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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!

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  • $\begingroup$ If $f$ is 'vibrational' as you put it, you might try using Dirichlet's Test. $\int_0^{\infty} f^2(x)\,dx$ might help show $\int_a^{\infty} f(x)\,dx$ is bounded. $\endgroup$ – Integrand Jul 3 '20 at 2:27
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    $\begingroup$ Just consider Cauchy-Schwarz on $ \int_{a}^{A}\left|\frac{f(x)}{x}\right|dx$ instead. (At least, for $a > 0$. It's not true for $a=0$) $\endgroup$ – Brian Moehring Jul 3 '20 at 2:31
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    $\begingroup$ The above is the simplest and most direct way. You can also derive it from your approach by noting that it holds for arbitrary $a,A$ so taking these to be large the RHS goes to $0$ and it converges by Cauchy's criterion. $\endgroup$ – Winther Jul 3 '20 at 3:03
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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.

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