Prove $\int\limits_{0}^{\infty} \mathrm{exp}(-ax^{2}-\frac{b}{x^{2}}) \mathrm{d} x = \frac{1}{2}\sqrt{\frac{\pi}{a}}\mathrm{e}^{-2\sqrt{ab}}$
A slick way is to exploit Glasser's master theorem. For any $c>0$,
$$\begin{eqnarray*} e^{2\sqrt{c}}\int_{-\infty}^{+\infty}\exp\left(-x^2-\frac{c}{x^2}\right)\,dx&=&\int_{-\infty}^{+\infty}\exp\left[-\left(x-\frac{\sqrt{c}}{x}\right)^2\right]\,dx\\&=&\int_{-\infty}^{+\infty}e^{-x^2}\,dx = \sqrt{\pi}. \end{eqnarray*}$$
$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \color{#f00}{\int_{0}^{\infty}\exp\pars{-ax^{2} - {b \over x^{2}}}\,\dd x} = \int_{0}^{\infty}\exp\pars{-\root{ab}\bracks{\root{a \over b}x^{2} + \root{b \over a}{1 \over x^{2}}}}\,\dd x \end{align}
With $\ds{\pars{a \over b}^{1/4}\,\,x = \expo{\theta}}$: \begin{align} &\color{#f00}{\int_{0}^{\infty}\exp\pars{-ax^{2} - {b \over x^{2}}}\,\dd x} = \int_{-\infty}^{\infty}\exp\pars{-2\root{ab}\cosh\pars{2\theta}} \pars{b \over a}^{1/4}\expo{\theta}\,\dd\theta \\[4mm] = &\ \pars{b \over a}^{1/4} \int_{-\infty}^{\infty}\exp\pars{-2\root{ab}\cosh\pars{2\theta}} \bracks{\cosh\pars{\theta} + \sinh\pars{\theta}}\,\dd\theta \\[5mm] = &\ 2\pars{b \over a}^{1/4} \int_{0}^{\infty}\exp\pars{-2\root{ab}\bracks{2\sinh^{2}\pars{\theta} + 1}} \cosh\pars{\theta}\,\dd\theta \\[5mm] \stackrel{t\ \equiv\ \sinh\pars{\theta}}{=}\,\,\,\,\,\,&\ 2\pars{b \over a}^{1/4}\exp\pars{-2\root{ab}} \int_{0}^{\infty}\exp\pars{-4\root{ab}t^{2}}\,\dd t \\[5mm] = &\ 2\pars{b \over a}^{1/4}\exp\pars{-2\root{ab}} \bracks{{1 \over \root{4\root{ab}}}\,{\root{\pi} \over 2}} = \color{#f00}{{\root{\pi} \over 2}\,{\expo{-2\root{ab}} \over \root{a}}} \end{align}