If $f$ continuous at $[1,\infty)$ and $\int _1^\infty\,f\left(x\right)dx$ converge, then $\int _1^\infty\frac{f\left(x\right)}{x}dx$ also converge? I need to prove or disprove this statement.
I didn't find any counter example. So I tried to prove it.
I know that $\int _1^\infty\,f\left(x\right)dx$ converge so the limit

$\lim _{t\to \infty }\int _1^tf\left(x\right)dx$ exists

I also know that $f$ is coninuous , so there's a function $F\left(x\right)\:=\:\int _1^xf\left(s\right)ds$   

Such that $F'(x) = f(x)$

That's all I know and I dont know how to put this together.
If $f$ is non-negative, then
$$
\int_1^x\frac{f(t)\,dt}{t}\le\int_1^x f(t)\,dt,
$$
and hence $\int_1^\infty\frac{f(t)\,dt}{t}$ converges.

In general, let $\,F(x)=\int_1^x f(t)\,dt$. Since $\lim_{x\to\infty} F(x)=a\in\mathbb R$, then $F$ is bounded. Say $\vert F(x)\rvert\le M$. Then
$$
\int_1^x\frac{f(t)\,dt}{t}=\frac{F(t)}{t}\,\Big|_1^x+\int_1^x \frac{F(t)\,dt}{t^2} \\
=\frac{F(x)}{x}+\int_1^x \frac{F(t)\,dt}{t^2}.
$$
Observe that
$$
\lim_{x\to\infty}\frac{F(x)}{x}=0
$$
while the limit
$$
\lim_{x\to\infty}\int_1^x \frac{F(t)\,dt}{t^2},
$$
exists, since $F$ is bounded.

Note. The fact that $\lim_{x\to\infty}\int_1^x f(t)\,dt$ exists does NOT imply that $f$ is integrable over $[1,\infty)$, i.e., $f\in L^1[1,\infty)$.
So you can use a comparison theorem which says that if $g(x) \leq h(x)$ and $\int_1^\infty h(x) \,dx$ converges then $\int_1^\infty g(x) \,dx$ also converges.  Since in your problem $x &gt;1$ then $\frac{f(x)}{x} \leq f(x)$, you know that $\int_1^\infty \frac{f(x)}{x} \,dx$ converges if $\int_1^\infty f(x) \, dx$ converges.