Suppose $f$ is positive and increasing on $[0, +\infty)$. Let $F(x) = \displaystyle\int_{0}^{x}f(t)dt$.
How do we show that $\displaystyle \int_{0}^{+\infty} \frac{dx}{f(x)}$ converges if and only if $\displaystyle\int_{0}^{+\infty} \frac{x}{F(x)}dx$ converges?
I was thinking of using the comparison test, but I can't really find a way to connect the two integrals.
This question has been asked yesterday, but I could not find the related post.
We see that $xf(x)\leq\displaystyle\int_{x}^{2x}f(t)dt\leq F(2x)$, so $\displaystyle\int_{0}^{\infty}\dfrac{1}{f(x)}dx\geq\dfrac{1}{2}\int_{0}^{\infty}\dfrac{2x}{F(2x)}dx=\dfrac{1}{4}\int_{0}^{\infty}\dfrac{x}{F(x)}dx$.
On the other hand, $F(x)\leq xf(x)$, so $\displaystyle\int_{0}^{\infty}\dfrac{x}{F(x)}dx\geq\int_{0}^{\infty}\dfrac{1}{f(x)}dx$.