= Solution
With $\langle\theta\rangle=(1+|\theta|^2)^{1/2}$, the <symbol class> $\operatorname{Sym}(X,\mathbb R^k;N)=S^N_{1,0}$ consists of complex-valued $a\in C^\infty(X\times\mathbb R^k)$ such that, for every compact $K\Subset X$ and every pair of <multi-indices> $\alpha,\beta$,
$$
\boxed{|\partial_x^\alpha\partial_\theta^\beta a(x,\theta)|
\leq C_{K,\alpha,\beta}\langle\theta\rangle^{N-|\beta|},\qquad x\in K.}
$$
An $x$ derivative preserves symbol order, while a $\theta$ derivative lowers it by one. The definition is local in $x$ and imposes estimates of every derivative order.
In the homogeneous convention, a <phase function> is real, smooth on $X\times(\mathbb R^k\setminus\{0\})$, positively homogeneous of degree one in $\theta$, and satisfies
$$
(\nabla_x\Phi,\nabla_\theta\Phi)\ne0\quad(\theta\ne0).
$$
Only the large-frequency behavior is relevant to the regularization argument; any prescribed smooth modification near $\theta=0$ contributes an ordinary convergent integral. If the phase is presented on all of $X\times\mathbb R^k$, homogeneity can be required outside a bounded frequency region. Equivalently for this construction, its large-frequency symbol bounds and the following uniform estimate are sufficient:
$$
|\nabla_x\Phi|^2+|\theta|^2|\nabla_\theta\Phi|^2\geq c_K|\theta|^2,\qquad x\in K,\ |\theta|\geq1.
$$
For a homogeneous <phase function>, this follows from compactness of $K\times S^{k-1}$. \b[Nonvanishing of the total gradient is the condition; $\nabla_\theta\Phi$ may vanish.] The displayed definition does not impose the stronger rank condition sometimes called a nondegenerate phase in Fourier-integral-operator theory.
We construct the <oscillatory integral> by <integration by parts> in both $x$ and $\theta$. For $|\theta|\geq1$, set
$$
q=|\nabla_x\Phi|^2+|\theta|^2|\nabla_\theta\Phi|^2,\qquad
L=\frac1{iq}\left(\nabla_x\Phi\cdot\nabla_x+|\theta|^2\nabla_\theta\Phi\cdot\nabla_\theta\right).
$$
Then $L(e^{i\Phi})=e^{i\Phi}$. The coefficients of the $x$ derivatives have symbol order $-1$ and those of the $\theta$ derivatives have order zero. Use the bilinear <formal transpose of a differential operator>, not the Hermitian adjoint:
$$
L^tb=-\sum_j\partial_{x_j}\left(\frac{\partial_{x_j}\Phi}{iq}b\right)
-\sum_\ell\partial_{\theta_\ell}\left(\frac{|\theta|^2\partial_{\theta_\ell}\Phi}{iq}b\right).
$$
The <symbol order reduction by a phase integration operator> follows from the product rule and the <symbol class> estimates: $L^t:S^s_{1,0}\to S^{s-1}_{1,0}$ locally in $x$. Differentiating the coefficients preserves precisely these orders because $q$ is elliptic of degree two on each compact $x$ set.
Let $\zeta\in C_c^\infty(\mathbb R^k)$ equal one on $|\theta|\leq1$ and vanish on $|\theta|\geq2$. The low-frequency part is an ordinary integral. For an integer $r>N+k$, define the high-frequency part by
$$
\begin{aligned}
\langle I_\Phi(a),\varphi\rangle={}&\iint e^{i\Phi}\zeta(\theta)a(x,\theta)\varphi(x)\,dx\,d\theta\\
&+\iint e^{i\Phi}(L^t)^r\big[(1-\zeta(\theta))a(x,\theta)\varphi(x)\big]\,dx\,d\theta.
\end{aligned}
$$
If the homogeneous phase is only defined off zero, its value at the single point zero is immaterial to the low-frequency integral; its modulus remains one. In the high-frequency term all coefficients are used away from zero. For $\operatorname{supp}\varphi\subset K$, repeated product rules give
$$
\left|(L^t)^r[(1-\zeta)a\varphi]\right|
\leq C_{K,a,\Phi,r}\langle\theta\rangle^{N-r}
\max_{|\alpha|\leq r}\sup_K|\partial^\alpha\varphi|.
$$
Since $N-r<-k$, both terms converge absolutely. They are linear in the <test function>, with the continuity bound
$$
\boxed{|\langle I_\Phi(a),\varphi\rangle|
\leq C_K\max_{|\alpha|\leq r}\sup_K|\partial^\alpha\varphi|.}
$$
Thus \b[$I_\Phi(a)\in\mathcal D'(X)$], with <order of a distribution> at most $r$ on each compact set.
To check that this construction is the intended <oscillatory integral> and does not depend on $r$, $\zeta$ or the large-frequency cutoff, let $\chi\in C_c^\infty(\mathbb R^k)$ equal one near zero and form
$$
J_R(\varphi)=\iint e^{i\Phi(x,\theta)}a(x,\theta)\varphi(x)\chi(\theta/R)\,dx\,d\theta.
$$
After $r$ applications of <integration by parts>, the term without a derivative on $\chi$ tends to the absolutely convergent high-frequency term by the <dominated convergence theorem>. Each extra term is supported in an annulus $|\theta|\asymp R$. A derivative of $\chi(\theta/R)$ contributes $R^{-1}$, so those terms have total absolute value at most $C_KR^{N-r+k}\max_{|\alpha|\leq r}\|\partial^\alpha\varphi\|_\infty\to0$. The low-frequency part is unchanged for large $R$. Hence $J_R(\varphi)$ tends to the displayed functional, independently of all cutoffs and of the permissible number of integrations. This proves the <cutoff independence of an oscillatory integral> as well as its continuity.
For the final <distribution>, the derivative convention $\langle\delta',g\rangle=-g'(0)$ identifies it as $-x_2\delta'(x_1)$. The <delta derivatives from polynomial oscillatory amplitudes> identity yields
$$
\boxed{\Phi(x,\theta)=x_1\theta,\qquad
a(x,\theta)=-\frac{i}{2\pi}x_2\theta,\qquad
u=I_\Phi(a)=-\frac{i x_2}{2\pi}\int_{\mathbb R}e^{ix_1\theta}\theta\,d\theta.}
$$
Here $k=1$, $a\in\operatorname{Sym}(\mathbb R^2,\mathbb R;1)$, and $\nabla_x\Phi=(\theta,0)$ is nonzero for $\theta\ne0$. Thus the phase is admissible even though $\partial_\theta\Phi=x_1$ vanishes on the <distribution>'s support.
For a direct proof of the sign and normalization, put $g(s)=\int_{\mathbb R}x_2\varphi(s,x_2)\,dx_2\in C_c^\infty(\mathbb R)$. The cutoff integral is
$$
J_R(\varphi)=-\frac{i}{2\pi}\int\theta\chi(\theta/R)\widehat g(-\theta)\,d\theta
=\frac{i}{2\pi}\int\xi\chi(-\xi/R)\widehat g(\xi)\,d\xi.
$$
The <Fourier transform> $\widehat g$ is rapidly decreasing, so the <dominated convergence theorem> and differentiated <Fourier inversion> give
$$
\lim_{R\to\infty}J_R(\varphi)=g'(0)
=\int_{\mathbb R}x_2\frac{\partial\varphi}{\partial x_1}(0,x_2)\,dx_2.
$$
This is exactly the required pairing, so the oscillatory representation gives the prescribed <distribution>, not its negative or a multiple of it.
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