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 $\varphi'(x\_0)=\varphi''(x\_0)= 0$, $\varphi'''(x\_0)\ne 0$? 



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


