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