If a function $f$ is bounded such that $|f|<M$, then can we say: $$ \left|\int_a^ \infty f(x)\, \mathrm{d}x \right| \le \int_a^ \infty |f(x)| \,\mathrm{d}x \le \int_a^ \infty M\, \mathrm{d}x$$ Is that correct?

Thank you!

  • $\begingroup$ But the last expression is just $\infty$ and the inequity does not provide useful information, I think? $\endgroup$
    – High GPA
    Jun 19, 2022 at 18:52

1 Answer 1


It is correct.

Since $ -|f| \leq f \leq |f|$: $$ -\int_a^\infty |f(x)|\,dx\leq \int_a^\infty f(x) \,dx \leq \int_a^\infty |f(x)| \,dx \\ \therefore \left| \int_a^\infty f(x) \,dx \right| \leq \int_a^\infty|f(x)| \,dx $$ The right one easily follows from the precondition.

$$ |f(x)| < M \Rightarrow |f(x)| \leq M \\ \int_a^\infty |f(x)|\, dx \leq \int_a^\infty M\,dx $$

Combine these two to get your desired inequality.

  • 1
    $\begingroup$ Suppose that we have a function $f=2$ which is surely bounded with a boundary $M\ge 2$, now we integrate $f$ over the interval $[a,\infty)$, which gives us infinity, i.e., the integral is not bounded. The problem in your proof is that the interval of integration is infinite. $\endgroup$
    – user142288
    Dec 9, 2021 at 3:00

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .