How to integrate $\int^{\infty}_{-\infty} e^{-2\pi^2/x^2} dx$?

Question

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.

Answer 1

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