Let $U\subset\mathbb{R}^{d}$ be open and $N\in\mathbb{N}$. Furthermore, let $a\in\mathcal{S}^{m}_{\rho,\sigma}(U\times\mathbb{R}^{N})$ be a symbol and $\Phi\in C^{\infty}(U\times (\mathbb{R}^{N}\backslash\{0\}))$ be a phase function (as in Definition 1.1. and 1.2. of Shubin's book on pseudodifferential operators). Then, the integral
$$I_{\Phi,a}(x):=\int_{\mathbb{R}^{N}}e^{i\Phi(x,\xi)}a(x,\xi)\,\mathrm{d}\xi$$
converges absolutely for every $x\in U$ whenever $m<-N$. Furthermore, for every $k\in\mathbb{N}$, we have that $I_{\Phi,a}\in C^{k}(U)$. In particular, this implies that $I_{\Phi,a}\in C^{\infty}(U)$ for a smoothing symbol $a\in\mathcal{S}^{-\infty}(U\times\mathbb{R}^{N})$. Now, one of the basic facts of oscillatory integrals is that there exists a unique continuous map $$\mathcal{S}^{\infty}(U\times\mathbb{R}^{N})=\bigcup_{m\in\mathbb{R}}\mathcal{S}^{m}_{\rho,\sigma}(U\times\mathbb{R}^{N})\ni a\mapsto I_{\Phi,a}\in\mathcal{D}^{\prime}(U)$$ where $\mathcal{D}^{\prime}(U)$ denotes the set of distributions, such that for $m<-N$ it coinides with the convergent integral above.
Now, on the other hand, the singular support of the distribution $I_{\Phi,a}\in\mathcal{D}^{\prime}(U)$ for some symbol $a\in\mathcal{S}^{m}_{\rho,\sigma}(U\times\mathbb{R}^{N})$ can be shown to satisfy
$$\mathrm{sing}\,\mathrm{supp}(I_{\Phi,a})\subset \{x\in U\mid \Phi^{\prime}_{x}(\xi)=0\text{ for some }\xi\in\mathbb{R}^{N}\}$$
where $\Phi^{\prime}_{\xi}$ denotes the gradient along $\xi$. What confuses me is that the set on the right-hand side is independent of $a$. For example, by my argument above, in the case $a\in\mathcal{S}^{-\infty}(U\times\mathbb{R}^{N})$ we should have that $I_{\Phi,a}$ is regular, i.e. $I_{\Phi,a}\in C^{\infty}(U)$, for every possible $\Phi$. Does someone have some intuition on this?
