= Solution
A linearly stable $f_0$ has no zeros of the causal <dielectric function> in $\operatorname{Re}p>0$. Hence the inverse <Laplace transform> contours can approach $\sigma=0^+$, retaining the causal boundary value at $p=-i\omega+0^+$. To obtain a stationary forced response, further assume that the excited collective perturbations actually damp. This is the decay condition in the question's long-time instruction.
A transparent evaluation of the double integral uses the stationary <power spectral density>. The <white noise> covariance $2A_k\delta(t-t')$ has temporal spectrum $2A_k$. Linear filtering by $1/\epsilon$ therefore gives
$$
S_{\varphi,k}(\omega)=\frac{2A_k}{|\epsilon(-i\omega+0^+,k)|^2},\qquad
\langle\varphi_k(t-\tau)\varphi_k(t)^*\rangle=\int_{-\infty}^{\infty}\frac{d\omega}{2\pi}\,S_{\varphi,k}(\omega)e^{i\omega\tau}.
$$
The causal solution for $\delta f_k$ in <2/a> gives, after the initial transients have decayed,
$$
D(v)=\frac{e^2}{m_e^2}\sum_k k^2\operatorname{Re}\int_0^\infty d\tau\,e^{-ikv\tau}\langle\varphi_k(t-\tau)\varphi_k(t)^*\rangle.
$$
The real part can be taken because the full $k$ sum pairs conjugates. Using the <distribution theory> identity $\operatorname{Re}\int_0^\infty e^{ia\tau}\,d\tau=\pi\delta(a)$, with a positive vanishing convergence factor, samples the temporal spectrum at $\omega=kv$:
$$
\boxed{D(v)=\frac{e^2}{m_e^2}\sum_k\frac{k^2A_k}{|\epsilon(-ikv+0^+,k)|^2}.}
$$
The factor $1/2$ from the one-sided time integral cancels the $2$ in the noise covariance. The result is nonnegative, as a stationary <velocity-space diffusion> coefficient should be.
\b[Qualification of stability.] Absence of growing modes alone does not guarantee this stationary limit. For example, the electron <waterbag distribution> $f_0(v)=n_e\mathbf1_{|v|<V}/(2V)$ has
$$
\epsilon(-i\omega,k)=1-\frac{\omega_{pe}^2}{\omega^2-k^2V^2},\qquad \omega^2=k^2V^2+\omega_{pe}^2
$$
as undamped collective roots. <White noise> drives their field variance to grow instead of settling. Thus the formula requires decay of excited modes, not merely absence of exponentially growing ones. For weak <Landau damping>, the waiting time can be much longer than $\omega_{pe}^{-1}$, and must still be short compared with evolution of $f_0$ for the frozen-background <quasilinear approximation> to apply.
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