Stationary phase method for $\varphi''(x_0)= 0$

Question

Stationary phase method (in the usual setup) gives asymptotic for $$ I(\lambda)=\int_{a}^{b} f(t) e^{i \lambda \varphi(t)} d t, $$ when at any stationary point $x_0$ ($\varphi'(x_0)=0$) second derivative does not vanishe ($\varphi''(x_0)\ne 0$). Is it possible to find in the literature asymptotic formula for $I(\lambda)$ with explicit error term (not like $1+o(1)$) in the case $\varphi'(x_0)=\varphi''(x_0)= 0$, $\varphi'''(x_0)\ne 0$?

Answer 1

Let me assume that $a=-\infty, b=+\infty, x_0=0$ and $f$ smooth and compactly supported near 0. Then after a suitable change of variable, you get that $ I(\lambda)=\int g(t) e^{i\lambda t^3/3} dt, $ with $g$ smooth and compactly supported and applying Plancherel formula you get $$ I(\lambda)=\int \hat g(\tau) A(\lambda ^{-1/3}\tau) d\tau \lambda ^{-1/3}=\lambda ^{-1/3}\psi(\lambda ^{-1/3}), $$ where $A$ is the Airy function (the inverse Fourier transform of $t\mapsto e^{it^3/3}$). Then you may apply what is known on the Airy function on the real line or simply use that $$ I(\lambda)=\lambda ^{-1/3} \psi(0)+O(\lambda^{-2/3}), \quad \psi (0)= g(0) A(0),\quad A(0)=3^{-1/6}\Gamma(1/3)/(2π). $$ To get an explicit error term, you may use the explicit expansion of the Airy function.