A simple oscillatory integral with a non-smooth phase

Let $$\phi\in C_c^\infty(\mathbb{R})$$ be an even function such that $$\chi_{(-1/2,1/2)}\le\phi\le \chi_{(-1,1)}$$, where $$\chi_{(a,b)}$$ stands for the indicator function of the interval $$(a,b)$$. For $$\lambda>0$$ consider the oscillatory integral $$I(\lambda)=\int_\mathbb{R} \phi(x)\, \exp \left(i\lambda(x+\epsilon|x|^{\sqrt{2}})\right)\, dx,$$ with some fixed (very small) positive constant $$\epsilon$$.

My question is: what is the asymptotic behavior of this integral as $$\lambda\rightarrow \infty$$? I can show, by essentially doing careful integration by parts, that the upper bound is $$\lesssim \lambda^{-\sqrt{2}}$$, but I wonder whether $$\lambda^{-\sqrt{2}}$$ is also a lower bound?

Note, that if the exponent $$\sqrt{2}$$ is replaced by $$2k$$ for some positive integer $$k$$, then the integral decays like $$\lambda^{-M}$$ for any $$M>0$$ due to the non-stationary phase estimate (the derivative of the function $$x+\epsilon x^{2k}$$ is $$\gtrsim 1$$).

I would appreciate any hints on how to approach this problem.

$$\newcommand{\R}{\mathbb R}\newcommand{\de}{\delta}\newcommand{\ep}{\epsilon}\newcommand{\tI}{\tilde I}$$ Take any $$a\in(1,2)$$ and then any nonzero $$\epsilon\in(-1/a,1/a)$$. Then \begin{align*} I(t)&:=\int_\mathbb{R} \phi(x)\, \exp(it(x+\epsilon|x|^a))\, dx \\ &\sim\frac{2\epsilon\,\Gamma(a+1)}{it^a}\,\sin\frac{\pi a}2 \tag{1} \end{align*} as $$t\to\infty$$. This asymptotics does not depend on $$\phi$$, as long as $$\begin{equation*} \phi\in C_c^\infty(\mathbb{R}) \tag{2} \end{equation*}$$ is a (not necessarily even) function such that $$\begin{equation*} \chi_{(-1/2,1/2)}\le\phi\le \chi_{(-1,1)}. \tag{3} \end{equation*}$$

Indeed, let $$\begin{equation*} g(x):=x+\ep|x|^a. \end{equation*}$$ Then $$g'(x)=1+\ep ax^{[a-1]}$$ for real $$x\ne0$$, where $$x^{[c]}:=|x|^{c-1}x$$. Therefore and because $$|\ep|<1/a$$, we see that $$2\ge g'\ge1-|\ep|a>0$$ on $$(-1,1)$$. So, there is a unique inverse $$h$$ of function $$g$$ on $$(-1,1)$$ such that for all $$x\in(-1,1)$$ and all $$y\in(g(-1),g(1))$$ we have $$\begin{equation*} y=g(x)\iff x=h(y). \tag{3.5} \end{equation*}$$ Also, $$g(-x)<0 for all $$x\in(0,1)$$.

Using now the substitution $$g(x)=y$$, using conditions (2)--(3) and integrating by parts, we have \begin{align*} I(t)&=\int_\R dx\,\phi(x)\, e^{itg(x)} \\ &=\int_{-1}^1 dx\,\phi(x)\, e^{itg(x)} \\ &=\int_{g(-1)}^{g(1)} dy\,h'(y)\phi(h(y))\, e^{ity} \\ &=-\frac1{it}\,(I_1+\tI_1), \tag{4} \end{align*} where \begin{align*} I_1&:=\int_{g(-1)}^{g(1)} dy\,e^{ity}h''(y)\phi(h(y)), \\ \tI_1&:=\int_{g(-1)}^{g(1)} dy\,e^{ity}h'(y)^2\phi'(h(y)). \end{align*} In view of (2)--(3), $$\phi'$$ is an (infinitely) smooth function supported on $$\begin{equation*} S:=[-1,1]\setminus(-1/2,1/2). \end{equation*}$$ Hence, $$(h')^2\,\phi'\circ h\in C^1(g(S))$$. So, integrating by parts, we have $$\begin{equation*} \tI_1\ll\frac1t; \tag{5} \end{equation*}$$ as usual we write $$u\ll v$$ to mean $$|u|\le Cv$$ for some real constant $$C>0$$. Next, in view of (3), $$\begin{equation*} I_1=I_2+\tI_2, \tag{6} \end{equation*}$$ where \begin{align*} I_2&:=\int_{g(-1/2)}^{g(1/2)} dy\,e^{ity}h''(y), \\ \tI_2&:=\int_{g(S)} dy\,e^{ity}h''(y). \end{align*} Note that $$h''\in C^1(g(S))$$. So, integrating by parts, we have $$\begin{equation*} \tI_2\ll\frac1t. \tag{7} \end{equation*}$$

