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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?

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  • $\begingroup$ 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

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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"?

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  • $\begingroup$ 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
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    $\begingroup$ @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

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