I posted this on Stackexchange but got no responses or comments.


Consider the following integral, for $\epsilon\ne 0:$


$$\displaystyle\frac{1}{(2\pi)^2\epsilon^4}\int\_{\Omega}yb\,e^{\frac{i}{\epsilon}[-ay+bx-yb]}\,dx\,da\,dy\,db\,,$$


where $\Omega$ is some compact neighborhood of the origin (you can assume the integral is over $\mathbb{R}^4$ and that there is a smooth cutoff function). The exponent $-ay+bx-yb$ has a nondegenerate and unique critical point at $(x,a,y,b)=(0,0,0,0)\,,$
and therefore the stationary phase approximation says that


$$\displaystyle\lim\limits\_{\epsilon\to 0}\frac{1}{(2\pi)^2\epsilon^2}\int\_{\Omega}yb\,e^{\frac{i}{\epsilon}[-ay+bx-yb]}\,dx\,da\,dy\,db=0\,.$$


My guess is that it's also true that
$$\lim\limits\_{\epsilon\to 0}\frac{1}{(2\pi)^2\epsilon^4}\int\_{\Omega}yb\,e^{\frac{i}{\epsilon}[-ay+bx-yb]}\,dx\,da\,dy\,db=0\,,$$


but I'm not sure how to show it. Does anyone have any suggestions?



How to compute the asymptotics of this oscillatory integral?
As Fedor Petrov suggested, you can diagonalize the quadratic form.


Writing $A = a+y$, $B = (a-y)$, $C = (b+x-y)$ and $D = (b-x +y)$, up to some negligible constants (and restoring the smooth cut-off) your integral is


$$ \int \varphi \cdot (C+D)(B-A) e^{\frac{i}{\epsilon} (- A^2 + B^2 + C^2 - D^2)} $$


This you can write as (up to negligible constants)
$$ \epsilon^2 \int \varphi \cdot (\partial\_C - \partial\_D)(\partial\_A + \partial\_B) e^{\frac{i}{\epsilon} (- A^2 + B^2 + C^2 - D^2)}$$


After integration by parts, the standard argument shows that the leading order term is, up to a proportionality constant, given by
$$ \epsilon^4 (\partial\_C - \partial\_D)(\partial\_A + \partial\_B)\varphi(0) $$


If your cut-off function is really just a cut-off function (so it is constant in a neighborhood of the origin), then in fact your integral is effectively supported away from the stationary point of the phase, and so should actually decay faster than any polynomial.


