TITLE: Evaluate $\int_{0}^{\infty} \mathrm{e}^{-x^2-x^{-2}}\, dx$
QUESTION [7 upvotes]: 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.

REPLY [5 votes]: 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}