Does $\int _1 ^\infty\frac {f(x)} x\,dx$ converge or diverge? Let $f(x)$ be continuous in $[1, \infty)$ and $\int_{1}^{\infty} f(x)\,dx$ converge. I need to prove or disprove this: $\int_{1}^{\infty}\frac{f(x)}{x}\,dx$ converge.
I think this is true but I don't know how to prove this. Someone can give me a hint how to prove this. 
 A: Abel's theorem: let $g = fh$ be defined on $[a, b)$ with $b \in (a, \infty]$. If it holds that:


*

*$h$ is continuous, positive, decreasing and $\lim_{x \to b} h(x) = 0$

*$f$ is locally integrable on $[a, b)$ (i.e. integrable in the neighbourhood of any point) and $F$ is bounded on $[a, b)$, where $F: [a,b) \to \Bbb R$,
$$F(x) = \int_a^x f(t) dt$$
Then, $\int_a^b g(x)dx$ is convergent. 
Use this with $a = 1$, $b = \infty$, $h(x) = 1/x$
A: Define $F(x):= \int_{1}^{x}f(t) \, dt$ and notice that $F$ is bounded and continuously differentiable on $[1,\infty)$.
Perform an integration by parts $$\int_{1}^{x}\frac{f(t)}{t} \, dt = \frac{F(x)}{x} + \int_{1}^{x}\frac{F(t)}{t^{2}} \, dt$$ What happens when you let $x\rightarrow \infty$?
A: Consider using the integral test. If you can prove that the series is always less than a convergent integral, then it must converge. You are given that $f(x)$ converges, and $\frac{f(x)}{x}$ will be less than $f(x)$ for all $x > 1$. Since the lower bound on your summation is $1$ ,$\frac{f(x)}{x} < f(x)$, and the series converges by the integral test.
