calculus - Evaluate $int_{0}^{infty} mathrm{e}^{-x^2-x^{-2}}, dx$

I have to find
$$I=\int_{0}^{\infty} \mathrm{e}^{-x^2-x^{-2}}\, dx $$
I think we could use

$$\int_{0}^{\infty} \mathrm{e}^{-x^2}\, dx = \frac{\sqrt \pi}{2} $$ But I don't know how.
Thanks.

Answer

Consider
\begin{align}
x^{2} + \frac{1}{x^{2}} = \left( x - \frac{1}{x} \right)^{2} +2
\end{align}
for which
\begin{align}
I = \int_{0}^{\infty} e^{-\left(x^{2} + \frac{1}{x^{2}}\right)} \, dx = e^{-2} \, \int_{0}^{\infty} e^{-\left(x - \frac{1}{x}\right)^{2}} \, dx.

\end{align}
Now make the substitution $t = x^{-1}$ to obtain
\begin{align}
e^{2} I = \int_{0}^{\infty} e^{- \left( t - \frac{1}{t} \right)^{2}} \, \frac{dt}{t^{2}}.
\end{align}
Adding the two integral form leads to
\begin{align}
2 e^{2} I = \int_{0}^{\infty} e^{- \left( t - \frac{1}{t} \right)^{2}} \left(1 + \frac{1}{t^{2}} \right) \, dt = \int_{-\infty}^{\infty} e^{- u^{2}} \, du = 2 \int_{0}^{\infty} e^{- u^{2}} \, du = \sqrt{\pi},
\end{align}
where the substitution $u = t - \frac{1}{t}$ was made. It is now seen that

\begin{align}
\int_{0}^{\infty} e^{-\left(x^{2} + \frac{1}{x^{2}}\right)} \, dx = \frac{\sqrt{\pi}}{2 e^{2}}.
\end{align}