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.



A simple oscillatory integral with a non-smooth phase
$\newcommand{\R}{\mathbb R}\newcommand{\de}{\delta}\newcommand{\ep}{\epsilon}\newcommand{\tI}{\tilde I}$
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, $g(-x)<0<g(x)$ 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.


