How to apply the method of stationary phase here?

Question

Consider the following oscillating integral $$ I(\xi;t) = \int\limits_{\mathbb R}\int\limits_{\mathbb R^n} e^{it(\xi y - \theta f(y))}a(y) \, dy \, d\theta, \quad \xi \in \mathbb R^n \setminus 0, \; t > 0, $$ where $a(y)$ is a smooth function with compact support and $f(y)$ is a smooth function with nowhere vanishing gradient on $\mathop{\mathrm{supp}} a$. I would like to obtain an asymptotic expansion of $I(\xi;t)$ with respect to $t \to +\infty$ assuming that there is a only one point $y^\ast \in \mathop{\mathrm{supp}} a$ with $\nabla f(y^\ast)$ collinear to $\xi$, i.e. $\theta^\ast \nabla f(y^\ast) = \xi$ for some $\theta^\ast \neq 0$ (that is, $(y^\ast,\theta^\ast)$ is the only stationary point of phase for $I(\xi;t)$). The classical method of stationary phase can't be applied here because the support of $a$ is not compact if we consider it as a function of $y$ and $\theta$. On the other hand, evaluating first the integral with respect to $\theta$, we have: $$ I(\xi;t) = \int\limits_{f(y) = 0}e^{it\xi y}a(y) \, \omega_y, \quad df \wedge \omega_y = dy. $$ The integrand in the latter integral doesn't have stationary points if $\xi \neq 0$. Does it mean that $I(\xi;t)$ decays faster than any power of $t$ as $t \to +\infty$?

Answer 1

There is a special version of the method of stationary phase for integrals of the type $$ I(\xi;t) = \int\limits_{f(y)=0} e^{it\xi y}a(y) \, \omega_y, $$ see e.g. [M. V. Fedoruk, Metod Perevala, Nauka, 1977 (in Russian)]. First, we find local extrema of $S(y) = \xi y$ on $Y = \{ y \colon f(y) = 0 \}$ (the point $y^\ast$ is called stationary point of second kind of function $S$ on $Y$). The conditions in the question imply that there is the only extremum $y^\ast$ of $S$ on $Y$.

Denote $y = (y_1,\ldots,y_n)$ and suppose that $y' = (y_1,\ldots,y_{n-1})$ can be taken as local coordinates on $Y$ near $y^\ast$ so that $y_n = g(y')$ and $y^\ast = (y'^\ast,g(y'^\ast))$. Denote $\widetilde S(y') := S(y',g(y'))$. Suppose that the Hesse matrix $\widetilde S_{y'y'}(y'^\ast)$ is nondegenerate. Then the following formula holds: $$ I(\xi;t) = (2\pi)^{\frac{n-1}{2}} t^{-\frac{n-1}{2}} e^{it\xi y^\ast+ i \frac \pi 4 \mathop{\mathrm{sgn}}\widetilde S_{y'y'}(y'^\ast)} |\det \widetilde S_{y'y'}(y'^\ast)|^{-\frac 1 2} \biggl( \frac{\partial f}{\partial y_n}(y^\ast) \biggr)^{-1} \bigl( a(y^\ast)+O(t^{-1}) \bigr), $$ when $t \to \infty$.