= Solution
\b[The step in the source must be at zero.] We use the <Heaviside step function> $H(x)=1$ for $x>0$, zero for $x<0$, with its value at zero immaterial here. The printed threshold $1$ is inconsistent with both $H'=\delta(x)$ and the stated <number density>. For example, measuring <velocities> in fixed units and taking $v_F=2$, the literal printed cutoff fills only $v<1$ and gives one eighth of the stated <number density>. Its <distributional derivative> is $\delta(x-1)$, not $\delta(x)$. The zero-threshold convention restores the stated <Fermi sphere> and its <normalization of a probability distribution>.
Let $k=|\mathbf k|>0$, choose $u=\mathbf v\cdot\mathbf k/k$, and abbreviate $V_\alpha=v_{\mathrm F\alpha}$. The three-dimensional <phase-space distribution function> is $f_{0\alpha}=C_\alpha\mathbf1_{v<V_\alpha}$, where
$$
C_\alpha=\frac{m_\alpha^3}{4\pi^3\hbar^3}=\frac{3n_\alpha}{4\pi V_\alpha^3}.
$$
Integrate over the transverse <velocity> disk to obtain the parallel <marginal distribution>
$$
F_\alpha(u)=\int f_{0\alpha}\,d^2v_\perp
=\frac{3n_\alpha}{4V_\alpha^3}(V_\alpha^2-u^2)\mathbf1_{|u|<V_\alpha},\qquad
F_\alpha'(u)=-\frac{3n_\alpha u}{2V_\alpha^3}\mathbf1_{|u|<V_\alpha}.
$$
There are no edge <Dirac delta functions> in $F_\alpha'$ because $F_\alpha$ itself vanishes continuously at the endpoints. In particular, the three-dimensional constant occupation does not imply a constant parallel <marginal distribution>.
Use time dependence $e^{pt+i\mathbf k\cdot\mathbf x}$, with $p=\gamma-i\omega$ and $\gamma<0$ for <damping>. The given <dielectric function> becomes
$$
\epsilon=1-\sum_\alpha\frac{\omega_{\mathrm p\alpha}^2}{n_\alpha k^2}
\int_{C_L}\frac{F_\alpha'(u)}{u-\omega/k-i\gamma/k}\,du.
$$
The <Landau contour> supplies causal continuation. For a real <phase velocity> $a=\omega/k>V_\alpha$, there is no resonant pole in the populated interval, and
$$
\begin{aligned}
\int\frac{F_\alpha'(u)}{u-a}\,du
&=-\frac{3n_\alpha}{2V_\alpha^3}\int_{-V_\alpha}^{V_\alpha}\frac{u}{u-a}\,du\\
&=-\frac{3n_\alpha}{2V_\alpha^3}\left[2V_\alpha+a\log\frac{a-V_\alpha}{a+V_\alpha}\right].
\end{aligned}
$$
Hence this species contributes
$$
\chi_\alpha=\frac{3\omega_{\mathrm p\alpha}^2}{k^2V_\alpha^2}
\left[1+\frac{s_\alpha}{2}\log\frac{s_\alpha-1}{s_\alpha+1}\right],\qquad
s_\alpha=\frac{\omega}{kV_\alpha}>1.
$$
For comparable species <number densities> and fixed charge ratios, $m_e\ll m_i$ implies $V_i/V_e=O(m_e/m_i)$ and $\omega_{\mathrm pi}^2/\omega_{\mathrm pe}^2=O(m_e/m_i)$. Since $\omega>kV_e$, the ions have $s_i\gg1$ and $\chi_i=-\omega_{\mathrm pi}^2/\omega^2+O(k^2V_i^2\omega_{\mathrm pi}^2/\omega^4)$. Their response is negligible to leading order in the <mass> ratio. The real <dispersion relation> is therefore
$$
\boxed{1+\frac{3\omega_{\mathrm pe}^2}{k^2V_e^2}\left[1+\frac{\omega}{2kV_e}\log\frac{\omega-kV_e}{\omega+kV_e}\right]=0.}
$$
This <degenerate-plasma dielectric response> treats a completely degenerate, nonrelativistic <Fermi gas> with classical particle trajectories. It does not include quantum recoil or a collision term. A finite ion response shifts the root slightly; the boxed equation is the stated leading electron approximation.
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