= Solution
An <oscillatory integral> defines a <distribution> by cancellation, even when its <oscillatory integral amplitude> is not integrable in frequency. We use <symbol class> $\operatorname{Sym}(X,\mathbb R^k;N)=S^N_{1,0}(X\times\mathbb R^k)$, where $X\subset\mathbb R^d$ is open and $N\in\mathbb R$. An <oscillatory integral amplitude> $a$ belongs to this class if it is smooth and, for every <compact set> $K\subset X$ and all <multi-indices> $\alpha,\beta$,
$$
|\partial_x^\alpha\partial_\theta^\beta a(x,\theta)|\leq C_{K,\alpha,\beta}\langle\theta\rangle^{N-|\beta|},
\qquad x\in K,\quad\langle\theta\rangle=(1+|\theta|^2)^{1/2}.
$$
In <symbol calculus>, spatial derivatives preserve the order and frequency derivatives lower it. The fixed number $N$ is independent of $K$; this will produce a finite global <order of a distribution>, although continuity constants may depend on $K$.
A <phase function> $\Phi$ is real and smooth on $X\times(\mathbb R^k\setminus\{0\})$, is positively homogeneous of degree one in $\theta$, and has nonzero total differential there:
$$
\Phi(x,r\theta)=r\Phi(x,\theta)\ (r>0),\qquad
(\nabla_x\Phi,\nabla_\theta\Phi)\ne0\quad(\theta\ne0).
$$
This <positive homogeneity> is required at nonzero frequency; arbitrary smooth low-frequency modifications give the equivalent version homogeneous only for large $|\theta|$. In particular, smoothness at zero is not an additional requirement on a general homogeneous <phase function>. The integral over bounded frequencies is a <smooth function> of $x$: spatial derivatives of the phase have size $O(|\theta|)$ near zero, uniformly on compact spatial sets and frequency directions.
Choose a <cutoff function> $\chi\in C_c^\infty(\mathbb R^k)$ equal to one near zero. The proposed meaning of the <oscillatory integral distribution> is
$$
\langle I_\Phi(a),\varphi\rangle
=\lim_{R\to\infty}\int_X\int_{\mathbb R^k}e^{i\Phi(x,\theta)}a(x,\theta)\varphi(x)\chi(\theta/R)\,d\theta\,dx,
\qquad\varphi\in\mathcal D(X).
$$
Existence and <cutoff independence of an oscillatory integral> require a proof. Split $a=a_0+a_\infty$ using a fixed frequency <cutoff function>, with $a_0$ supported in a bounded ball and $a_\infty$ zero for $|\theta|\leq1$. The low-frequency part is already absolutely integrable. On nonzero frequency set
$$
q=|\nabla_x\Phi|^2+|\theta|^2|\nabla_\theta\Phi|^2,\qquad
L=\frac1{iq}\left(\nabla_x\Phi\cdot\nabla_x+|\theta|^2\nabla_\theta\Phi\cdot\nabla_\theta\right).
$$
The <positive homogeneity> and nonvanishing total differential of the <phase function> imply $q\geq c_K|\theta|^2$ for $x\in K$: normalize to the compact <unit sphere> in frequency. Direct differentiation gives $Le^{i\Phi}=e^{i\Phi}$.
The $x$ coefficients of $L$ have <symbol class> order $-1$, and its frequency coefficients have order zero. If these coefficients are $A_j$ and $B_\ell$, the <formal transpose of a differential operator> is
$$
L^tb=-\sum_j\partial_{x_j}(A_jb)-\sum_\ell\partial_{\theta_\ell}(B_\ell b).
$$
Thus the <symbol order reduction by a phase integration operator> is $L^t:S^s_{1,0}\to S^{s-1}_{1,0}$: the first sum has an order-$-1$ coefficient, and the second contains a frequency derivative. In applying this to $a_\infty\varphi$, every application also differentiates $\varphi$ at most once. Repeated <integration by parts>, in both $x$ and $\theta$, consequently gives for an integer $m>N+k$
$$
\int\!\int e^{i\Phi}a_\infty\varphi\chi(\theta/R)\,d\theta\,dx
=\int\!\int e^{i\Phi}(L^t)^m\big(a_\infty\varphi\chi(\theta/R)\big)\,d\theta\,dx.
$$
There are no spatial boundary terms because $\varphi$ has <compact support>, and the frequency cutoff removes frequency boundary terms. Derivatives of $\chi(\theta/R)$ are bounded uniformly by constants times $\langle\theta\rangle^{-|\beta|}$ on their annular support. The transformed integrand is therefore bounded in absolute value by
$$
C_K\langle\theta\rangle^{N-m}\max_{|\alpha|\leq m}\|\partial^\alpha\varphi\|_\infty.
