Hello I need to examine the convergence of this followintg improper integral:

$\int_{0}^{\infty} \frac{1-e^{-a^{2} x^{2}}}{x^{2}}$

If I understand it correctly, I have to examine it in points where it is not continuous which is 0 and infinity. In infinity I can compare it with: $\frac{1}{x^{2}}$ and that is convergent for all values of parameter "a".

In my materials it is written that it has $\lim _{x \rightarrow 0}$ = -a^2 but I do not understand why this is sufficient? It would help me greatly if someone could explain to me what are other options for examining covergence other than comparisons.

Thank you very much.


1 Answer 1



Let $$f(x)=\frac{1-e^{-a^2x^2}}{x^2}$$

$ f $ is continuous and locally integrable at $ (0,+\infty)$.

near 0

$$\lim_{x\to 0^+}f(x)=a^2 \implies$$ $$\int_0^1f(x)dx \;\;converges$$ near infinity

$$\lim_{x\to+\infty} x^2f(x)=\color{red}{1}\implies$$

with $ \epsilon=\frac 12 $ and for $ x $ great enough, $$\color{red}{1}-\epsilon<x^2f(x)<\color{red}{1}+\epsilon=\frac 32$$


$$0<f(x)<\frac{3}{2x^2}$$ thus, $$\int_1^{+\infty}f(x)dx\;\; converges$$

we conclude that $$\int_0^{+\infty}f(x)dx\;\;\text{ is convergent}$$

  • $\begingroup$ I really thank you for your solution. The problem is that I do not know the individual steps leading to it. Giving me a hint on how to get to that solution would be a great help to me. Or more precisely, I am not sure how to find the function used in the comparison test. Also I somehow understand that part with limit near infinity, but what I do not understand why it is convergent when near 0. Thank you thousand times. $\endgroup$
    – FilipM
    Jan 24, 2021 at 18:08
  • $\begingroup$ @FilipM It is known that if $ a$ 8s a real and $ \lim_{a^+}f(x)\in\Bbb R$ then $\int_a f $ is convergent. $\endgroup$ Jan 24, 2021 at 19:07

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