Can somebody help me solve the proof about improper integrals?

Question

If $f(x) > 0$ is continuous at $[0, +\infty]$ and $\displaystyle \int_0^{+\infty} \frac{1}{f(x)} dx$ is convergent, please prove $\displaystyle \lim_{\lambda \to \infty} \frac{1}{\lambda} \int_0^\lambda f(x) dx = +\infty$.

However, I think if we can prove $\displaystyle \lim_{x \to \infty} f(x) = \infty$, and then we can prove the question. But I don't know how to prove it.

Can somebody help me solve it? Or can somebody give me some hints? thank you!

Answer 1

Hint: Use the Cauchy-Schwarz inequality.

details:

indeed,

$$ \sqrt{T} = \int_0^T \sqrt{\frac{f(x)}T}\frac 1{\sqrt{f(x)}} dx \le \left( \frac 1T\int_0^T f(x) dx\right)^{1/2} \left(\int_0^T \frac 1{f(x)} dx \right)^{1/2}\\ \le M^{1/2}\left(\frac 1T \int_0^T \frac 1{f(x)} dx \right)^{1/2} $$

Answer 2

Assume f is bounded above by M ==> f(x) < M, then 1/f(x) > 1/M, then Int(0 to n)1/f(x)dx > n/M. Taking limit as n --> infinity we get Int(0 to infinity)1/f(x)dx = infinity- contradiction. So f is unbounded, and we can take it from here.