The integral I want to calculate is defined as $$ P(s)=\int_{-\infty}^{\infty}{\rm d}x\int_{-\infty}^{\infty}{\rm d}y\ \delta\left(\frac{(x+y)^2+4x^2y^2}{(x+y)^2+(x+y)^4}-s\right)e^{-\left(x^2+y^2\right)/2a^2} $$ This integral seems no explicit solution (?) since the quartic terms in the $\delta$ function. So I just want to find the asymptotic behavior for small and positive $s$, i.e $P(s)$ for $s\to0^+$. I am trying to find the answer by the changing of variables as $x=p+q,y=p-q$, then I have $$ P(s)=2\int_{-\infty}^{\infty}{\rm d}p\int_{-\infty}^{\infty}{\rm d}q\ \delta\left(\frac{p^2+(p^2-q^2)^2}{p^2+4p^4}-s\right)e^{-\left(p^2+q^2\right)/a^2}. $$ To get small $s$, the variable $q$ should closed to $\pm p$, and $p$ should be as large as possible. So I directly let $q=p$, neglect the integral over $q$, and I have $$ P(s)\propto\int_{-\infty}^{\infty}{\rm d}p\ \delta\left(\frac{1}{4p^2}-s\right)e^{-p^2/a^2}, $$ where I used the assumption $p\gg1$ so that $1+4p^2\approx 4p^2$. Using the properties of $\delta$ function, I finally obtain $$ P(s)\propto s^{-2}e^{-1/(2a^2s)}. $$ I am probably wrong, especially for the step that I neglect the integration over $q$. But this is all I can get. Any suggestion will be helpful.
- $\begingroup$ You can solve exactly for $q$ in terms of $p$ in your second display (4 values) and then expand as a series in $s$. $\endgroup$Brendan McKay– Brendan McKay2023-08-19 08:53:13 +00:00Commented Aug 19, 2023 at 8:53
1 Answer
$$P(s)=2\int_{-\infty}^{\infty}{\rm d}p\int_{-\infty}^{\infty}{\rm d}q\ \delta\left(\frac{p^2+(p^2-q^2)^2}{p^2+4p^4}-s\right)e^{-\left(p^2+q^2\right)/a^2}$$ $$\qquad=2\int_{0}^{\infty}\frac{{\rm d}x}{\sqrt{x}}(x+4x^2)\int_{0}^{\infty}\frac{{\rm d}y}{\sqrt{y}}\ \delta\left(x+(x-y)^2-s(x+4x^2)\right)e^{-\left(x+y\right)/a^2}$$ $$\qquad=2\int_{0}^{\infty}\frac{{\rm d}x}{\sqrt{x}}e^{-x/a^2}\frac{x+4x^2}{z_+-z_-}\left[\frac{\theta(z_+)}{\sqrt{z_+}}e^{-z_+/a^2}+\frac{\theta(z_-)}{\sqrt{z_-}}e^{-z_-/a^2}\right]\theta\bigl(x-(1-s)/4s\bigr)$$ $$\qquad=\int_{(1-s)/4s}^{\infty}dx\,e^{-x/a^2}\frac{1+4x}{\sqrt{4 s x+s-1}}\left[\frac{1}{\sqrt{z_+}}e^{-z_+/a^2}+\frac{1}{\sqrt{z_-}}e^{-z_-/a^2}\right],$$ with $z_\pm=x\pm\sqrt{x (4 s x+s-1)}$ and $\theta(z)$ the unit step function. In the final equality I assumed $0<s<1/4$.
For $s\ll 1$ this approximates to $$P(s)\approx\frac{4}{\sqrt{s}}\int_{1/4s}^{\infty}dx\,e^{-2x/a^2}\frac{1}{\sqrt{4 s x-1}}=\sqrt{2\pi}(a/s)\exp\left(-\frac{1}{2a^2s}\right).$$
- 1$\begingroup$ Very nice answer! $\endgroup$Guoqing– Guoqing2023-08-19 13:11:36 +00:00Commented Aug 19, 2023 at 13:11
- $\begingroup$ What does your $\approx$ sign mean? $\endgroup$Iosif Pinelis– Iosif Pinelis2023-08-20 01:27:23 +00:00Commented Aug 20, 2023 at 1:27
- $\begingroup$ corrections to $\log P(s)$ are of order $s$. $\endgroup$Carlo Beenakker– Carlo Beenakker2023-08-20 06:24:25 +00:00Commented Aug 20, 2023 at 6:24
- $\begingroup$ (i) If you meant here the asymptotic of $\ln P(s)$, why didn't you clearly state that? Also, why do you then retain the constant factor $\sqrt{2\pi}\,a$? (ii) How do you prove this asymptotic? (iii) What are the meanings of the integrals with the delta function, and how do you justify the equalities involving those integrals? $\endgroup$Iosif Pinelis– Iosif Pinelis2023-08-20 14:14:19 +00:00Commented Aug 20, 2023 at 14:14