Suppose $f : [1, \infty) \to \mathbb{R}$ is such that $g(x) : = x^2f(x)$ is a bounded function. Prove that $\int_1^\infty f$ converges.

Question

Suppose $f : [1, \infty) \to \mathbb{R}$ is such that $g(x) : = x^2f(x)$ is a bounded function. Prove that $\int_1^\infty f$ converges.

I think that $f(x)$ should be $1/x^p$ for $p >2$ otherwise $\lim_{x \to \infty}g(x)$ diverges. In this case, the integral of $f$ converges by p-test for integrals. But, I do not know how to formally prove the question. I appreciate if you give some help.

Answer 1

Here’s the first hint: If $g$ is bounded, then there exists a constant $0 \leq M < \infty$ such that $|g(x)| \leq M$ for all $x$. This means $$|f(x)| \leq M / x^2 ,$$ and $\int_1^\infty 1/x^2 \mathrm{d} x$ converges.