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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.

user153330
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Conrado Costa
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    http://mathworld.wolfram.com/GlassersMasterTheorem.html – Simply Beautiful Art Aug 04 '17 at 18:26
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    Integral's value is $\dfrac{\sqrt{\pi } e^{-2 a}}{2 \sqrt{a}}$ computed with $Mathematica$. The indefinite integral is $F(s)=-\dfrac{\sqrt{\pi } e^{-2 a} \left(\text{erfc}\left(\frac{\sqrt{a} \left(s^2-1\right)}{s}\right)+e^{4 a} \text{erfc}\left(\frac{\sqrt{a} \left(s^2+1\right)}{s}\right)-2\right)}{4 \sqrt{a}}\quad$ I don't know if it is useful to you... maybe the result can give you some clue how to proceed – Raffaele Aug 04 '17 at 18:57

2 Answers2

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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 $f(x)=\exp(-a x^2)$.

kimchi lover
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  • Nice answer and nice to cite George Boole which is the right name... (+1) – Olivier Oloa Aug 04 '17 at 21:30
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    @OlivierOloa Thanks! I remember this from an old Putnam exam, in the reign of Charlemagne or so. I guess the standard modern reference is https://link.springer.com/article/10.1007/BF02756706 (Adler & Weiss, Israel Journal, 1973). Boole's original paper (http://rstl.royalsocietypublishing.org/content/147/745.full.pdf) is hard for me to understand. It is a wonderfully flashy trick, when applicable. – kimchi lover Aug 04 '17 at 21:42
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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.

Jack D'Aurizio
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