Determine if
$$\int_{0}^{\infty}\frac{e^{-1/x^2}}{x^2}dx$$
is convergent or not.
Since the function is discontinous at $x=0$, I cannot apply comparison theorems for improper integrals. I have tried taking the integral to evaluate the limits but I couldn't do it because there is a problem with exponential of $e$. Thanks for your help.
On
As Lucian hints in commments, with the change of variables $$t = \frac{1}{x}, dt = -\frac{dx}{x^2}$$ this integral becomes $$\int_0^\infty e^{-t^2}\, dt$$ which does converge, and in fact equals $\frac{\sqrt{\pi}}{2}$. There is a standard trick to evaluating integrals of this sort: for the more usual example $\int_{-\infty}^\infty e^{-x^2}\, dx$, write $$\left( \int e^{-x^2}\, dx\right)^2 = \int_{-\infty}^\infty e^{-x^2}\, dx \int_{-\infty}^\infty e^{-y^2}\, dy = \iint_{\mathbf{R}^2} e^{-x^2 - y^2}\, dx\, dy$$ and then change to polar coordinates: $$\iint_{\mathbf{R}^2} e^{-x^2 - y^2}\, dx\, dy = \int_0^{2\pi} \int_0^\infty e^{-r^2} (r\, dr)\, d\theta$$ and this last integral can be done easily.
On
Briefly: For convergence we only need to check that $\int_0^1 \frac{e^{-1/x^2}}{x^2}\, dx$ and $\int_1^\infty \frac{e^{-1/x^2}}{x^2}\, dx$ both converge. In the first integral, the integrand $\to 0$ at $0.$ So the integrand is bounded on $(0,1]$ and there is no trouble at all. For the second integral, the integrand is positive and bounded above by $\frac{1}{x^2}.$ Since $\int_1^\infty \frac{1}{x^2}\, dx <\infty,$ we're OK here as well. It foll0ws that $\int_0^\infty \frac{e^{-1/x^2}}{x^2}\, dx$ converges.
Perform the substitution $u = \frac{1}{x}$ on $\int_{0}^{\infty}\dfrac{e^{-\frac{1}{x^2}}}{x^2}\;dx$. So $-du = \frac{1}{x^2}dx$ and $-\int_{\infty}^{0}e^{-u^2}\;du$. We recognize this is just $\int_{0}^{\infty}e^{-u^2}\;du$. We know that $\int_{-\infty}^{\infty}e^{-u^2}\;du = \sqrt{\pi}$ and by the symmetry of $e^{-u^2}$ that the area under x-axis is the same on either side, thus $\int_{0}^{\infty}e^{-u^2}\;du = \dfrac{\sqrt{\pi}}{2}$.