I am interested in the improper integral: $$I=\int_{-\infty}^\infty 1-e^{-\frac{1}{x^2}}{\rm d}x=2\int_{0}^\infty 1-e^{-\frac{1}{x^2}}{\rm d}x$$ which I am fairly sure converges.

I broke the integral into one over $1$ and one over the exponential and then tried to evaluate this through a change to polar coordinates similar to the way to evaluate the Gaussian Integral.

However, when I attempt to introduce the variable change $\frac{1}{x}=u$ to apply the polar coordinates, I am left with the bounds being $\lim_{\epsilon\to0}[\epsilon,-\epsilon]$. I am not sure of how to convert these bounds into polar form in terms of $r$ and $\theta$. If anyone can give me a hint of where to continue, what I am doing wrong, or if there is a better way of evaluating this integral it would be greatly appreciated.

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    $\begingroup$ @Simply Beautiful Art Are you sure? $1-e^{-0}=0$. $\endgroup$ – Professor Vector Dec 14 '17 at 15:34
  • $\begingroup$ @ProfessorVector Oops, my bad. $\endgroup$ – Simply Beautiful Art Dec 14 '17 at 15:35
  • $\begingroup$ $I=2\sqrt{\pi}$, I think, but I'll have to write that up later. $\endgroup$ – Professor Vector Dec 14 '17 at 15:37
  • $\begingroup$ @ProfessorVector That makes sense, I expect the answer to be of the form $\sqrt{\pi}$ due to the similarity to the Gaussian Integral. $\endgroup$ – aleden Dec 14 '17 at 15:39

As I said I would, I'll add my two cents (it's more or less self-explanatory, I hope): \begin{align}\int^\infty_0(1-e^{-1/x^2})\,dx&=\int^\infty_0\frac{1-e^{-x^2}}{x^2}\,dx \\&=\int^\infty_0\int^1_0e^{-ax^2}\,da\,dx \\&=\int^1_0\int^\infty_0e^{-ax^2}\,dx\,da \\&=\int^1_0\frac{\sqrt{\pi}}{2\sqrt{a}}\,da \\&=\sqrt{\pi} \end{align}


Notice that

$$\int e^{-\frac{1}{x^{2}}}{\rm d}x=\int 1\cdot e^{-\frac{1}{x^{2}}}{\rm d}x=$$

$$=xe^{-\frac{1}{x^{2}}}-\int \frac{2}{x^{2}}e^{-\frac{1}{x^{2}}}{\rm d}x$$

using integration by parts. For the last integral, define $u=\frac{1}{x}$ and ${\rm d}u=-\frac{1}{x^{2}}{\rm d}x$ to get

$$-\int \frac{2}{x^{2}}e^{-\frac{1}{x^{2}}}{\rm d}x=2\int e^{-u^{2}}{\rm d}u=\sqrt{\pi}{\rm erf}\left(u\right)=\sqrt{\pi}{\rm erf}\left(\frac{1}{x}\right)$$

Can you take it from here?

  • $\begingroup$ Thank you, I will get back to you after I evaluate. $\endgroup$ – aleden Dec 14 '17 at 15:57

With Laplace transform properties, let $\dfrac{1}{x^2}=u$ so you have \begin{align} J &= \int_{0}^\infty 1-e^{-\frac{1}{x^2}}{\rm d}x \\ &= \dfrac12\int_0^\infty\dfrac{1-e^{-u}}{u\sqrt{u}}du \\ &= \dfrac12\int_0^\infty\dfrac{\Gamma\left(\dfrac12\right)}{s^\frac12}-\dfrac{\Gamma\left(\dfrac12\right)}{(s+1)^\frac12}ds \\ &= \dfrac12\Gamma\left(\dfrac12\right)\left(2\sqrt{s}-2\sqrt{s+1}\right)_{0}^\infty \\ &= \color{blue}{\sqrt{\pi}} \end{align}

  • $\begingroup$ I think the area should be positive, right? $\endgroup$ – aleden Dec 14 '17 at 15:58
  • $\begingroup$ You should consider explaining the equality between Lines 2 and 3. $\endgroup$ – Mark Viola Dec 14 '17 at 16:00
  • $\begingroup$ @MyGlasses You should have $$\Big[2\sqrt{s}-2\sqrt{s+1}\Big]_{0}^{\infty}=0-(-2)=2$$ It explain your wrong minus sign. $\endgroup$ – eranreches Dec 14 '17 at 16:01
  • $\begingroup$ @eranreches Thanks. I see your comment now! $\endgroup$ – Nosrati Dec 14 '17 at 16:06

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