Decay of oscillatory integral for non-analytic phase function

Question

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 $\alpha>0$.

Answer 1

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.

Answer 2

Change variables from $x$ to $y = e^{-{1 \over x^2}}$, so that $x = (-\ln y)^{-{1 \over 2}}$ and $dx = {1 \over 2y (-\ln y)^{3 \over 2}}dy$. So the integral becomes $${1 \over 2} \int_0^{e^{-{1 \over \delta^2}}} e^{i\lambda y} {1 \over y (-\ln y)^{3 \over 2}}dy$$ You can apply the usual stationary phase method here, dividing the integral into $0$ to $f(\lambda)$ and $f(\lambda)$ to $e^{-{1 \over \delta^2}}$ portions for an appropriately chosen $f(\lambda)$. The idea is that $f(\lambda)$ is chosen as small as possible so that integrating by parts in the second integral gives you some improvement if you integrate the $e^{i\lambda y}$ factor and differentiate the ${1 \over y (-\ln y)^{3 \over 2}}$ factor. Since the differentiation gives you a factor of magnitude $C{1 \over |y \ln y|}$ and the integration gives you a factor of magnitude ${1 \over \lambda}$, a natural choice of $f(\lambda)$ is the $y$ satisfying $|y \ln y| = \lambda^{-1}$. This can be described in terms of the Lambert $W$ function if desired.

Breaking up the integral into two parts according to this formula, you can bound the first integral by taking absolute values of the integrand and integrating, and bound the second integral by doing the integration by parts, and then taking absolute values of the integrand and integrating. The result of the two integrals should be the same (I think), namely a constant times $$\int_0^{f(\lambda)} {1 \over y (-\ln y)^{3 \over 2}}\,dy$$ In other words, you get a bound of $C(-\ln f(\lambda))^{-{1 \over 2}}$ for the overall integral. Since $f(\lambda)$ is between $\lambda^{-2}$ and $\lambda^{-1}$ for example, this is the same as a bound of $C(\ln \lambda)^{-{1 \over 2}}$ as in the earlier answer.

I don't know if this bound is optimal since the standard ways of showing such things that I know of don't apply. But this is a nontrivial result given by stationary phase.