Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $$\lim_{s\to +\infty}\frac{\int_0^{s}f(x)dx}{s^2}=+\infty?$$

share|cite|improve this question
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
up vote 6 down vote accepted

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 $\lim_{s\to \infty}\int_s^{2s}\frac{d\xi}{f(\xi)}=0$.

share|cite|improve this answer
Very slick, +1. – David Mitra Feb 4 '12 at 17:30

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.