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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.

David
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rackne
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3 Answers3

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

Tyg13
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  • In the untransformed integral, integrand is positive on $[0,\infty).$ Yet your calculation gives the value as negative... – coffeemath May 08 '15 at 01:47
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    Fixed, screwed up the u-sub in the beginning so the sign was off. – Tyg13 May 08 '15 at 01:57
  • Where do we know that $ \int_{-\infty}^{\infty}e^{-u^2};du = \sqrt{\pi}$? – rackne May 08 '15 at 02:01
  • 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 08 '15 at 02:05
  • 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 08 '15 at 02:09
  • @rackne, once you perform the desired substitution, we can use regular methods to determine whether or not the transformed integral converges. If it converges, we know the original integral converges as well. Transforming the integral removes the discontinuity at 0, so we can use standard methods to check if it converges. See this – Tyg13 May 08 '15 at 02:11
  • @Tyg13 now that is nice solution to me thanks – rackne May 08 '15 at 02:17
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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.

Connor Harris
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  • 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 08 '15 at 02:04
  • 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 08 '15 at 02:07
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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.

zhw.
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