= Solution
In the periodic cube, $H_0=|\mathbf v|^2/2=(2\pi/L)^2|\mathbf J|^2/2$, so $\boldsymbol\Omega=(2\pi/L)\mathbf v$. The supplied <action-angle variables> are canonical: $d^3\theta\,d^3J=d^3x\,d^3v$. For $\mathbf k=2\pi\mathbf n/L$,
$$
\mathbf n\cdot\boldsymbol\Omega=\mathbf k\cdot\mathbf v,\qquad \mathbf n\cdot\partial_{\mathbf J}f_0=\mathbf k\cdot\partial_{\mathbf v}f_0.
$$
Consequently <3/c> becomes
$$
\widetilde{\delta f}_{\mathbf k}(\mathbf v,\omega)=\frac{\mathbf k\cdot\partial_{\mathbf v}f_0}{\mathbf k\cdot\mathbf v-\omega}\widetilde{\delta\Phi}_{\mathbf k}^{\mathrm{tot}}(\omega).
$$
For each nonzero <wavenumber>, the <Poisson equation for Newtonian gravity> gives
$$
-k^2\widetilde{\delta\Phi}_{\mathbf k}=4\pi G\int\widetilde{\delta f}_{\mathbf k}\,d^3v.
$$
Substitute the preceding response and add the externally imposed <Newtonian gravitational potential>:
$$
\widetilde{\delta\Phi}_{\mathbf k}^{\mathrm{tot}}=\widetilde{\delta\Phi}_{\mathbf k}^{\mathrm{ext}}-\frac{4\pi G}{k^2}\widetilde{\delta\Phi}_{\mathbf k}^{\mathrm{tot}}\int d^3v\,\frac{\mathbf k\cdot\partial_{\mathbf v}f_0}{\mathbf k\cdot\mathbf v-\omega}.
$$
Therefore
$$
\boxed{\widetilde{\delta\Phi}_{\mathbf k}^{\mathrm{tot}}=\frac{\widetilde{\delta\Phi}_{\mathbf k}^{\mathrm{ext}}}{\epsilon_{\mathbf k}(\omega)},\qquad \epsilon_{\mathbf k}(\omega)=1+\frac{4\pi G}{k^2}\int d^3v\,\frac{\mathbf k\cdot\partial_{\mathbf v}f_0}{\mathbf k\cdot\mathbf v-\omega}.}
$$
This <gravitational dielectric function> has the opposite susceptibility sign to the electron <dielectric function>, reflecting attractive rather than repulsive collective interactions. The homogeneous mean field is explicitly subtracted in this fictitious periodic model: the <Poisson equation for Newtonian gravity> acts on density fluctuations, and the $\mathbf k=0$ mode is omitted. Otherwise a nonzero uniform positive <mass density> would be incompatible with a periodic potential. This is the convention often called the <Jeans swindle>.
For <analytic continuation>, resolve <velocity> parallel to $\mathbf k$: let $u=\widehat{\mathbf k}\cdot\mathbf v$ and $F(u)=\int f_0(u,\mathbf v_\perp)\,d^2v_\perp$. Then
$$
\epsilon_{\mathbf k}(\omega)=1+\frac{4\pi G}{k^2}\int\frac{F'(u)}{u-\omega/k}\,du,\qquad k=|\mathbf k|.
$$
Start at $\operatorname{Im}\omega>0$ and deform the <Landau contour> continuously below the moving pole. At real $\omega$, the integral is
$$
\operatorname{PV}\int\frac{F'(u)}{u-\omega/k}\,du+i\pi F'(\omega/k).
$$
For $\operatorname{Im}\omega<0$, this causal continuation equals the undeformed real-axis integral plus $2\pi iF'(\omega/k)$ wherever $F'$ has the necessary analytic extension. It must not be replaced by the undeformed integral alone.
Zeros of $\epsilon_{\mathbf k}$ are self-consistent collective <Landau modes>: with no external forcing, a nonzero <Newtonian gravitational potential> is possible at such a root. For time dependence $e^{-i\omega t}$, upper-half-plane roots grow, lower-half-plane roots describe <Landau damping>, and real roots are neutral. Damped <Landau modes> are poles of the continued response; they need not be regular discrete eigenfunctions of the original collisionless operator.
\b[Regularity qualification.] Continuation throughout the complex plane presupposes suitable analytic $F$. Arbitrary smooth $F$ need not allow it. For example, $F(u)=C\exp[-1/(1-u^2)]$ for $|u|<1$ and zero elsewhere is smooth but not analytic at $u=\pm1$; its Cauchy integral cannot be holomorphically continued across those endpoints. The <Landau contour> gives the standard causal continuation on the domains where the equilibrium permits it, with possible singularities retained.
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