# Improper integral with discontinuity

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.

• Discontinuous at $0$? Why do you say that? – zhw. May 8 '15 at 1:00
• @zhw. because the denominator is $x^2?$ – rackne May 8 '15 at 1:03
• Isn't it possible that the integrand has a nice finite limit at $0$? – zhw. May 8 '15 at 1:16
• Yes it can but I still don't get it. – rackne May 8 '15 at 1:23
• Hint: Let $t=\dfrac1x$ – Lucian May 8 '15 at 1:32

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 $\int_{0}^{\infty}e^{-u^2}\;du = \dfrac{\sqrt{\pi}}{2}$.

• In the untransformed integral, integrand is positive on $[0,\infty).$ Yet your calculation gives the value as negative... – coffeemath May 8 '15 at 1:47
• Fixed, screwed up the u-sub in the beginning so the sign was off. – Tyg13 May 8 '15 at 1:57
• Where do we know that $\int_{-\infty}^{\infty}e^{-u^2}\;du = \sqrt{\pi}$? – rackne May 8 '15 at 2:01
• Connor Harris in his answer gives a basic idea on how to find the integral. The trick is to square the result and then interpret it at as a double integral. You can then re-express the region in polar coordinates and then evaluate that integral, which is very simple, and then take the square root of the result. The answer is $\sqrt{\pi}$ and is hard to evaluate without a knowledge of multivariable calculus. – Tyg13 May 8 '15 at 2:05
• Thanks @Tyg13 but the question is asked in a single variable course they wouldn't expect us to find the answer in this way. – rackne May 8 '15 at 2:09

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.

• We haven't learnt double integrals yet. I don't understand after $\int_0^\infty e^{-t^2}\, dt$. There must be an easy way – rackne May 8 '15 at 2:04
• Unfortunately $e^{-t^2}$ has no antiderivative that is expressible with elementary functions, so we cannot find a general expression for the integral so we can evaluate it. – Tyg13 May 8 '15 at 2:07

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.