Consider the following integral
$$
I\_\delta(\lambda)=\int\_0^\delta e^{i\lambda \exp(-x^{-2})}dx.
$$
Here, $\phi(x)=\exp(-x^{-2})$ is the phase function. I would like to study the rate of decay of $I(\lambda)$ as $\lambda\to \infty$.


In Stein's *Harmonic Analysis*, the case where the phase function has finite order of vanishing was discussed. More precisely, if $\phi^{(j)}(0)=0$ for all $0\leq j\leq k$ but $\phi^{(k+1)}(0)\neq 0$, then there is $\delta>0$ such that
$$
I\_\delta(\lambda)=c \lambda^{-(k+1)^{-1}}+O(\lambda^{-(k+2)^{-1}}), \quad \text{as $\lambda\to \infty$}.
$$
where $c$ is a nonzero constant. Now since the phase function has infinite order of decay, I expect that $I\_\delta(\lambda)$ will have very slow decay, probably slower than $\lambda^{-\alpha}$ for any $\alpha>0$.



Decay of oscillatory integral for non-analytic phase function
Let me fix $\delta=1$ for simplicity. Let us use a Van der Corput method. We have for $\epsilon\in (0,1)$ to be chosen later, with $\phi(x)= e^{-x^{-2}}, $ noting that $\phi'(x)=\phi(x) 2 x^{-3}$
$$
I(\lambda)=\underbrace{\int\_0^{\epsilon} e^{i\lambda \phi(x)} dx}\_{O(\epsilon)}+\underbrace{\int\_{\epsilon}^1
\frac{d}{dx}\bigl(e^{i\lambda \phi(x)} \bigr) \frac{x^3dx}{i\lambda 2\phi(x)}}\_{J(\lambda)}.
$$
We have 
$$
2i\lambda J(\lambda)=\Bigl[e^{i\lambda \phi(x)}\frac{x^3}{\phi(x)}\Bigr]^{x=1}\_{x=\epsilon}
-\int\_{\epsilon}^1 e^{i\lambda \phi(x)}\left(\frac{3x^2}{\phi(x)}-\frac{ 2 }{\phi(x)}\right) dx.
$$
As a result, we get
$$
2i\lambda J(\lambda)=O(1)+O(\epsilon^3 e^{\epsilon^{-2}})+O(e^{\epsilon^{-2}})=O(e^{\epsilon^{-2}}),
$$
and thus
$
\vert I(\lambda)\vert\le \epsilon+O(\lambda^{-1}e^{\epsilon^{-2}}).
$
We choose $\epsilon=2(\ln \lambda)^{-1/2}$ and we have 
$$
\epsilon=2(\ln \lambda)^{-1/2}\ge \lambda^{-1}e^{\epsilon^{-2}}=\lambda^{-1+\frac14}
=\lambda^{-3/4},
$$
providing
$$
\vert I(\lambda)\vert\le C(\ln \lambda)^{-1/2}.
$$
It is quite likely that the exact solution of the equation $ \epsilon=\lambda^{-1}e^{\epsilon^{-2}}$ will give a slightly better estimate.


