= Solution
Set $\kappa=|k|$ and first work with a representative positive $k=\kappa$. For $\kappa\lambda_{\mathrm{De}}\ll1$, the thermal core is strongly screened. However, the causal <dielectric function> has weakly damped <Langmuir wave> roots $\omega_s+i\gamma_s$, with $s=\pm1$, $\omega_s\simeq s\omega_{pe}$ and $\gamma_s<0$. Their <phase velocities> $v_s=\omega_s/\kappa$ are in the tail. Both roots must be retained for each spatial <Fourier series> harmonic.
Write $\epsilon(-i\omega+0^+,\kappa)=\epsilon_R(\omega)+i\epsilon_I(\omega)$. The supplied approximation gives
$$
\epsilon_R\simeq1-\frac{\omega_{pe}^2}{\omega^2},\qquad \epsilon_I\simeq-\frac{\pi\omega_{pe}^2}{\kappa^2n_e}f_0'(\omega/\kappa),\qquad \epsilon_R'(\omega_s)\simeq\frac2{\omega_s}.
$$
Expanding $\epsilon(\omega_s+i\gamma_s)=0$ to first order gives the <Landau damping> rate
$$
\gamma_s=-\frac{\epsilon_I(\omega_s)}{\epsilon_R'(\omega_s)}\simeq\frac{\pi\omega_s\omega_{pe}^2}{2\kappa^2n_e}f_0'(v_s).
$$
In particular, stable damping means $\omega_sf_0'(v_s)<0$. Near either root,
$$
|\epsilon(-i\kappa v+0^+,\kappa)|^2\simeq\frac4{\omega_{pe}^2}\left[(\kappa v-\omega_s)^2+\gamma_s^2\right].
$$
The narrow <Lorentzian resonance approximation> then gives the leading resonant contribution
$$
\frac1{|\epsilon(-i\kappa v+0^+,\kappa)|^2}\simeq\frac{\pi\omega_{pe}^2}{4}\sum_{s=\pm1}\frac{\delta(\kappa v-\omega_s)}{|\gamma_s|}.
$$
This is a formula under the <integral> over <velocity>; it is not a pointwise approximation away from the <resonances>. Insert it into the heating formula in <2/c>. For one spatial harmonic, the contribution from one root is
$$
\begin{aligned}
Q_{k,s}
&=-\frac{e^2}{m_e}k^2A_k\frac{\pi\omega_{pe}^2}{4\kappa}\frac{v_sf_0'(v_s)}{|\gamma_s|}\\
&=\frac{e^2n_e}{2m_e}k^2A_k,
\end{aligned}
$$
because
$$
\frac{v_sf_0'(v_s)}{|\gamma_s|}=-\frac{2\kappa n_e}{\pi\omega_{pe}^2}.
$$
Thus the tiny tail slope cancels the tiny <Landau damping> rate: weak absorption is compensated by a large noise-driven wave intensity. The two temporal roots supply two equal halves of the answer. Negative $k$ gives the same result by the conjugate response of a real <electric field>, without altering the original convention that the spatial sum includes both signs. Therefore
$$
\boxed{\frac{d\mathcal K}{dt}=\sum_k\sum_{s=\pm1}Q_{k,s}\simeq\frac{e^2n_e}{m_e}\sum_k k^2A_k.}
$$
As in <2/c>, equality means the leading result in the stated asymptotic limit. This calculation assumes nonzero damping, observation times $t\gg\max_s|\gamma_s|^{-1}$, and forcing weak enough that $f_0$ changes negligibly during that relaxation. Taking zero damping before the long-time limit invalidates the stationary calculation.
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