How to integrate $\displaystyle 1-e^{-1/x^2}$? 
How to integrate $\displaystyle 1-e^{-1/x^2}$ ?
as hint is given: $\displaystyle\int_{\mathbb R}e^{-x^2/2}=\sqrt{2\pi}$

If i substitute $u=\dfrac{1}{x}$, it doesn't bring anything:
$\,\displaystyle\int\limits_{-\infty}^{\infty}\left(1-e^{-1/x^2}\right)dx=\int\limits_{-\infty}^{0}\left(1-e^{-1/x^2}\right)dx+\int\limits_{0}^{\infty}\left(1-e^{-1/x^2}\right)dx\overset{?}=2\int\limits_{0}^{\infty}\left(1-\frac{e^{-u^2}}{-u^2}\right)du$
$2\displaystyle\int\limits_{0}^{\infty}\left(1-\frac{e^{-u^2}}{-u^2}\right)du=?$
How to continue ?
$\textbf{The original exercise was}$:
If a probability has a density $f(x)=C(1-e^{-1/x^2})$ then determine the value of constant $C$
Since $\displaystyle\int f\overset{!}=1$, i thought first to calculate the expression above.
($\textbf{ATTENTION:}$ Question edited from integrating $e^{-1/x^2}$ to integrating
$1-e^{-1/x^2}$)

$$\int_0^\infty\bigg(1-e^{^{-\tfrac1{x^2}}}\bigg)dx=\sqrt\pi.$$ This can relatively easily be proven by letting $t=e^{^{-\tfrac1{x^2}}}$, and then using Euler's initial logarithmic expression for the $\Gamma$ function. But you said that this is apparently not allowed and I'm afraid that I don't know any other approaches either. :-(
