Improper integral convergence with parameter

Question

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.

Answer 1

hint

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, $$\color{red}{1}-\epsilon<x^2f(x)<\color{red}{1}+\epsilon=\frac 32$$

or

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

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