inequality of oscillatory intergral I just see the inequality below on the paper of A. Moyua, A. Vargas, and L. Vega concerning about the Schrodinger maximal operator.
$$\left|\int_{\mathbb{R}^{2}} e^{it|\xi|^{2}+2 \pi i x\xi} \frac{d \xi}{|\xi|}\right| \leq \frac{C}{|x|}$$
The authors say that it can be proved by the stationary phase lemma and scaling arguments.
I try to decomposition $\mathbb{R}^{2}$ by $\{x:2^k<|x|<2^{k+1}\}_{k\in \mathbb{Z}}$,but failed.
And I also try to calculate the result concretely with the Bessel function, but the estimate is not good enough.
Any idea will be helpful. Thanks.
Remark of Jason's answer:
$g(\xi)=\frac{\Psi(\xi)}{\xi}$
\begin{eqnarray*}
  \int e^{i|\xi|^2}e^{2\pi i x\xi}\frac{d\xi}{\xi}&=&\sum_k\int e^{i|\xi|^2+2\pi i x\xi}\Psi(2^k\xi)\frac{d\xi}{\xi}\\
    &=&\sum_k2^{-k}\int e^{i2^{-2k}|xi|^2+2\pi ix2^{-k}\xi}g(\xi)d\xi\\
    (\lambda=\sqrt{2^{-4k}+2^{-2k}|x|^2})&=&\sum_k2^{-k}\int e^{i\lambda(|\xi|^2\frac{2^{-2k}}{\lambda}+2\pi \xi\frac{2^{-k}x}{\lambda})}g(\xi)d\xi\\
    \mbox{(oscillatory intergral)}&\leq&\sum_k2^{-k}\frac{1}{\lambda}\\
    &=&\sum_k\frac{1}{\sqrt{2^{-2k}+|x|^2}}
\end{eqnarray*}
 A: Here's a sketch of how to get the estimate. I'm not sure if it's the simplest way or what the authors had in mind, but it does work.
First of, via the change of variable $\eta = t^{1/2} \xi$, it suffices to establish the estimate for $t = 1$.
Next, let $1 \equiv \sum_{k \in \mathbb{Z}} \Psi(2^k \xi)$ be a smooth, dyadic partition of unity on $\mathbb{R}^2 \setminus 0$. We split the integral in question as follows:
$$
\begin{align*}
\int e^{i | \xi |^2} e^{2 \pi i x \cdot \xi} \, \frac{d \xi}{| \xi |}
= &\sum_{k = 1}^{\infty}
\int e^{2 \pi i x \cdot \xi} \Psi(2^k \xi) \, \frac{d \xi}{| \xi |} \\
&+ \int e^{2 \pi i x \cdot \xi} \, \frac{e^{i | \xi |^2} - 1}{| \xi |} \,
\left( \sum_{k = 1}^{\infty} \Psi(2^k \xi) \right) d \xi \\
&+ \sum_{k \leq 0}
\int e^{i | \xi |^2} e^{2 \pi i x \cdot \xi} \Psi(2^k \xi)
\, \frac{d \xi}{| \xi |}.
\end{align*}$$
The integrals in the first sum can be viewed as inverse Fourier transforms of rescaled versions of a single function, namely $g(\xi) = \Psi(\xi) / |\xi|$. In light of this, the full sum can be shown to be $O(|x|^{-1})$. (We can actually get $O((1+|x|)^{-1})$.)
The second integral can be viewed as the inverse Fourier transform of a single $C_c^{\infty}$ function, so its contribution as a function of $x$ is well controlled.
Since the Fourier transform converts convolution to multiplication, the integrals in the final sum can be viewed as the convolution of two functions: $2^{-k} \check{g}(2^{-k} \cdot)$, where $g$ is as defined above; and $h(x) = C e^{-i \pi^2 |x|^2}$, i.e. the inverse Fourier transform of $e^{i |\xi|^2}$. We can estimate this convolution evaluated at a point in two ways. The first, simply using Young's inequality, gives an estimate that is $O(2^{k})$. The other, using integration by parts (twice), gives an estimate that is $O(2^{-k} |x|^{-2})$. For small $x$ the first estimate is always better, and results in an estimate for the full sum that is $O(1)$. For large $x$ the two estimates are each useful for certain $k$, and the estimate for the full sum is $O(|x|^{-1})$. Putting together the different $x$ regimes gives an estimate that is $O(|x|^{-1})$ (or, again, we can get $O((1+|x|)^{-1})$).
Putting the three pieces together then gives the desired $O(|x|^{-1})$ estimate.

Edit:
Let's clarify about integrating by parts in the convolution.
(1) First we'll write out the convolution integral in question. Ignoring some constants in the definition of $h$ for the sake of simplifying calculations, we're looking at
$$f(x) = \int 2^{-k} e^{i |x-y|^2} \check{g}(2^{-k} y) \, dy.$$
We can rewrite this as
$$f(x)
= 2^{-k} e^{i |x|^2}
\int e^{-2 i x \cdot y} e^{i |y|^2} \check{g}(2^{-k} y) \, dy.$$
(2) Now we get to the integration by parts. For this, we use that
$$e^{-2 i x \cdot y}
= (-4 |x|^2)^{-1} \Delta_y (e^{-2 i x \cdot y}).$$
Therefore
$$
\begin{align*}
f(x)
&= 2^{-k} e^{i |x|^2}
\int (-4 |x|^2)^{-1} \Delta_y (e^{-2 i x \cdot y}) e^{i |y|^2} \check{g}(2^{-k} y) \, dy \\
&= 2^{-k} e^{i |x|^2} (-4 |x|^2)^{-1}
\int e^{-2 i x \cdot y} \Delta_y( e^{i |y|^2} \check{g}(2^{-k} y) ) \, dy.
\end{align*}$$
(3) From the Laplacian, the largest-order term that dictates the subsequent estimate is $2^{-2k} e^{i |y|^2} \Delta \check{g}(2^{-k} y)$. In other words, the main term is now
$$2^{-3k} e^{i |x|^2} (-4 |x|^2)^{-1}
\int e^{-2 i x \cdot y} e^{i |y|^2} \Delta \check{g}(2^{-k} y) \, dy.$$
(4) Finally, we take the absolute value and push it through so that all the oscillatory factors go away. This leaves
$$2^{-3k} (-4 |x|^2)^{-1}
\int | \Delta \check{g}(2^{-k} y) | \, dy
= 2^{-3k} (-4 |x|^2)^{-1} \cdot O(2^{2k})
= O(2^{-k} |x|^{-2}).$$
