Let $f: [1,\infty)\rightarrow \mathbb{R}$ a continous periodic function, none negative, which is not the zero function. Show that the integral $\displaystyle\int_1^{\infty}\frac{f(x)}{x}dx$ diverges.
This is my solution but I am not sure if it is true, I feel there are some details I miss.
We know that $f$ is not the zero function, let $I=[a,b]$ be the periodic part of the function, so we know that we have for sure $[x_1,x_2]\subseteq I$ that $f(x)>0$ for every $x\in [x_1,x_2]$. Thus, there exist $M>0$ s.t $M\leq f(x)$ Hence we can say that $\frac{M}{x}\leq \frac{f(x)}{x}$ for $x\in [x_1,x_2]$.
From here I applied the Convergence tests with the divergence integral $\frac{1}{x}$ and I showed the limit is $M$ which is a finite number. I feel this is not good enough, or should I say not a professional proof?
I would like appreciate for notes and reviews.
Assume that $f$ is continuous with period $t>0$. Then $$A:=\int_1^{1+t}f(x)\text{d}x>0.$$ Fix $n\in \mathbb{N}^*$ and we have \begin{align*} I_n& =\int_1^{1+nt}\frac{f(x)}{x}\text{d}x=\sum_{k=1}^n\int_{1+(k-1)t}^{1+kt}\frac{f(x)}{x}\text{d}x\\ & =\sum_{k=1}^n\int_1^{1+t}\frac{f(x)}{x+(k-1)t}\text{d}x>\sum_{k=1}^n\frac{A}{1+kt}. \end{align*} Thus $I_n\to +\infty,n\to \infty$. The original integral $\int_1^{\infty}\frac{f(x)}{x}\text{d}x$ diverges.