that the improper integral $\int_{1}^{\infty}{\frac{f(x)}{x}}dx$ diverges.
I tried using integration by parts ($f$ is continuous so it has an anti-derivative) and then determining convergence/divergence separately ($\int_{1}^{\infty}{\frac{f(x)}{x}}dx$ = $\lim_{b\to\infty}(\frac{F(b)}{b}-F(1)+\int_{1}^{b}{\frac{F(x)}{x^2}}dx)$, but I can't seem to find what to do from here. Am I on the right track or am I missing something?
WLOG $f$ is periodic with period $1$. Since $f$ is non-negative, not identically zero, and continuous, there is an interval $I$ of length $\delta>0$ in $(1,2)$ on which $f(x)>0$. This implies $\int_{I} f(x)\,dx = k>0$. Then $$ \int _1^{\infty} \frac{f(x)}{x}\,dx = \sum_{n=1}^{\infty} \int_{n}^{n+1}\frac{f(x)}{x}{dx} \geq\sum_{n=1}^{\infty}\frac{1}{n} \int_{1}^{2}f(x){dx} $$ $$ \geq\sum_{n=1}^{\infty}\frac{1}{n} \int_{I}f(x){dx}=\sum_{n=1}^{\infty}\frac{k}{n} = \infty $$So the integral diverges. If $f$ is continuous and periodic with period $1$, a necessary and sufficient condition that $\int_1^{\infty} \frac{f(x)}{x}\,dx$ converge is that $\int_1^2 f(x)\,dx = 0$.