Improper Integral of a periodic function

Question

I need to proof the following statement:

let f(x) be a continuous periodic function that $0\leq\ f(x)$ and not equal to zero. prove that $\int_1^\infty\frac{f(x)}{x}dx$ diverge.

all I secceed so far is to prove that $\int_1^\infty\ f(x)dx$ diverges. any suggestion?

Answer 1

Let $T$ be a period of $f$ and $\int_1^{T+1} f = P$. Then \begin{align*} \int_1^\infty \; \frac{f(x)}{x} \,\mathrm{d}x & = \sum_{k=0}^\infty \int_{kT+1}^{(k+1)T+1} \; \frac{f(x)}{x} \,\mathrm{d}x \\ & \geq \sum_{k=0}^\infty \int_{kT+1}^{(k+1)T+1} \; \frac{f(x)}{\max ( [kT+1,(k+1)T+1] ) } \,\mathrm{d}x \\ & = \sum_{k=0}^\infty \int_{kT+1}^{(k+1)T+1} \; \frac{f(x)}{(k+1)T+1} \,\mathrm{d}x \\ & = \sum_{k=0}^\infty \frac{\int_{kT+1}^{(k+1)T+1} \; f(x) \,\mathrm{d}x}{(k+1)T+1} \\ & = \sum_{k=0}^\infty \frac{\int_{1}^{T+1} \; f(x) \,\mathrm{d}x}{(k+1)T+1} \\ & = \sum_{k=0}^\infty \frac{P}{(k+1)T+1} \text{.} \end{align*}

Answer 2

Suppose $T\geq 1$ (if $T<1$ start the sum at $N$ big enough to have $TN\geq 1$). $$\int_1^\infty \frac{f(x)}{x}\,\mathrm d x\geq \sum_{k=1}^\infty \int_{kT}^{(k+1)T}\frac{f(x)}{x}\,\mathrm d x\geq \sum_{k=1}^\infty \frac{\alpha }{(k+1)T}=\infty $$ where $\alpha =\int_0^T f(x)\,\mathrm d x>0$. By the way, no need continuity as far as $f\in L^1(0,T)$.