$$
This is integrable in $k$ frequency dimensions. Pointwise the transformed integrand tends to $(L^t)^m(a_\infty\varphi)$, so the <dominated convergence theorem> establishes
$$
\langle I_\Phi(a),\varphi\rangle
=\int\!\int e^{i\Phi}a_0\varphi\,d\theta\,dx
+\int\!\int e^{i\Phi}(L^t)^m(a_\infty\varphi)\,d\theta\,dx.
$$
The limit is independent of the expanding cutoff, the fixed splitting cutoff, and the admissible $m$, since each formula is the limit of the same truncated integral. Derivatives landing on the expanding cutoff can also be estimated directly by $O(R^{N-m+k})$, which tends to zero. In particular,
$$
\boxed{|\langle I_\Phi(a),\varphi\rangle|\leq C_K\max_{|\alpha|\leq m}\|\partial^\alpha\varphi\|_\infty,
\quad m\in\mathbb Z_{\geq0},\quad m>N+k.}
$$
This proves continuity on the <space of test functions> and the <finite order of an oscillatory integral distribution>. The same $m$ works for all <compact sets>. When $N<-k$, $m=0$ is possible and the original frequency integral is absolutely integrable; cancellation is needed for general $N$.
The <singular support> theorem states that
$$
\boxed{\operatorname{sing\,supp}I_\Phi(a)\subset
\{x\in X:\text{some }\theta\ne0\text{ satisfies }\nabla_\theta\Phi(x,\theta)=0\}.}
$$
By <positive homogeneity>, frequencies can be restricted to the <unit sphere>, so this projected set is closed locally in $X$. More precisely, the <stationary-direction bound for singular support> restricts the frequency directions to the closed <conic support of an oscillatory amplitude>; using a closed directional support avoids losing limits occurring at arbitrarily high frequency. The simpler displayed bound is sufficient here.
To prove the bound, take $x_0$ outside the displayed stationary set. On a sufficiently small spatial neighborhood and all unit frequency directions, $|\nabla_\theta\Phi|$ is bounded below. The frequency-only operator
$$
L_\theta=\frac{\nabla_\theta\Phi\cdot\nabla_\theta}{i|\nabla_\theta\Phi|^2}
$$
satisfies $L_\theta e^{i\Phi}=e^{i\Phi}$, and its <formal transpose> lowers <symbol class> order by one. A spatial derivative of order $|\alpha|$ of $e^{i\Phi}a$ has amplitude order at most $N+|\alpha|$. Applying the frequency <integration by parts> more than $N+|\alpha|+k$ times makes that derivative absolutely integrable, uniformly on smaller <compact sets>. Every spatial derivative therefore exists and is continuous there. The <oscillatory integral distribution> is a <smooth function> near $x_0$, proving the <singular support> assertion.
For the <linear transport equation>, use spacetime $X=\mathbb R^n\times(0,\infty)$, frequency $\theta\in\mathbb R^n$, and
$$
\Phi(x,t,\theta)=(x-ct)\cdot\theta,\qquad a=(2\pi)^{-n}.
$$
This is a valid <phase function>, because $\nabla_x\Phi=\theta\ne0$ at nonzero frequency; the <oscillatory integral amplitude> is in <symbol class> order zero. The <Fourier representation of the Dirac delta function> gives the <transport of a Dirac point mass>:
$$
\boxed{u(x,t)=\frac1{(2\pi)^n}\int e^{i(x-ct)\cdot\theta}\,d\theta=\delta_0(x-ct).}
$$
Its rigorous spacetime <distribution> pairing is $\langle u,\varphi\rangle=\int_0^\infty\varphi(ct,t)\,dt$. Consequently
$$
\langle(\partial_t+c\cdot\nabla_x)u,\varphi\rangle
=-\int_0^\infty\frac d{dt}\varphi(ct,t)\,dt=0,
$$
where the endpoints vanish since a spacetime <test function> has <compact support> in $t>0$. At each fixed $t$, $\langle u(t),\psi\rangle=\psi(ct)$, so <weak convergence of distributions> gives $u(t)\to\delta_0$ as $t\downarrow0$. This is the required initial trace.
Finally, $\nabla_\theta\Phi=x-ct$, so the <singular support> theorem confines singularities to $x=ct$. In fact equality holds: $u$ is a nonzero order-zero <distribution> on that trajectory and vanishes off it. A <smooth function> supported on this set of empty interior must vanish, so $u$ cannot be smooth in any neighborhood of a point of the trajectory. Thus
$$
\boxed{\operatorname{sing\,supp}u=\{(ct,t):t>0\}.}
$$
The <method of characteristics> also proves uniqueness among solutions with a distributional initial trace: the <change of variables> $y=x-ct$ turns the <linear transport equation> into $\partial_t w=0$. Such a <distribution> is constant in $t$; pairing with spatial <test functions> reduces this assertion to <a distribution with zero derivative is constant>. Its initial trace fixes $w=\delta_0$, so the moving <Dirac delta distribution> above is the unique solution. There is transport of the singularity along the characteristic and no smoothing.
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