Prove $\int_1^\infty \frac{f(x)}{x}$ diverges 
Let $f$ be a continuous periodic function on $\mathbb{R},$ such that $0 \not\equiv f \geq 0.$
Prove $\int_1^\infty \frac{f(x)}{x}$ diverges.

My thoughts:
I tried applying the Limit Comparison Test with $\frac{1}{x}:$
$$\lim_{x\to \infty}\frac{\frac{f(x)}{x}}{\frac{1}{x}}=\lim_{x\to \infty} f(x)$$
Continuity and periodicity of $f$ implies $\lim_{x\to \infty}f(x)$ does not exist, so I got stuck.
Next I tried the Comparison Test with $\frac{1}{x},$ but couldn't manipulate the inequality to achieve $\frac{1}{x}\leq \frac{f(x)}{x}.$
Any help appreciated.
 A: If $f$ has period $1$ then
$$\int_1^\infty\frac{f(x)}x\,dx=
\sum_{n=1}^\infty\int_n^{n+1}\frac{f(x)}x\,dx
=\sum_{n=1}^\infty\int_0^1\frac{f(t+1)}{n+t}\,dt
=\int_0^1 f(t+1)\sum_{n=1}^\infty\frac1{n+t+1}\,dt$$
and this is a divergent series. With rather more
fiddling around, one can do general period in a similar
fashion.
A: hint 
If $T $ is the period then
$$\int_{nT}^{(n+1)T}\frac {f (x)}{x}dx=\int_0^T\frac {f (t)}{t+nT} dt\ge \frac {1}{(n+1)T} \int_0^Tf(t)dt$$
A: Let $T > 0$ be a period of $f$, and let $C := \int_0^T f(x)\, dx$. By the assumptions on $f$ we have that $C > 0$.
Let $n\in\mathbb{N}$ and let us compute
$$
\int_1^{1+nT} \frac{f(x)}{x}\, dx =
\sum_{k=1}^n \int_{1+(k-1)T}^{1+kT} \frac{f(x)}{x}\, dx
\geq \sum_{k=1}^n \int_{1+(k-1)T}^{1+kT} \frac{f(x)}{1+(k-1)T}\, dx
= \sum_{k=1}^n \frac{C}{1+(k-1)T}.
$$
As $n\to +\infty$, the last term goes to $+\infty$ (since the corresponding series is divergent).
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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Lets $\ds{T > 0}$ the $\ds{\mrm{f}}$-period.

With $\ds{R > 0}$:
\begin{align}
\int_{1}^{1 + R}{\mrm{f}\pars{x} \over x}\,\dd x & =
\int_{1}^{1 + \left\lfloor{R/T}\right\rfloor T + \braces{R/T}T}{\mrm{f}\pars{x} \over x}\,\dd x
\\[5mm] & =
\int_{1}^{1 + T}{\mrm{f}\pars{x} \over x}\,\dd x +
\int_{1 + T}^{1 + \left\lfloor{R/T}\right\rfloor T}{\mrm{f}\pars{x} \over x}\,\dd x +
\int_{1 + \left\lfloor{R/T}\right\rfloor T}^{1 + \left\lfloor{R/T}\right\rfloor T + \braces{R/T}T}{\mrm{f}\pars{x} \over x}\,\dd x
\\[5mm] & =
\bracks{\int_{1}^{1 + T}{\mrm{f}\pars{x} \over x}\,\dd x +
\int_{1}^{1 + \braces{R/T}T}{\mrm{f}\pars{x} \over x  + \left\lfloor{R/T}\right\rfloor T }\,\dd x} +
\bbox[15px,#ffe]{\ds{\int_{1 + T}^{1 + \left\lfloor{R/T}\right\rfloor T}{\mrm{f}\pars{x} \over x}\,\dd x}} 
\end{align}

\begin{align}
\bbox[15px,#ffe]{\ds{\int_{1 + T}^{1 + \left\lfloor{R/T}\right\rfloor T}{\mrm{f}\pars{x} \over x}\,\dd x}} & =
\sum_{n = 1}^{\left\lfloor{R/T}\right\rfloor - 1}\int_{1 + nT}^{1 + nT + T}{\mrm{f}\pars{x} \over x}\,\dd x =
\sum_{n = 1}^{\left\lfloor{R/T}\right\rfloor - 1}\int_{1}^{1 + T}{\mrm{f}\pars{x} \over x + nT}\,\dd x
\\[5mm] & =
{1 \over T}\int_{1}^{1 + T}\mrm{f}\pars{x}
\sum_{n = 0}^{\left\lfloor{R/T}\right\rfloor - 2}{1 \over n + 1 + x/T}\,\dd x
\\[5mm] & =
{1 \over T}\int_{1}^{1 + T}\mrm{f}\pars{x}
\sum_{n = 0}^{\infty}\pars{%
{1 \over n + 1 + x/T} - {1 \over n + \left\lfloor{R/T}\right\rfloor + x/T}}
\,\dd x
\\[5mm] & =
{1 \over T}\int_{1}^{1 + T}\mrm{f}\pars{x}
\pars{H_{\left\lfloor{R/T}\right\rfloor + x/T - 1} - H_{x/T}}\,\dd x\qquad
\pars{~\substack{\ds{H_{z}:\ Harmonic}\\ \ds{Number}}~}
\end{align}


Note that
  $\ds{H_{\left\lfloor{R/T}\right\rfloor + x/T - 1} \sim \ln\pars{\left\lfloor{R \over T}\right\rfloor + {x \over T} - 1}\quad\text{as}\quad R \to \infty}$. What do you conclude with the above relations ?.

