Prove $\int_1^\infty \frac{f(x)}{x}$ diverges

Question

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.

Answer 1

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).

Answer 2

hint If $T $ is the period then

$$\int_{nT}^{(n+1)T}\frac {f (x)}{x}dx=\int_0^T\frac {f (t)}{t+nT} dt\ge \frac {1}{(n+1)T} \int_0^Tf(t)dt$$

Answer 3

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.

Answer 4

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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 ?.