calculus - Evaluating $int_{-infty}^infty 1-e^{-frac{1}{x^2}}{rm d}x$

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.

Answer

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}