How to integrate $\int^{\infty}_{-\infty} e^{-2\pi^2/x^2} dx$? I am wondering how can i integrate this quantity above?
Here it is again,
$$\int^{\infty}_{-\infty} e^{-2\pi^2/x^2}dx.$$
Thanks a lot.

Since the function is even then
\begin{align}
 I = \int_{ - \infty }^\infty  {\exp \left( { - \frac{{2\pi ^2 }}{{x^2 }}} \right)dx}  = 2\int_0^\infty  {\exp \left( { - \frac{{2\pi ^2 }}{{x^2 }}} \right)dx}  
\end{align}
Setting
 \begin{align}
 x = \frac{{\sqrt 2 }}{{\sqrt u }}\pi ,\,\,\,\text{i.e.,}\,\,\,u = \frac{{2\pi ^2 }}{{x^2 }} \Rightarrow du =  - \frac{{4\pi ^2 }}{{x^3 }}dx \Rightarrow dx &=  - \frac{1}{{4\pi ^2 }}\left( {\frac{{\sqrt 8 }}{{\sqrt {u^3 } }}\pi ^3 } \right)du \\ 
 dx &=  - \frac{\pi }{{\sqrt 2 \sqrt {u^3 } }}du,
\end{align}
$x = 0,u \to \infty ;x = \infty ,u = 0$ 
\begin{align} 
 I = 2\int_\infty ^0 {\exp \left( { - u} \right)\frac{{ - \pi }}{{\sqrt 2 \sqrt {u^3 } }}du}  = \pi \sqrt 2 \int_0^\infty  {u^{-3/2} \exp \left( { - u} \right)du}   
\end{align}
But since $ \exp \left( { - u} \right) \ge 1-u$ then $ \int_0^\infty  {u^{-3/2} \exp \left( { - u} \right)du} \ge \int_0^\infty  {(1-u)u^{-3/2} du} $ and this is  diverge
