A sharp estimate for an oscillatory integral with a simple phase

Question

Let $\alpha>1$ not necessarily an integer, and let $\psi:\mathbb{R}\rightarrow \mathbb{R}$ be a smooth function with compact support. Consider the oscillatory integral $$I(\lambda):=\int_{0}^{\infty}\int_{0}^{1} \psi(x)\,e^{\dot{\imath} \lambda x t^{\alpha}}dtdx.$$

My question is about the asymptotic behavior of $I(\lambda)$ as $\lambda \to \infty$. I am interested in finding a sharp estimate for $I(\lambda)$ in terms of negative powers of $\lambda$ rather than the asymptotic expansion. But the latter is sometimes essential to obtain such an estimate.

A solution for a special case: If $\alpha=k$ where $k$ is an integer, and $\psi$ is supported away from $x=0$, then the phase $\phi(t,x):=x t^{\alpha}$ satisfies $\partial^{k}_{t}\phi>0$ and we have the estimate

$$\left|\int_{0}^{1} \,e^{\dot{\imath} \lambda x t^{\alpha}}dt\right|\leq \frac{C}{\lambda^{\frac{1}{k}}}.$$ From here we get $|I(\lambda)|=O(\lambda^{-1/k})$. But this is possibly not the best estimate for $I(\lambda)$, because we did not benefit from the oscillation (the cancellation) that comes from integration w.r.t the variable $x$.

Another approach is to interchange order of integration and look first at the oscillatory integral $$\int_{0}^{\infty} \psi(x)\,e^{\dot{\imath} \lambda x t^{\alpha}}dx.$$ The problem with this approach is that the phase $x \to \phi(t,x)$ has degenerate stationary points on the line $t=0$.

Intuitively, when the phase is small enough, there is no oscillation. If $\delta<1$ is fixed and $xt^{\alpha}<\delta/\lambda$ then $e^{i \lambda xt^{\alpha}}=1+O(\delta)$. And we have the estimate

$${\int\int}_{xt^{\alpha}<\delta/\lambda} e^{i \lambda xt^{\alpha}} dt \psi(x)dx\geq \frac{C_{\delta}}{{\lambda}^{1/\alpha}} \int_{0}^{\infty}\frac{\psi(x)}{x^{1/\alpha}} dx.$$ Nevertheless, there may be some interaction between this portion of $I(\lambda)$ and another portion.

Answer 1

$\newcommand{\la}{\lambda}\newcommand{\Ga}{\Gamma}$Write \begin{equation*} I(\la)=\int_0^\infty dx\,f(x)J(\la x), \end{equation*} where \begin{equation*} J(z):=\int_0^1 dt\,e^{i z t^a}=\frac b{z^b}K(z), \end{equation*} \begin{equation*} K(z):=\int_0^z du\,u^{b-1}e^{i u}, \end{equation*} $b:=1/a\in(0,1)$. Note that \begin{equation*} K(z)\to\int_0^\infty du\,u^{b-1}e^{i u}=(-i)^{-b}\Ga(b) \tag{0}\label{0} \end{equation*} as $z\to\infty$. (The equality in \eqref{0} can be obtained in a number of ways; in particular, it follows immediately from formulas 3.761.4 and 3.761.9 of Gradshteyn and Ryzhik, 7th Edition.)
Also, since $u^{b-1}$ decreases to $0$ as $u$ increases from $0$ to $\infty$, we have $|K(z)|\le C$ for some real $C>0$ and all real $z\ge0$.

So, letting $\la\to\infty$, by dominated convergence we get
\begin{equation*} \la^b I(\la)=\int_0^\infty dx\,f(x)\frac b{x^b}K(\la x) \\ \to (-i)^{-b}\Ga(b+1)\int_0^\infty dx\,f(x)/x^b, \end{equation*} so that \begin{equation*} I(\la)\sim R(\la):=(-i\la)^{-b}\,\Ga(b+1)\int_0^\infty dx\,f(x)/x^b. \tag{1}\label{1} \end{equation*}

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Here are the graphs $\{(\la,\Re\frac{I(\la)}{R(\la)})\colon0<\la\le20\}$ (black) and $\{(\la,\Im\frac{I(\la)}{R(\la)})\colon0<\la\le20\}$ (blue) for $a=2.1$ and $f(x)=\exp(-\frac1{x(1-x)})\,1(0<x<1)$, which confirm \eqref{1}: