# Showing the limit is $+\infty$

Let $f(x)$ be continuous and positive on $[0,+\infty)$. Suppose $$\int_0^{+\infty}\frac{1}{f(x)}dx<+\infty.$$ How can one show that $$\lim_{s\to +\infty}\frac{\int_0^{s}f(x)dx}{s^2}=+\infty?$$

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Hint: If $g(x) > 0$ everywhere and $\int_0^\infty g(x)\ dx$ is finite, then $g(x) = O(x^n)$ for what $n$? –  Rahul Feb 4 '12 at 16:57

For $t>0$, we have $$t=\int_t^{2t}\frac 1{\sqrt{f(\xi)}}\sqrt{f(\xi)}d\xi\leq \left(\int_t^{2t}\frac 1{f(\xi)}d\xi\right)^{1/2}\left(\int_t^{2t}f(\xi)d\xi\right)^{1/2}$$ so $$s^2\leq \int_s^{2s}\frac{dx}{f(x)}\cdot \int_s^{2s}f(\xi)d\xi$$ and so $$\frac{\int_0^{2s}f(\xi)d\xi}{s^2}\geq\frac{\int_s^{2s}f(\xi)d\xi}{s^2}\geq \frac 1{\int_s^{2s}\frac{d\xi}{f(\xi)}},$$ and we can conclude, since the fact that $\int_0^{+\infty}\frac{d\xi}{f(\xi)}$ converges implies that $\lim_{s\to \infty}\int_s^{2s}\frac{d\xi}{f(\xi)}=0$.