= Solution
Use the <Fourier transform> convention $\widehat f(\xi)=\int e^{-ix\cdot\xi}f(x)\,dx$ and $D_j=-i\partial_{x_j}$. Thus $D_j$ has <Fourier symbol> $\xi_j$. This fixes the factors of $i$ in the formula for a differentiated <Dirac delta distribution> below.
With the <Japanese bracket> $\langle\theta\rangle=(1+|\theta|^2)^{1/2}$, the <symbol class> $\operatorname{Sym}(X,\mathbb R^k;N)=S^N_{1,0}(X\times\mathbb R^k)$ consists of <smooth functions> $a$ such that, for every <compact set> $K\subset X$ and all <multi-indices> $\alpha,\beta$,
$$
\boxed{|\partial_x^\alpha\partial_\theta^\beta a(x,\theta)|
\leq C_{K,\alpha,\beta}\langle\theta\rangle^{N-|\beta|}\quad(x\in K).}
$$
Here $N$ is one fixed finite real order; the constants may depend on $K$. Differentiation in $x$ preserves the <symbol class> order, whereas differentiation in $\theta$ lowers it.
A <phase function> is real-valued and smooth on $X\times(\mathbb R^k\setminus\{0\})$, is a <positively homogeneous function> of degree one in $\theta$, and has nonzero total differential:
$$
\Phi(x,t\theta)=t\Phi(x,\theta)\quad(t>0),\qquad
(\nabla_x\Phi,\nabla_\theta\Phi)\ne0\quad(\theta\ne0).
$$
The nonvanishing condition concerns both sets of variables, not just $\nabla_\theta\Phi$. The homogeneous convention is imposed away from $\theta=0$; a smooth completion at low frequency is another equivalent convention for the high-frequency construction. Low-frequency changes contribute a <smooth function> of $x$.
Choose a <cutoff function> $\rho(\theta)$ equal to one near zero. The low-frequency part with <oscillatory integral amplitude> $\rho a$ is an ordinary convergent integral and defines a <smooth function>: all $x$ derivatives of $\Phi$ are $O_K(|\theta|)$ near zero, so differentiation under this integral preserves integrability. For the remaining part define, when $\theta\ne0$,
$$
G=|\nabla_x\Phi|^2+|\theta|^2|\nabla_\theta\Phi|^2,
\qquad
L=\frac1{iG}\left(\sum_j\partial_{x_j}\Phi\,\partial_{x_j}
+|\theta|^2\sum_\ell\partial_{\theta_\ell}\Phi\,\partial_{\theta_\ell}\right).
$$
Then $Le^{i\Phi}=e^{i\Phi}$. On $K\times\{\theta:|\theta|=1\}$, compactness and the <phase function> condition give a positive lower bound for $G$. Homogeneity consequently gives $G\geq c_K|\theta|^2$. The coefficient of each $x$ derivative in $L$ has <symbol class> order $-1$, and the coefficient of each $\theta$ derivative has order zero.
If $L=\sum b_j\partial_{x_j}+\sum c_\ell\partial_{\theta_\ell}$, its <formal transpose> is
$$
L^t f=-\sum_j\partial_{x_j}(b_jf)-\sum_\ell\partial_{\theta_\ell}(c_\ell f).
$$
This is the bilinear transpose for <integration by parts>, without <complex conjugation>. It lowers the <symbol class> order by one. For a <test function> $\varphi$ supported in $K$, choose a nonnegative integer $r>N+k$ and set
$$
\begin{aligned}
\langle I_\Phi(a),\varphi\rangle={}&\int_X\int_{\mathbb R^k}e^{i\Phi}\rho a\varphi\,d\theta\,dx\\
&+\int_X\int_{\mathbb R^k}e^{i\Phi}(L^t)^r\bigl((1-\rho)a\varphi\bigr)\,d\theta\,dx.
\end{aligned}
$$
Both integrals are absolutely convergent, because the last <oscillatory integral amplitude> has order $N-r<-k$ and <compact support> in $x$.
