= Solution
The <static Lindhard dielectric function> describes the real, zero-frequency <linear response> of a uniform <free electron gas>. Here $\epsilon_1=\operatorname{Re}\epsilon$, $\mathbf r$ is the position operator, and $\mathbf q$ is the perturbation <wavevector>, $q=|\mathbf q|$, $V$ the normalization volume, $e>0$ the elementary charge, and $\epsilon_0$ the vacuum <permittivity>. The normalized one-electron states are $\psi_{\mathbf k}=V^{-1/2}e^{i\mathbf k\cdot\mathbf r}$ with $E_{\mathbf k}=\hbar^2k^2/(2m_e)$; $f(E)$ is their <Fermi-Dirac distribution> occupation. Both momentum sums exclude spin; the prefactor two counts the two equal <spin> populations. The density-operator matrix element is $\delta_{\mathbf k',\mathbf k+\mathbf q}$. Coincident energy denominators are interpreted by the static limit or <Cauchy principal value>.
A weak total <electric potential> $\phi_{\rm tot}$ perturbs an electron by $-e\phi_{\rm tot}$. First-order state mixing produces $\delta n=e\Pi_0(q)\phi_{\rm tot}$, where
$$
\Pi_0(q)=\frac2V\sum_{\mathbf k}\frac{f(E_{\mathbf k})-f(E_{\mathbf k+\mathbf q})}{E_{\mathbf k+\mathbf q}-E_{\mathbf k}}.
$$
The induced charge is $-e\delta n$. Self-consistent <Poisson equation> screening then gives $\phi_{\rm ext}=\epsilon\phi_{\rm tot}$, with $\epsilon=1+e^2\Pi_0/(\epsilon_0q^2)$. This is the <random phase approximation>: use the independent-electron response to the self-consistent total field.<Lecture Notes for Quantum Matter by Steven H. Simon>[Chapter8,§8.3, printed pp.123–126: density response and the Lindhard function]{ref}
At zero temperature $f(E_{\mathbf k})=\Theta(k_F-k)$. Shifting the integration variable in the second occupation term gives
$$
\epsilon-1=\frac{4m_ee^2}{\epsilon_0\hbar^2q^2}\int_{k<k_F}\frac{d^3k}{(2\pi)^3}\left(\frac1{q^2+2\mathbf k\cdot\mathbf q}+\frac1{q^2-2\mathbf k\cdot\mathbf q}\right).
$$
The two angular integrals are equal. Using the supplied integral, with $x=q/(2k_F)$, gives
$$
\boxed{\epsilon(q,0)=1+\frac{\kappa^2}{q^2}F(x),\qquad
\kappa^2=\frac{m_ee^2k_F}{\epsilon_0\pi^2\hbar^2},\qquad
F(x)=\frac12+\frac{1-x^2}{4x}\log\left|\frac{1+x}{1-x}\right|.}
$$
The limits are $F(0)=1$, $F(1)=1/2$ and $F(x)\sim1/(3x^2)$ for large $x$. The first recovers <Thomas-Fermi screening>; the logarithm produces the <Kohn singularity of static Lindhard screening> at $q=2k_F$.
Use the <Fourier transform> convention $\rho(\mathbf q)=\int d^3r\,e^{-i\mathbf q\cdot\mathbf r}\rho(\mathbf r)$. A point charge $Ze$ has total screened charge $Ze/\epsilon(q)$ and induced charge $-Ze[1-\epsilon(q)^{-1}]$. Angular inversion therefore gives the <radial induced-charge screening kernel>
$$
\rho_{\rm ind}(r)=-\frac{Ze}{r}\int_0^\infty dq\,g(q)\sin(qr),\qquad
\boxed{g(q)=\frac{q}{2\pi^2}\left[1-\frac1{\epsilon(q)}\right]=\frac{q\kappa^2F(q/2k_F)}{2\pi^2[q^2+\kappa^2F(q/2k_F)]}.}
$$
In particular, $\rho_{\rm ind}(\mathbf q=0)=-Ze$: the total screening charge is negative although its local density can oscillate in sign.
Integrate the sine transform by parts twice; $g(0)=g(\infty)=0$ removes the boundary terms. For the <Friedel density tail from a derivative singularity>, write $Q=2k_F$ and $g''(q)\sim C/(q-Q)$. Then
$$
\rho_{\rm ind}(r)=\frac{Ze}{r^3}\int_0^\infty g''(q)\sin(qr)\,dq
\sim\frac{ZeC\pi}{r^3}\cos(Qr).
$$
Indeed, near $Q$ the <principal value> transform is $\int_{-\infty}^{\infty}\sin[(Q+s)r]\,ds/s=\pi\cos(Qr)$ for $r>0$. Smooth terms do not remove this leading oscillatory long-range contribution. Thus \b[$\rho_{\rm ind}\propto\cos(2k_Fr)/r^3$].
For a three-dimensional, single-valley gas with two <spin> states, the <Friedel oscillation wavelength from carrier density> is
$$
\boxed{k_F=(3\pi^2n)^{1/3}=3.90\times10^8\,\mathrm{m^{-1}},\qquad
\lambda=\frac{\pi}{k_F}=8.06\,\mathrm{nm}.}
$$
Low-temperature <scanning tunneling spectroscopy> near exposed dopants can map the oscillatory <local density of states>, with a Fourier feature near $2k_F$ at the <Fermi energy>. An energy-resolved tunnelling image measures local spectral weight, not directly the bulk energy-integrated charge: comparing occupied-bias maps and accounting for the surface geometry is necessary. In actual <GaAs>, the background <relative permittivity> and band <effective mass> change screening strengths; they do not change this wavelength at the stated density.
\Image[/past-exam-of-the-advanced-quantum-condensed-matter-physics-course-of-the-university-of-cambridge/2013/paper-1-aqc2013-screening-and-tail.png]
{title=Static Lindhard screening has a singular slope at twice the Fermi wavevector and an oscillatory far-field density tail}
{source=paper-1-aqc2013-screening-and-tail.py}
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