= Solution
Write $m=k_\phi$, $B_{\mathbf k}=\mathbf k\cdot\partial_{\mathbf J}f_0$, and let the detuning from the rotating perturbation be
$$
\Delta_{\mathbf k}(\mathbf J)=\mathbf k\cdot\boldsymbol\Omega(\mathbf J)-m\Omega_{\mathrm p}.
$$
Substitute $\delta\Phi_{\mathbf k}=\Psi_{\mathbf k}e^{-im\Omega_{\mathrm p}t}$ into the <linear response in action-angle variables>, then set $\tau=t-t'$:
$$
\delta f_{\mathbf k}=iB_{\mathbf k}\Psi_{\mathbf k}e^{-im\Omega_{\mathrm p}t}\int_0^te^{-i\Delta_{\mathbf k}\tau}\,d\tau
=B_{\mathbf k}\Psi_{\mathbf k}e^{-im\Omega_{\mathrm p}t}\frac{1-e^{-i\Delta_{\mathbf k}t}}{\Delta_{\mathbf k}}.
$$
At exact <resonance> the last quotient is interpreted by its continuous limit $it$. Inserting this response in <3/c> gives
$$
\mathcal T(t)=i(2\pi)^3\sum_{\mathbf k}m\int B_{\mathbf k}|\Psi_{\mathbf k}|^2\frac{1-e^{-i\Delta_{\mathbf k}t}}{\Delta_{\mathbf k}}\,d^3J.
$$
Pair $\mathbf k$ with $-\mathbf k$. Because $B_{-\mathbf k}=-B_{\mathbf k}$, $\Delta_{-\mathbf k}=-\Delta_{\mathbf k}$, and $|\Psi_{-\mathbf k}|^2=|\Psi_{\mathbf k}|^2$, the imaginary cosine terms cancel. The finite-time <torque> is
$$
\boxed{\mathcal T(t)=-(2\pi)^3\sum_{\mathbf k}m\int B_{\mathbf k}|\Psi_{\mathbf k}|^2\frac{\sin(\Delta_{\mathbf k}t)}{\Delta_{\mathbf k}}\,d^3J.}
$$
This is a leading, second-order response in the perturbation amplitude.
For the long-time limit, assume a continuum of occupied <action variables>, sufficiently smooth integrable weights, convergent harmonic sums, and regular <resonance> surfaces $\nabla_{\mathbf J}\Delta_{\mathbf k}\ne0$ wherever a contributing surface intersects the distribution. Assume also that the perturbation is weak enough to neglect nonlinear trapping and secular changes in $f_0$, and that the prescribed <stellar bar> pattern and its coefficients stay fixed. Then <phase mixing> removes nonresonant transients in the integrated response. The relevant <long-time resonance kernel> is the <distributional convergence>
$$
\frac{\sin(tx)}x\longrightarrow\pi\delta(x)\qquad(t\to+\infty).
$$
It is not a pointwise limit: at $x=0$ the kernel equals $t$. To justify its use under the <integral> over <action variables>, resolve the integral into detuning surfaces; the <coarea formula> gives
$$
\int W(\mathbf J)\delta(\Delta_{\mathbf k}(\mathbf J))\,d^3J
=\int_{\Delta_{\mathbf k}=0}\frac{W(\mathbf J)}{|\nabla_{\mathbf J}\Delta_{\mathbf k}|}\,dS.
$$
Smoothness and decay of this surface weight reduce the limit to the one-dimensional <long-time resonance kernel>. Equivalently, a causal convergence factor $e^{-\eta\tau}$ gives $\operatorname{Re}\int_0^\infty e^{-\eta\tau-i\Delta\tau}\,d\tau=\eta/(\eta^2+\Delta^2)\to\pi\delta(\Delta)$ as $\eta\downarrow0$.
Therefore the <resonant torque from a rotating perturbation> is
$$
\boxed{\mathcal T=-\pi(2\pi)^3\sum_{\mathbf k}k_\phi\int|\Psi_{\mathbf k}|^2\,
\mathbf k\cdot\frac{\partial f_0}{\partial\mathbf J}\,
\delta\bigl(\mathbf k\cdot\boldsymbol\Omega-k_\phi\Omega_{\mathrm p}\bigr)\,d^3J.}
$$
Only orbital <resonances> contribute to this steady integrated <torque>. The limit is a formal weak-perturbation limit, or an intermediate-time approximation after <phase mixing> but before substantial secular redistribution. It does not prove an unchanged halo can absorb <angular momentum> at this rate forever. A discrete set of orbital <frequencies>, singular <resonance> surfaces, or strong trapping can invalidate the stated steady limit.
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