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, $g(-x)<0<g(x)$ 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.

  • $\begingroup$ That's a wonderful answer! Thanks so much, Professor Pinelis! $\endgroup$
    – Tony419
    Jan 3 at 17:14
  • $\begingroup$ 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? $\endgroup$ Jan 3 at 22:18
  • $\begingroup$ @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. $\endgroup$ Jan 3 at 22:44
  • $\begingroup$ 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. $\endgroup$ Jan 3 at 23:21
  • $\begingroup$ @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. $\endgroup$ Jan 4 at 0:35

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