Asymptotic analysis for integrals

Question

I am a physics student doing some integrals of the form $$\lim_{\rho\to\infty} \int dx \int dy \text{ }f(x,y) e^{-\rho (x y)^{3/2}}$$ with $x,y \geq 0$, and $f(x,y)$ is a polynomial ($a_0+a_{x1} x + a_{y1} y + a_{x2} x^2+a_{x1y1} x y...$). I have realised that either $x$ or $y$ has to be zero, so that the integral can be non-zero (have non-exponentially-suppressed terms). However, I do not understand exactly how the contribution can be calculated over specific shapes of $x,y$. For example, I am interested in the integral $$\lim_{\rho\to\infty} \int_{\frac{T}{\sqrt2}}^{{\sqrt2}T} \int_{\sqrt2 T-x}^{\frac{T}{\sqrt2}} f(x,y)e^{-\rho (x y)^{3/2}} \,dy\,dx$$ where $$f(x,y)=(y-x)(L\pi +\sqrt2 (y-x))(-2T+\sqrt{2}(x+y))$$ and $T,L>0$. I tried to use Mathematica to integrate this expression but Mathematica does not evaluate the integral. This shape has a point touching the x-axis, but I do not know the exact contribution of this point. I guess the contribution is dependent on the angles of the touching point but I do not have enough knowledge to proceed. If anyone can help me on this integral (even better, this type of integral), or at least the highest power of $\rho$ from this integral (numerical work suggest the power is zero?), it would be greatly appreciated.

Answer 1

More interestingly and to the point for your question about behavior, consider instead the limit

$$\lim_{\rho\to\infty}\rho^{\frac{4}{3}}\int_{\frac{T}{\sqrt2}}^{{\sqrt2}T} \int_{\sqrt2 T-x}^{\frac{T}{\sqrt2}} f(x,y)e^{-\rho (x y)^{3/2}} \,dy\,dx = \lim_{\rho\to\infty}\rho^{\frac{4}{3}}\int_0^{\frac{T}{\sqrt2}} \int_{\sqrt2 T-y}^{\sqrt2 T} f(x,y)e^{-\rho (x y)^{3/2}} \,dx\,dy$$

which has an additional factor of $\rho^\frac{4}{3}$. Under the variable interchange $x\leftrightarrow \sqrt2 T - x$ we have

$$\lim_{\rho\to\infty}\rho^{\frac{4}{3}}\int_0^{\frac{T}{\sqrt2}} \int_0^y f(\sqrt2 T - x,y)e^{-\rho (\sqrt2 Ty - xy)^{3/2}} \,dx\,dy$$

Now use the substitution $(u,v) = \rho^\frac{2}{3}(x,y)$ and take the limit

$$\lim_{\rho\to\infty}\int_0^{\frac{\rho^\frac{2}{3}T}{\sqrt2}} \int_0^v f\left(\sqrt2 T - \frac{u}{\rho^\frac{2}{3}},\frac{v}{\rho^\frac{2}{3}}\right)e^{- \left(\sqrt2 Tv - \frac{uv}{\rho^\frac{2}{3}}\right)^{3/2}} \,du\,dv $$ $$ \longrightarrow \int_0^\infty\int_0^vf(\sqrt2 T, 0)e^{-(\sqrt2Tv)^\frac{3}{2}}dudv = f(\sqrt2 T, 0)\int_0^\infty ve^{-(\sqrt2Tv)^\frac{3}{2}}dv$$

$$= \frac{f(\sqrt2 T, 0)}{3T^2}\int_0^\infty t^\frac{1}{3}e^{-t}dt = \boxed{\frac{\Gamma\left(\frac{4}{3}\right)}{3T^2}f(\sqrt2 T, 0)}$$

by dominated convergence. This also means that the limit in your original problem is $0$.