I am supposed to solve the following question:

For what function classes it makes sense to talk about the Improper Riemann Integral?

I know that we can talk now about bounded functions defined on unbounded interval or unbounded functions defined on bounded interval.

But is there any more specific answer?

  • $\begingroup$ Bounded is not enough. The function $f(x)=1$ is bounded, but the improper Riemann integral over the entire number line doesn't exist. Also, continuity must be discussed. $\endgroup$
    – Arthur
    Oct 13, 2019 at 14:48
  • $\begingroup$ @Arthur so when we include continuity and differentiability, will it be correct? $\endgroup$
    – Peter F.
    Oct 13, 2019 at 14:58
  • $\begingroup$ As I said, $f(x)=1$ is continuous, differentiable and bounded, but the improper integral doesn't exist. On the other hand, there are discontinuous functions with existing improper integrals. $\endgroup$
    – Arthur
    Oct 13, 2019 at 15:00

1 Answer 1


For the first case, you can assume that the interval is in the form $[a,+\infty)$ for some $a$ (the $(-\infty,a]$ case us similar). Then the integral of $f$ on $[a,+\infty)$ is defined as $$\int_a^{+\infty} f := \lim_{b \to +\infty } \int_a^b f$$ And if $f$ is "unbounded at $a$", then the integral on $(a,b]$ is defined as $$\int_a^b f = \lim_{x \to a+0} \int_x^b f$$ What do we need for these definitions to "make sense"?

  • $\begingroup$ We need to have continous functions? just that? $\endgroup$
    – Peter F.
    Oct 13, 2019 at 17:57
  • $\begingroup$ @PeterF. No, we don't need continuity. For the first case, we need that $\forall b >a$, $f$ is integrable on $[a,b]$, because we could not take the limit otherwise. And we also need the limit to be finite. What do you think about the $(a,b]$ case? $\endgroup$
    – Botond
    Oct 13, 2019 at 18:03
  • $\begingroup$ We do not need limit to be finite, but still we need integrability on interval (a,b] ? $\endgroup$
    – Peter F.
    Oct 13, 2019 at 18:08
  • $\begingroup$ @PeterF. The finiteness of the limit is correct, but how do you define the integrability on $(a,b]$? $\endgroup$
    – Botond
    Oct 13, 2019 at 18:11
  • 1
    $\begingroup$ @PeterF. In the case of $[a, +\infty)$, we wanted integrability on $[a,b]$ for all $b>a$. In the case of $(a,b]$, we need something similar: Integrability on $[c,b]$ for all $c\in (a,b)$. $\endgroup$
    – Botond
    Oct 13, 2019 at 18:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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