Limit ($x \to \infty$) of integral of positive and continuous function (in $\mathbb{R}$ and centered at $0$) always finite?

Question

For all functions $f(x)$ that are positive and continuous everywhere in the real numbers. If $\lim\nolimits_{x\to \infty}f(x)=0$, is $\lim\nolimits_{x\to \infty}F(x)$ convergent for all possible $f(x)$? $F(x)$ denotes the indefinite integral of $f(x)$ centered at $0$.

I first thought so until I found a possible counterexample: $$ f(x) =\begin{cases} \frac{1}{x-2}+2 & \text{if }\; x<1,\\ \frac{1}{x} & \text{if }\; x\geq 1.\end{cases} $$

Edit: I would also like to know / possibly see a proof if my counterexample is actually valid or not.

Answer 1

A counterexample is $f(x) = \frac{1}{\sqrt{x^2+1}}$

Answer 2

Here's an example where everything is even fully differentiable on the real line. Define $f$ on $\mathbb{R}$ by $$ f(x) = \begin{cases} -x+2, & x \leq 1\\ \frac{1}{x}, & x > 1. \end{cases} $$ This function is continuous everywhere (in fact, differentiable) and has $\lim_{x\to +\infty} f(x)=0$. However, the antiderivative $$F(x) = \begin{cases} -\frac{x^2}{2}+2x-\frac{3}{2}, & x \leq 1\\ \ln x, & x > 1. \end{cases} $$ will have $\lim_{x \to +\infty} F(x) = +\infty$.

Edit: with your updated question, yes, your counterexample is right.