Prove $\int\limits_{0}^{\infty} \mathrm{exp}(-ax^{2}-\frac{b}{x^{2}}) \mathrm{d} x = \frac{1}{2}\sqrt{\frac{\pi}{a}}\mathrm{e}^{-2\sqrt{ab}}$ This is Example 8.4.1 from Chapter 8 (The Normal Integral) of Irresistible Integrals by Boros and Moll. The authors outlined a solution method, which I will provide in full here. My question: is there another way to obtain this result?
Here is the solution of Boros and Moll with their notation.
Let
\begin{equation}
\mathrm{L}(a,b) := \int\limits_{0}^{\infty} \mathrm{exp}(-ax^{2}-\frac{b}{x^{2}}) \mathrm{d} x
\label{eq:iic8-1}
\tag{1}
\end{equation}
Making the substitution $t=x\sqrt{a}$ yields
\begin{equation}
\int\limits_{0}^{\infty} \mathrm{exp}(-ax^{2}-\frac{b}{x^{2}}) \mathrm{d} x = \frac{1}{\sqrt{a}} \int\limits_{0}^{\infty} 
\mathrm{exp}(-t^{2}-\frac{ab}{t^{2}}) \mathrm{d} t
\label{eq:iic8-2}
\tag{2}
\end{equation}
Letting $ab=c$ we call the integral in equation \eqref{eq:iic8-2} $f(c)$,
\begin{equation}
f(c) = \int\limits_{0}^{\infty} \mathrm{exp}(-t^{2}-\frac{c}{t^{2}}) \mathrm{d} t
\label{eq:iic8-3}
\tag{3}
\end{equation}
so that 
\begin{equation}
\mathrm{L}(a,b) = \frac{f(ab)}{\sqrt{a}}
\label{eq:iic8-4}
\tag{4}
\end{equation}
In equation \eqref{eq:iic8-3} we let $y=\sqrt{c}/t$
\begin{equation}
f(c) = \sqrt{c} \int\limits_{0}^{\infty} \mathrm{exp}(-y^{2}-\frac{c}{y^{2}}) y^{-2} \mathrm{d} y
\label{eq:iic8-5}
\tag{5}
\end{equation}
Combining equations \eqref{eq:iic8-3} and \eqref{eq:iic8-5}, we have
\begin{equation}
f(c) = \frac{1}{2} \int\limits_{0}^{\infty} \mathrm{exp}(-t^{2}-\frac{c}{t^{2}}) \left(1+\frac{\sqrt{c}}{t^{2}} \right) \mathrm{d} t
\label{eq:iic8-6}
\tag{6}
\end{equation}
Now we let $s = t - \sqrt{c}/t$
\begin{equation}
f(c) = \frac{\mathrm{e}^{- 2\sqrt{c}}}{2} \int\limits_{-\infty}^{\infty} \mathrm{exp}(-s^{2}) \mathrm{d} s = \frac{\sqrt{\pi}}{2} \mathrm{e}^{- 2\sqrt{c}} \lim_{z \to \infty} \mathrm{erf}(z) = \frac{\sqrt{\pi}}{2} \mathrm{e}^{- 2\sqrt{c}}
\label{eq:iic8-7}
\tag{7}
\end{equation}
Combining equations \eqref{eq:iic8-1}, \eqref{eq:iic8-4}, and \eqref{eq:iic8-7} yields our result.
 A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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\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}
A: A slick way is to exploit Glasser's master theorem. For any $c>0$,
$$\begin{eqnarray*} e^{2\sqrt{c}}\int_{-\infty}^{+\infty}\exp\left(-x^2-\frac{c}{x^2}\right)\,dx&=&\int_{-\infty}^{+\infty}\exp\left[-\left(x-\frac{\sqrt{c}}{x}\right)^2\right]\,dx\\&=&\int_{-\infty}^{+\infty}e^{-x^2}\,dx = \sqrt{\pi}. \end{eqnarray*}$$
A: Some of the steps in @FelixMarin's answer were not obvious. Here I expand upon his answer so that it will be easier to follow.


*

*$\mathrm{e}^{z} = \cosh(z) + \sinh(z)$

*$$\int\limits_{-\infty}^{\infty} \cosh(z) + \sinh(z) \mathrm{d} z = \int\limits_{-\infty}^{0} \cosh(z) + \sinh(z) \mathrm{d} z 
+ \int\limits_{0}^{\infty} \cosh(z) + \sinh(z) \mathrm{d} z$$
In the first integral, let $z=-y$,
$$\int\limits_{-\infty}^{0} \cosh(z) + \sinh(z) \mathrm{d} z = \int\limits_{0}^{\infty} \cosh(y) - \sinh(y) \mathrm{d} y$$
Addition yields,
$$\int\limits_{-\infty}^{\infty} \cosh(z) + \sinh(z) \mathrm{d} z = 2\int\limits_{0}^{\infty} \cosh(z) \mathrm{d} z$$

