How to prove that $\int\limits_0^{+\infty}(e^{-1/x^2} - e^{-4/x^2})\,dx$ converges?
I know Abel's and Dirihle's rules, but here it doesn't help. Also, i tried to use equivalence, and use only one integral.
Maybe we can say that $ e^{\frac{-1}{x^2}} - e^{\frac{-4}{x^2}} \leq $some function? How to prove integrals like this?
It is straightforward to show that the integral converges (the integrand is $\displaystyle O(x^{-2})$ as $x \displaystyle \to \infty$). In fact, we can evaluate this integral in closed form. To this latter end we proceed.
Enforcing the substitution $x\to 1/x$ reveals
$$\int_0^L \left(e^{-1/x^2}-e^{-4/x^2}\right)\,dx=\int_{1/L}^\infty \frac{e^{-x^2}-e^{-4x^2}}{x^2}\,dx \tag 1$$
Integrating by parts the integral on the right-hand side of $(1)$ with $u=e^{-x^2}-e^{-4x^2}$ and $v=-\frac{1}{x}$ yields
$$\int_0^L \left(e^{-1/x^2}-e^{-4/x^2}\right)\,dx=\left.\left(\frac{e^{-x^2}-e^{-4x^2}}{x}\right)\right|_{1/L}^\infty-2\int_{1/L}^\infty \left(e^{-x^2}-4e^{-4x^2}\right)\,dx\tag 2$$
Letting $L\to \infty$ we find that the first term on the right-hand side of $(2)$ vanishes and we obtain
$$\int_0^L \left(e^{-1/x^2}-e^{-4/x^2}\right)\,dx=-2\int_{0}^\infty \left(e^{-x^2}-4e^{-4x^2}\right)\,dx=\sqrt \pi$$
Therefore, the integral not only converges, but is equal to $\sqrt \pi$.
