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.
 A: Let 
\begin{equation}
J=\int_{0}^{\infty}e^{-x^2-x^{-2}}dx\\
 =\frac{1}{e^2}\int_{0}^{\infty}e^{-(x-\frac{1}{x})^2}dx\\
\end{equation}
Let $x=\frac{1}{t}$, then we have
\begin{equation}
J=\frac{1}{e^2}\int_{0}^{\infty}e^{-(t-\frac{1}{t})^2}\frac{1}{t^2}dt\\
 =\frac{1}{e^2}\int_{0}^{\infty}e^{-(x-\frac{1}{x})^2}\frac{1}{x^2}dx
\end{equation}
So taking the average we have
\begin{equation}
J=\frac{1}{2e^2}\int_{0}^{\infty}e^{-(x-\frac{1}{x})^2}(1+\frac{1}{x^2})dx
\end{equation}
Letting $z=x-\frac{1}{x}$, we have
\begin{equation}
J=\frac{1}{2e^2}\int_{0}^{\infty}e^{-z}dz=\frac{\sqrt{\pi}}{2e^2}
\end{equation}
A: First recall that, for any 'nice' function, we have

$$
\int_{-\infty}^{+\infty}f\left(x-\frac{s}{x}\right)\mathrm{d}x=\int_{-\infty}^{+\infty} f(x)\: \mathrm{d}x, \quad s>0. \tag1
$$

Apply it to $f(x)=e^{-x^2}$, you get
$$
\int_{-\infty}^{+\infty}e^{-(x-s/x)^2}\mathrm{d}x=\int_{-\infty}^{+\infty} e^{-x^2} \mathrm{d}x=\sqrt{\pi}, \quad s>0. \tag2
$$
Thus
$$
\int_{-\infty}^{+\infty}e^{-x^2-s^2/x^{2}}\mathrm{d}x=\sqrt{\pi}\:e^{-2s}\tag3
$$ then put $s=1$ and use parity to obtain your integral.
A: 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}
