How to compute $\int_0^\infty e^{-a(s^2+1/s^2)}\, ds$ How do I integrate
$$\int_0^\infty e^{-a(\frac{1}{s^2} + s^2)}\, ds \tag{*}$$
Context: At page 602 of the paper "Reaction-Diffusion equations for Interacting Particle Systems" from De Masi Ferrari and Lebowitz one reads:

$$\tilde{c}_0(q-q';t)=8\gamma\int_0^tds(4\pi s)^{-1/2}\exp[-(q-q')^2/4s]\cdot\exp[-4(1-\gamma/\gamma_c)s]\tag{2.32a}$$ For $\gamma<\gamma_c$, $$\tilde{c}_0(q-q';t)\underset{t\to\infty}{\xrightarrow{\quad\!\!\!\!\qquad}}\dfrac{\gamma}{(\gamma_c-\gamma)^{1/2}}\exp[-2|q-q'|(\gamma_c-\gamma)^{1/2}]\tag{2.32b}$$

So using $C$, $\Delta$ and $A$ to denote constants that are in our way, we rewrite more neatly:
$$C \int_0^t \frac{1}{\sqrt s} e^{\frac{-\Delta}{s}} e^{-A s}\, ds $$
Now consider a series of change of variables:
1) $ s = \Delta y$ ( $ds = \Delta dy$) yields
$$ C(\Delta)^{1/2}\int_0^t \frac{1}{\sqrt y} e^{\frac{-1}{y}} e^{-A\Delta y}\, dy $$
2) $y = x^2$ $dy = 2x dx$ yields
$$2C(\Delta)^{1/2}\int_0^{\sqrt{t}}  e^{\frac{-1}{x^2}} e^{-A\Delta x^2}\, dx  $$
3) Finally $x = \lambda z$ yields
$$2C(\Delta)^{1/2}\lambda\int_0^{\frac{\sqrt{t}}{\lambda}}  e^{\frac{-1}{\lambda^2x^2}} e^{-A\Delta\lambda^2 x^2}\, dx  $$
Choose $\lambda$ such that 
$$\frac{1}{\lambda^2} = A\Delta\lambda^2$$ 
That is, choose $\lambda = \frac{1}{(A\Delta)^{1/4}}$
So we arrive at
$$2C(\Delta)^{1/2}\frac{1}{(A\Delta)^{1/4}}\int_0^{\frac{\sqrt{t}}{\lambda}}  e^{\frac{-(A\Delta)^{1/2}}{x^2}} e^{-(A\Delta)^{1/2} x^2}\, dx  $$
Therefore, to conclude this integral, need to compute $(*)$ with $a =(A\Delta)^{1/2}$.
but I am stuck.
 A: I think this drops out of George Boole's result that $x\mapsto x-1/x$ is measure-preserving.  That is, $\int_{\mathbb R }f(x)\,dx = \int_{\mathbb R }f(x-1/x)\,dx.$ Apply this to $f(x)=\exp(-a x^2)$.
A: By Glasser's Master Theorem for any $a>0$ we have
$$\begin{eqnarray*} \int_{0}^{+\infty}\exp\left[-a\left(s^2+\frac{1}{s^2}\right)\right]\,ds&\stackrel{\text{parity}}{=}&\frac{e^{-2a}}{2}\int_{-\infty}^{+\infty}\exp\left[-a\left(s-\frac{1}{s}\right)^2\right]\\&\stackrel{\text{GMT}}{=}&\frac{e^{-2a}}{2}\int_{-\infty}^{+\infty}e^{-as^2}\,ds=\color{blue}{\frac{\sqrt{\pi}}{2e^{2a}\sqrt{a}}}.\end{eqnarray*} $$
As mentioned by the previous answer, the full generality of $\text{GMT}$ is not really needed, it is enough to prove Boole's statement

If $f(s)$ and $g(s)=f\left(s-\frac{1}{s}\right)$ are integrable
  function over the real line, they have the same integral.

Indeed,
$$ \int_{-\infty}^{0}f\left(s-\frac{1}{s}\right)\,ds\stackrel{s\mapsto\frac{t-\sqrt{4+t^2}}{2}}{=}\int_{-\infty}^{+\infty}f(t)\left(\frac{1}{2}-\frac{t}{2\sqrt{4+t^2}}\right)\,dt $$
$$ \int_{0}^{+\infty}f\left(s-\frac{1}{s}\right)\,ds\stackrel{s\mapsto\frac{t+\sqrt{4+t^2}}{2}}{=}\int_{-\infty}^{+\infty}f(t)\left(\frac{1}{2}+\frac{t}{2\sqrt{4+t^2}}\right)\,dt $$
and the claim simply follows by adding the left hand sides and the right hand sides of these identities.
