Suppose $f$ is positive and increasing on $[0, +\infty)$. Suppose that $F(x)=$$\int_0^x f(t)dt$. Prove that $$\int_0^{+\infty} \frac {1}{f(x)} dx$$ converges if and only if $$\int_0^{+\infty} \frac {x}{F(x)} dx$$ converges.

My thoughts, and attempt at solution -

Since $f$ is increasing, $F(x)=\int_0^xf(t)\,dt\le\int_0^xf(x)\,dt=xf(x).$ We can rearrange that to $\frac1{f(x)}\le\frac{x}{F(x)}.$ Hence if the integral of $\frac{x}{F(x)}$ converges, then so does the integral of $\frac1{f(x)}.$

However, the above working seems incomplete as it bounds the integral from the upper side only. In fact, it shows that if the second integral converges, then the first one may converge, but not vice versa. How do I place a lower bound on the integral, and go about the rest of the solution?

Haven't been able to proceed from here, please help.


For the lower bound, for $x>0$, we have that $f(x)>0$, $F(x)>0$ and, since $f$ is increasing, it follows that $$xf(x)=(2x-x)f(x)\leq \int_x^{2x}f(t)dt=F(2x)-F(x)\leq F(2x)\implies \frac{x}{F(2x)}\leq \frac{1}{f(x)}.$$ Hence $$\frac{1}{4}\int_{0}^{\infty}\frac{t}{F(t)}dt=\int_{0}^{\infty}\frac{x}{F(2x)}dx\leq \int_{0}^{\infty}\frac{1}{f(x)}dx.$$

P.S. Note that $F(0)=0$ and by L'Hopital $$\lim_{x\to 0^+} \frac{x}{F(x)}=\lim_{x\to 0^+} \frac{1}{f(x)}=\frac{1}{f(0)}$$ which implies that $x/F(x)$ is integrable in a right neighbourhood of $0$.


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