and (\[ibp2\]) by the directional derivative $\theta\cdot\partial_{\eta}$, then the same result will remain true (uniformly in $\theta$).

First consider the case when we have a directional derivative, and assume that it is $\partial_{1}$. We will integrate by parts in $\eta_{1}$ a total of $N$ times, where $N$ is a large integer to be determined. Let the differential operator $D$ be defined by$$Du=\frac{\partial_{1}u}{\partial_{1}\Phi};\,\,\,\,\,\,\,D(e^{{i}K\Phi})={i}Ke^{{i}K\Phi},$$ then its dual $D'$ would be $$D'u=-\partial_{1}\bigg(\frac{u}{\partial_{1}\Phi}\bigg).$$ Therefore, integrating by parts, we obtain $$\bigg|\int_{\mathbb{R}^{3}}e^{{i}K\Phi(\eta)}h(\eta)\,\mathrm{d}\eta\bigg|\leq K^{-N}\big\|(D')^{N}h\big\|_{L^{1}}.$$ In order to bound $(D')^{N}h$, we will use the following explicit formula, namely $$\label{induct2}(D')^{N}h=\sum_{r=0}^{N}\sum_{\alpha_{0}+\cdots+\alpha_{r}=N-r}\Omega(n,r;\alpha_{0},\cdots,\alpha_{r})\cdot\partial_{1}^{\alpha_{0}}h\cdot\frac{\partial_{1}^{\alpha_{1}+2}\Phi\cdots\partial_{1}^{\alpha_{r}+2}\Phi}{(\partial_{1}\Phi)^{r+N}},$$ where the coefficient $$|\Omega|\leq 3^{N}(N+1)!.$$ The formula (\[induct2\]) is easily proved by induction.

Now, using (\[ibp2\]), (\[ibp1\]) and (\[induct2\]), we have$$\label{boundp}K^{-N}\big\|(D')^{N}h\big\|_{L^{1}}\lesssim ((3C+3)N)!\cdot K^{-N}\lambda^{\alpha_{0}}\cdot\epsilon_{1}^{-N-r}\prod_{j=0}^{n-2}\epsilon_{j+2}^{r_{j}}.$$ Here we denote by $r_{j}$ the number of $q\geq 1$ such that $\alpha_{q}=j$. Let$$\label{prepare}\rho:=r-\sum_{j=0}^{n-2}r_{j}\geq 0;\,\,\,\,\,\,\,\alpha_{0}+\sum_{j=0}^{n-2}jr_{j}\leq N-r-(n-1)\rho.$$

We now denote $K\epsilon_{1}/\lambda=M_{0}$ and $K\epsilon_{1}\epsilon_{j}/\epsilon_{j+1}=M_{j}$, where $\epsilon_{n+1}=1$. We could then solve that$$\epsilon_{j}=(M_{1}\cdots M_{j-1}K)^{\frac{j-n-1}{n+1}}(M_{j}\cdots M_{n})^{\frac{j}{n+1}};$$$$\lambda=(M_{1}\cdots M_{n}K)^{\frac{1}{n+1}}M_{0}^{-1}.$$Plugging into (\[boundp\]) we obtain$$\begin{aligned}
K^{-N}\big\|(D')^{N}h\big\|_{L^{1}}&\lesssim&((3C+3)N)!K^{-N}(M_{1}\cdots M_{n}K)^{\frac{\alpha_{0}}{n+1}} M_{0}^{-\alpha_{0}}\times(M_{1}\cdots M_{n})^{\frac{-N-r}{n+1}}\times\nonumber\\
&\times&K^{\frac{n(N+r)}{n+1}}\prod_{j=0}^{n-2}(M_{1}\cdots M_{j+1}K)^{\frac{r_{j}(j-n+1)}{n+1}}(M_{j+2}\cdots M_{n})^{\frac{r_{j}(j+2)}{n+1}}\nonumber\\
&=&((3C+3)N)!K^{\sigma}\prod_{j=0}^{n}M_{j}^{\tau_{j}},\end{aligned}$$ where $$\sigma=-N+\frac{\alpha_{0}+n(N+r)}{n+1}+\sum_{j=0}^{n-2}\frac{r_{j}(j-n+1)}{n+1}\leq 0,$$ and $\tau_{0}=-\alpha_{0}\leq 0$, and$$\tau_{j}=\frac{\alpha_{0}-N-r}{n+1}-\sum_{i=j-1}^{n-2}r_{i}+\frac{1}{n+1}\sum_{i=0}^{n-2}r_{i}(i+2)\leq -\rho\leq 0$$for $1\leq j\leq n$, and $$\begin{aligned}
\sum_{j=0}^{n}\tau_{j}&=&\frac{-\alpha_{0}-n(N+r)}{n+1}+\frac{n}{n+1}\sum_{i=0}^{n-2}r_{i}(i+2)-\sum_{i=0}^{n-2}r_{i}(i+1)\nonumber\\
&=&-\frac{nN}{n+1}-\frac{1}{n+1}\big(\alpha_{0}+\sum_{i=0}^{n-2}r_{i}(i+1)+n\rho\big)\leq-\frac{n}{n+1}N\nonumber,\end{aligned}$$ all the inequalities being consequences of (\[prepare\]). This then implies that$$K^{-N}\big\|(D')^{N}h\big\|_{L^{1}} \lesssim ((3C+3)N)!\min(M_{0},\cdots,M_{l})^{-\frac{nN}{n+1}}.$$If we now choose $N$ appropriately, an easy computation will show that $$\bigg|\int_{\mathbb{R}^{3}}e^{{i}K\Phi(\eta)}h(\eta)\,\mathrm{d}\eta\bigg|\lesssim e^{-\gamma 