Writing, for brevity, $$x$$ for $$h(y)$$ (cf. (3.5)), we have $$\begin{equation*} h'(y)=\frac1{1+\ep ah(y)^{[a-1]}}=\frac1{1+\ep ax^{[a-1]}} \end{equation*}$$ and hence \begin{align*} h''(y)&=-\frac{\ep a(a-1)|x|^{a-2}}{(1+\ep ax^{[a-1]})^2},\\ h'''(y)&\ll|x|^{a-3}+|x|^{2a-4}\ll|x|^{a-3}\ll|y|^{a-3} \tag{8} \end{align*} for $$|x|\le1/2$$, that is, for $$y\in[g(-1/2),g(1/2)]$$. Next, for $$y=g(x)\in[g(-1/2),g(1/2)]$$ we have $$y=x(1+\ep x^{[a-1]})$$ and hence \begin{align*} x=h(y)&=\frac y{1+\ep x^{[a-1]}} \\ &=\frac y{1+\ep y^{[a-1]}(1+O(|y|^{a-1}))} \\ &=\frac y{1+\ep y^{[a-1]}} \, (1+O(|y|^{2a-2}) \\ &=y\, (1+O(|y|^{a-1})) \end{align*} and \begin{align*} h''(y)&=-\frac{\ep a(a-1)|h(y)|^{a-2}}{(1+\ep ah(y)^{[a-1]})^2}\\ &=-\ep a(a-1)|y|^{a-2}[1+O(|y|^{a-1})] \\ &=-\ep a(a-1)|y|^{a-2}+O(|y|^{2a-3}). \tag{9} \end{align*} Further,
$$\begin{equation*} I_2=I_3+\tI_3, \tag{10} \end{equation*}$$ where \begin{align*} I_3&:=\int_{|y|\le\de} dy\,e^{ity}h''(y), \\ \tI_3&:=\int_{[g(-1/2),g(1/2)]\setminus[-\de,\de]} dy\,e^{ity}h''(y), \\ \de&:=t^{-3/4}. \end{align*} Integrating by parts and using (8) and (9), we have \begin{align*} \tI_3&\ll\frac{|h''(\de)|+|h''(-\de)|+O(1)}t +\frac1t\, \int_{|y|>\de} dy\,|y|^{a-3}, \\ &\ll \frac{\de^{a-2}}t=t^{1/2-3a/4}=o(t^{1-a}). \end{align*} Using (9) again, we have $$\begin{equation*} I_3=-\ep a(a-1)I_4+O(\tI_4), \tag{11} \end{equation*}$$ where \begin{align*} I_4&:=\int_{|y|\le\de} dy\,e^{ity}|y|^{a-2}, \\ \tI_4&:=\int_{|y|\le\de} dy\,|y|^{2a-3}. \end{align*} Next, $$\begin{equation*} \tI_4\ll\de^{2a-2}=t^{-(2a-2)3/4}=o(t^{1-a}), \tag{12} \end{equation*}$$ \begin{align*} I_4&=t^{1-a}\int_{|z|\le t\de} dz\,e^{iz}|z|^{a-2} \\ &=t^{1-a}\int_{|z|\le t^{1/4}} dz\,e^{iz}|z|^{a-2} \\ &\sim t^{1-a}\int_\R dz\,e^{iz}|z|^{a-2} =2 t^{1-a}\Gamma(a-1)\sin\frac{\pi a}2. \tag{13} \end{align*}

Collecting the pieces (4)--(7) and (10)--(13), we get the result.

• That's a wonderful answer! Thanks so much, Professor Pinelis! Jan 3 at 17:14
• I do not see the $i$ at the denominator in (1). $I(t)$ is real if $\phi$ is even and this is in any case true in the asymptotics. Am I wrong? Jan 3 at 22:18
• @GiorgioMetafune : Thank you for this sanity check. However, I do not see an argument for $I(t)$ to be real even when $\phi$ is even (pun not intended). That would be the case if $g$ were odd. But $g$ is odd only if $\epsilon=0$, in which case the expression after $\sim$ in (1) is $0$ and hence real, even though (1) then loses meaning. Also, I do not see a mistake (do you?) in this proof, except that I now see that I forgot to require $\epsilon$ to be nonzero -- going to fix this now. Jan 3 at 22:44
• Yes, you are right, no need to be real. But then I do not see why it is purely imaginary! I get almost the same but with $\exp{i a \pi/2}$ instead of $i^{-1}\sin{\pi a/2}$. I did the computations on $(0, \infty)$ instead of the whole line, but this should only mean a factor 2. Jan 3 at 23:21
• @GiorgioMetafune : (i) The complete explanation of why it is purely imaginary is the entire proof, which is pretty long. A short explanation may be this: If $\epsilon=0$, then $g$ is odd and hence $I(t)$ is real, and then $g$ is also smooth and hence $I(t)\to0$ faster than any (say negative) power of $t$. Now, $\epsilon$ is the measure of both the non-oddness and non-smoothness of $g$. Therefore and because $\epsilon$ comes with the factor $i$, it seems natural to expect $I(t)$ to be asymptotically purely imaginary. Jan 4 at 0:35