= Solution
The required time separation is $\tau\ll t_{\mathrm{sec}}$, where $t_{\mathrm{sec}}$ is the evolution timescale of $f_0$, together with $t_{\mathrm{peak}}\gg\tau$ and an observation time well after the pulse. The first condition permits freezing $f_0(\mathbf J,t')$ at its pre-pulse value; its neglected change would enter beyond the retained second order. The second makes the original initial time lie in a negligible tail of the <Gaussian function>, permitting replacement of the lower time limits by $-\infty$. Short duration alone would not justify that replacement for a pulse centered at the initial time.
Integrate the <continuity equation> in <3/c> over the whole pulse. For fixed $\mathbf J$, the time factor is the <ordered Gaussian response integral>
$$
\mathcal I(\omega)=\int_{-\infty}^{\infty}dt\int_{-\infty}^{t}dt'\,e^{-i\omega(t-t')}g(t)g(t').
$$
A common translation by $t_{\mathrm{peak}}$ removes the peak time. Pair the positive and negative azimuthal harmonics: $(-m,-k_R)$ is paired with $(m,k_R)$. Both $k_\varphi$ and $\mathbf k\cdot\partial_{\mathbf J}f_0$ change sign, so their product does not; $\omega_{-\mathbf k}=-\omega_{\mathbf k}$ makes the two time factors complex conjugates. Hence the pair contributes $2\operatorname{Re}\mathcal I$.
For completeness, the real part of the integrand is symmetric under $t\leftrightarrow t'$. The triangular time domain therefore gives half of the full plane, and the <Fourier transform> of a <Gaussian function> gives
$$
\operatorname{Re}\mathcal I(\omega)=\frac12\left|\int_{-\infty}^{\infty}g(t)e^{-i\omega t}\,dt\right|^2=\pi\tau^2e^{-\omega^2\tau^2}.
$$
This derives the supplied identity rather than requiring a separate time-ordering calculation.
Define
$$
D_{mk_R}=m\frac{\partial f_0}{\partial J_\varphi}+k_R\frac{\partial f_0}{\partial J_R}.
$$
Using the paired harmonics and $\Delta F_0=-\partial_{J_\varphi}\int Q_0\,dt$ now yields
$$
\Delta F_0=4\pi^2m\tau^2\frac{\partial}{\partial J_\varphi}\sum_{k_R}\int_0^\infty dJ_R\,|u_{mk_R}|^2D_{mk_R}e^{-\omega_{mk_R}^2\tau^2}.
$$
By the stated assumption, $\omega_{mk_R}$ is independent of $J_R$, so its exponential can be taken outside the radial <integral>. Put $\Gamma=(\sqrt2\tau)^{-1}$ and use
$$
\mathcal R_\Gamma(\omega)=\frac{e^{-\omega^2/(2\Gamma^2)}}{\sqrt{2\pi}\Gamma},\qquad 4\pi^2\tau^2e^{-\omega^2\tau^2}=\frac{\sqrt{8\pi^5}}{\Gamma}\mathcal R_\Gamma(\omega).
$$
Therefore
$$
\boxed{\Delta F_0=\frac{\sqrt{8\pi^5}\,m}{\Gamma}\frac{\partial}{\partial J_\varphi}\sum_{k_R}\mathcal R_\Gamma(\omega_{mk_R})\int_0^\infty dJ_R\,|u_{mk_R}|^2\left(m\frac{\partial f_0}{\partial J_\varphi}+k_R\frac{\partial f_0}{\partial J_R}\right).}
$$
The outer derivative acts on the <Gaussian resonance broadening> factor as well as on the radial <integral>. The result describes <quasilinear diffusion in action space> weighted by the squared pulse amplitude, with a detuning width $\Gamma$. A shorter pulse has broader frequency selectivity, while a longer pulse selects narrower <orbital resonances>. If the action-space boundary flux vanishes, integrating $\Delta F_0$ over $J_\varphi$ gives zero, as required by <conservation of mass>.
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