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Improper integral of unbounded function over bounded interval

Mathematics Asked by Wykk on December 15, 2020

I’m asked to prove one of the properties of improper integrals for unbounded functions over bounded integrals. The problem is stated at follows:


Let f,g:[0,a] $ to mathbb R$ be two functions s.t
$$0 le f(x) le g(x) forall xin[0,a]$$ and $$lim_{xto 0^+} f(x) = lim_{xto 0^+}
g(x) = infty $$

Then, if $int_0^a g(x)$ exists and $int_epsilon^a f(x)$ exists $forall epsilon > 0$

Prove that $int_0^a f(x)$ exists


Well my first doubt is about the functions I reckon they should be defined as functions over (0,a] and not [0,a], or they wouldn’t approach infinity as the x goes to zero, right? Anyway it could be just a mistake.

I was thinking of using $int_0^a g(x)$ as an upper bound for $int_0^a f(x)$ if it exists

Then $$int_0^a f(x) = int_0^epsilon f(x) + int_epsilon^a f(x)$$
then by taking the limit as $epsilon to 0$ the $ int_0^epsilon f(x) $ part goes to zero and we have that the other expression exists and is bounded above by $int_0^a g(x)$, however when I try to write it formally I am not sure of how to proceed, any help would be appreciated

One Answer

To show that $lim_{epsilon to 0+} int_{epsilon}^{a} f(x)dx$ exists it is enough to show that $lim_{epsilon, delta to 0+ } int_{epsilon}^{delta} f(x)dx=0$. To show this use the fact that $lim_{epsilon, delta to 0+ } int_{epsilon}^{delta} g(x)dx=0$ and use Sequeeze Theorem.

Answered by Kavi Rama Murthy on December 15, 2020

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