If $\int\limits_0^{\infty}f^2(x) dx$ is convergent, prove $\int\limits_a^{\infty}\frac{f(x)}x dx$ is convergent for any $a\ge 0$
I use the Cauchy-Schwarz Inequality to get : $$\left( \int\limits_a^A \frac{f(x)}{x}dx \right)^2 \leq \left( \int\limits_a^A f^2(x)dx \right) \left( \int\limits_a^A \frac{1}{x^2}dx\right)\le +\infty$$, so for every $A$, the $\int\limits_a^A \frac{f(x)}{x} dx$ is bounded, but then I can not prove the $\int\limits_a^A \frac{f(x)}x dx$ is not the vibration one such as $(-1)^n$ since it is not given monotone. So I can not address this. Could you give some way to solve it?(Or could you have any other method to solve it?) Thank you！
Note that the same argument works for $|f|$ instead of $f$, and then the integrand becomes non-negative. $$\lim_{A\to\infty}\int_a^A\dfrac {|f(x)|}{x}dx\le\sqrt{\dfrac 1a\times\int_0^\infty f^2(x)dx}<\infty$$ Now use the fact that absolutely convergence imply conditional convergence.