# Improper Riemann Integral

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?

• 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. Oct 13, 2019 at 14:48
• @Arthur so when we include continuity and differentiability, will it be correct? Oct 13, 2019 at 14:58
• 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. Oct 13, 2019 at 15:00

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"?
• @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? Oct 13, 2019 at 18:03
• @PeterF. The finiteness of the limit is correct, but how do you define the integrability on $(a,b]$? Oct 13, 2019 at 18:11
• @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 $c\in (a,b)$. Oct 13, 2019 at 18:22