To verify that this defines the intended <oscillatory integral>, take any $\chi\in C_c^\infty(\mathbb R^k)$ equal to one near zero and insert $\chi(\varepsilon\theta)$ in the original integral. Repeated <integration by parts> gives the preceding expression with $(L^t)^r$ applied also to this <cutoff function>. Its derivatives satisfy uniform <symbol class> bounds: on the annulus where they are nonzero, $\varepsilon^{|\beta|}\leq C\langle\theta\rangle^{-|\beta|}$. The transformed integrands are bounded by an integrable multiple of $\langle\theta\rangle^{N-r}$. The <dominated convergence theorem> therefore proves
$$
\boxed{\langle I_\Phi(a),\varphi\rangle
=\lim_{\varepsilon\downarrow0}\int_X\int_{\mathbb R^k}
e^{i\Phi(x,\theta)}a(x,\theta)\varphi(x)\chi(\varepsilon\theta)\,d\theta\,dx.}
$$
It also proves independence of the <cutoff function>, the chosen $r$, and the integration-by-parts representation.
At most $r$ derivatives fall on $\varphi$, so the same estimates give
$$
\boxed{|\langle I_\Phi(a),\varphi\rangle|
\leq C_K\max_{|\alpha|\leq r}\|\partial^\alpha\varphi\|_\infty,
\qquad I_\Phi(a)\in\mathcal D'(X).}
$$
This proves linearity and continuity on the <space of test functions>. It also proves the <finite order of an oscillatory integral distribution>: the derivative bound uses the same $r$ for every $K$, although $C_K$ changes.
For $x_0\in X$, take $k=n$, $\Phi=(x-x_0)\cdot\theta$, and $a=(2\pi)^{-n}\theta^\alpha$. This <phase function> is valid because $\nabla_x\Phi=\theta\ne0$ for nonzero $\theta$, and the <oscillatory integral amplitude> is in <symbol class> $S^{|\alpha|}_{1,0}$. The required identity is
$$
\boxed{D^\alpha\delta_{x_0}=(2\pi)^{-n}\int_{\mathbb R^n}
e^{i(x-x_0)\cdot\theta}\theta^\alpha\,d\theta.}
$$
Indeed, extending a <test function> by zero outside $X$ and integrating in $x$ first gives the absolutely convergent expression
$$
(2\pi)^{-n}\int e^{-ix_0\cdot\theta}\theta^\alpha\widehat\varphi(-\theta)\,d\theta
=(2\pi)^{-n}\int e^{ix_0\cdot\xi}(-\xi)^\alpha\widehat\varphi(\xi)\,d\xi
=i^{|\alpha|}\partial^\alpha\varphi(x_0).
$$
Here <Fourier inversion> applies because $\varphi$ is a <Schwartz function>. By the definition of a <distributional derivative>, the last expression is precisely $\langle D^\alpha\delta_{x_0},\varphi\rangle$. If $D^\alpha$ is instead used for plain $\partial^\alpha$, the <oscillatory integral amplitude> in the boxed formula is $(2\pi)^{-n}(i\theta)^\alpha$, and the pairing is $(-1)^{|\alpha|}\partial^\alpha\varphi(x_0)$.
\b[Not every <distribution> is one <oscillatory integral> of the stated class.] On $X=\mathbb R$, consider the locally finite <distribution of unbounded order>
$$
T=\sum_{j=1}^\infty D^j\delta_j.
$$
Only finitely many differentiated <Dirac delta distributions> contribute to each <test function>, so $T$ is a <distribution>. Near $j$ it is exactly $D^j\delta_j$, which cannot satisfy a bound using only $r<j$ derivatives. Explicitly choose $\eta\in C_c^\infty((-1/4,1/4))$ with $\eta^{(j)}(0)\ne0$ and use
$$
\varphi_\varepsilon(x)=\varepsilon^r\eta((x-j)/\varepsilon).
$$
The derivatives through order $r$ remain bounded for $0<\varepsilon\leq1$, whereas
$$
|\langle T,\varphi_\varepsilon\rangle|
=\varepsilon^{r-j}|\eta^{(j)}(0)|\longrightarrow\infty.
$$
Any single <oscillatory integral> with finite $k$ and a fixed finite <symbol class> order $N$ has the uniform order bound just proved, so it cannot equal $T$. The same construction works on any nonempty <open set> in positive dimension by choosing points that leave every <compact set> and taking locally finite differentiated <Dirac delta distributions> there. This obstruction concerns a single fixed-order <oscillatory integral>, rather than local representations or an infinite sum of them.
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