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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 $\alpha>0$.

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    $\begingroup$ Just ask yourself for how long the phase stays essentially fixed (say, between $0$ and $1/2$) to get the order of magnitude. $\endgroup$ – fedja Dec 17 '19 at 7:13
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    $\begingroup$ will it decay at all? I would estimate $I\simeq e^{i\lambda} (\delta-\frac{1}{\sqrt{\ln \lambda}})$ $\endgroup$ – Carlo Beenakker Dec 17 '19 at 7:16
  • $\begingroup$ The outputs of NIntegrate[Exp[I*10000*Exp[-x^(-2)]], {x, 0, 1}] $$0.318318\, +0.0260854 i $$ and NIntegrate[Exp[I*1000000*Exp[-x^(-2)]], {x, 0, 1}] $$ 0.263123 + 0.0144692 I$$ and NIntegrate[Exp[I*10000000*Exp[-x^(-2)]], {x, 0, 1}] $$0.244425\, +0.011566 i $$may be useful. $\endgroup$ – user64494 Dec 17 '19 at 12:39
  • $\begingroup$ NIntegrate[Exp[I*10^300*Exp[-x^(-2)]], {x, 0, 1}] $$ 0.03125\, -1.840616403608205 \cdot 10^{-184} i$$ $\endgroup$ – user64494 Dec 17 '19 at 13:00
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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.

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    $\begingroup$ Thanks! I think $(\ln \lambda)^{-1/2}$ is already optimal up to an absolute constant, since the solution of $\epsilon=\lambda^{-1}e^{-\epsilon^{-2}}$ has solution comparable to $\epsilon=(\ln \lambda)^{-1/2}$. It seems that this analysis also gives the same lower bound. $\endgroup$ – Thomas Yang Dec 17 '19 at 23:01
  • $\begingroup$ Sorry. After some computation, I found that the lower bound seems to be much harder. It is very difficult to balance the lower bounds given by the term $O(\epsilon)$ and $J(\lambda)$. $\endgroup$ – Thomas Yang Dec 18 '19 at 2:10
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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.

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  • $\begingroup$ Thanks! Do you know if there is a way to find a lower bound for the integral? $\endgroup$ – Thomas Yang Dec 18 '19 at 2:11
  • $\begingroup$ Not sure. The only thing that comes to mind is that the singularity at $y = 0$ might be strong enough that if one breaks up the integral at $y = c\lambda^{-1}$ for an appropriate $c > 0$ then the phase in the left-hand integral is constant enough that the magnitude of this term is bounded below by some $c (\ln \lambda)^{-{1 \over 2}}$, and after doing the integration by parts the magnitude of the right-hand integral is bounded by $c'(\ln \lambda)^{-{1 \over 2}}$ for $c' << c$. But I haven't examined the integrals in enough detail to make this kind of determination. $\endgroup$ – Zarrax Dec 18 '19 at 3:46

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