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

Intuitively I think it seems reasonable. As $x^2$ is increasing and unbounded above, $f(x)$ must be a decreasing function with higher absolute slope than $x^2$ so as $g(x)$ can be bounded. Not sure how to proceed further from here.


1 Answer 1


Let $|g(x)|\le M$ (assumption it is bounded). Then $|f(x)|\le \frac{M}{x^2}$ so $\int_1^\infty|f(x)|dx\le M\int_1^\infty\frac{dx}{x^2}=M$. Therefore $f(x)$ is integrable.

  • 5
    $\begingroup$ +1, but just to make it explicit: the OP's intuitive reasoning about "increasing and unbounded above" is not correct, because the claim becomes false if we replace $g(x) := x^2 f(x)$ with $g(x) := x f(x)$ (even though $x$ is "increasing and unbounded above" just as $x^2$ is). $\endgroup$
    – ruakh
    Feb 4, 2019 at 2:29

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