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If an integral isn't improper, and there is a domain given, is there a comparison test (like the basic one used for improper integrals) that can show whether the integral is convergent or divergent? Or are all proper integrals convergent?

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  • $\begingroup$ What is a proper integral? $\endgroup$
    – Compacto
    Feb 10, 2022 at 1:54
  • $\begingroup$ @Compacto I'm not entirely sure if it's correct but I named it proper integral just because it wasn't an improper integral. I may be wrong on my definition of improper integral but isn't that just an integral with singularities? So the type of integral I'm talking about is one without singularities on its interval. $\endgroup$
    – gino
    Feb 10, 2022 at 2:19

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What do you mean by singularity? As far as I know, for continuous functions, "regular" (not improper) Riemann integrals are always defined and finite whenever the domain is sufficiently nice (for example, bounded domains whose boundary has measure zero).

When the function is bounded but not continuous, it may not be "integrable" in the sense that its Riemann sums don't approach anything, but this is a subtle difference between convergence of Riemann sums (which are finite sums taken over all partitions of the domain) and convergence of improper integrals (sums which can be infinite either because the function goes to infinity somewhere or because the interval is not bounded).

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