I'm working through a derivation of the Kohn Effect, as presented in Ziman's 'Principles of the Theory of Solids.' However, I find myself getting a somewhat different result. The book states that calculating
$$\epsilon(\mathbf{q}, \omega) = 1+ \frac{4\pi e^2}{q^2}\sum_{\mathbf{k}} \frac{f_0(\mathbf{k})-f_0(\mathbf{k+q})}{E_{k+q}-E_{k}-\hbar\omega + i\hbar\alpha}$$
at zero temperature (and $\omega = 0$) grants
$$\epsilon(\mathbf{q}, 0) = 1+\frac{4\pi e^2}{q^2}\frac{n}{\frac{2}{3}\mathcal{E}_F}\left[\frac{1}{2}+\frac{4k_F^2-q^2}{8k_Fq}\ln|{\frac{2k_F+q}{2k_F-q}}|\right]$$
Here is my work: $\epsilon(\mathbf{q}, 0) = 1+ \frac{4\pi e^2}{\frac{\hbar^2q^2}{2m}}\sum_{\mathbf{k}} \frac{f_0(\mathbf{k})-f_0(\mathbf{k+q})}{2\mathbf{k\cdot q}+q^2}$
$$\epsilon(\mathbf{q}, 0) = 1+ \frac{4\pi e^2}{\frac{\hbar^2q^2}{2m}}\sum_{\mathbf{k}}\left[ \frac{f_0(\mathbf{k}))}{2\mathbf{k\cdot q}+q^2}- \frac{f_0(\mathbf{k+q})}{2\mathbf{k\cdot q}+q^2} \right]$$
We may exchange $\sum_\mathbf{k} = \frac{1}{(2\pi)^3}\int\,d^3\mathbf{k}$ and define $\alpha = \frac{8\pi me^2}{\hbar^2q^2}$
$$\epsilon(\mathbf{q}, 0) = 1+ \frac{\alpha}{(2\pi)^3}\left[ \int\,d^3k\frac{f_0(\mathbf{k}))}{2\mathbf{k\cdot q}+q^2}- \int\,d^3k\frac{f_0(\mathbf{k+q})}{2\mathbf{k\cdot q}+q^2} \right]$$ Given the fact that $f_0(\mathbf{k})$ is zero outside the Fermi sphere, and one inside it allows us to select bounds on the integrals. For the second integral, a change of variables to $u = k+q$ allows us to center the Fermi sphere. \begin{align} \epsilon(\mathbf{q}, 0) &= 1+ \frac{\alpha}{(2\pi)^3}\left[\int d\Omega\int_0^{k_F}\,\frac{dk\;k^2\;f_0(\mathbf{k}))}{2\mathbf{k\cdot q}+q^2}-\int d\Omega\int_0^{k_F}\,\frac{du\;u^2\;f_0(\mathbf{u})}{2\mathbf{(u-q)\cdot q}+q^2} \right]\\ &= 1+ \frac{\alpha}{(2\pi)^3}\left[ \int d\Omega\int_0^{k_F}\,\frac{dk\;k^2}{2\mathbf{k\cdot q}+q^2}-\int d\Omega\int_0^{k_F}\,\frac{du\;u^2}{2\mathbf{(u\cdot q}-q^2} \right] \end{align} If we relabel our dummy integration variable $u$ as $k$, we may combine back into a single integral $$\epsilon(\mathbf{q}, 0) = 1+ \frac{\alpha}{(2\pi)^3}\int d\Omega\int_0^{k_F}\,k^2\,dk\left[\frac{1}{2\mathbf{k\cdot q}+q^2}-\frac{1}{2\mathbf{k\cdot q}-q^2} \right]$$
We may select our coordinate system so that $\mathbf{q}$ lies along the $z$ axis. \begin{align} \epsilon(\mathbf{q}, 0) -1 &= \alpha A\int d\Omega\int_0^{k_F}\,k^2\,dk\left[\frac{1}{2kq\cos\theta+q^2}-\frac{1}{2kq\cos\theta-q^2} \right]\\ &= \frac{\alpha}{(2\pi)^3}\int d\Omega\int_0^{k_F}\,k^2\,dk\left[\frac{2kq\cos\theta-q^2}{(2kq\cos\theta)^2-q^4}-\frac{2kq\cos\theta+q^2}{(2kq\cos\theta)^2)-q^4} \right]\\ &= \frac{\alpha}{(2\pi)^3}\int d\Omega\int_0^{k_F}\,k^2\,dk\left[\frac{-2q^2}{(2kq\cos\theta)^2-q^4} \right]\\ &= \frac{2\alpha}{(2\pi)^3}\int d\Omega\int_0^{k_F}\,dk\left[\frac{-k^2}{(2k\cos\theta)^2-q^2} \right]\\ & = \frac{4\pi\alpha}{(2\pi)^3}\int_1^{-1}\int_0^{k_F}\,d\cos\theta\,dk\left[\frac{k^2}{(2k\cos\theta)^2-q^2} \right]\\ &= \frac{4\pi\alpha}{(2\pi)^3}\int_1^{-1}\int_0^{k_F}\,d\cos\theta\,dk\left[\frac{k^2}{(2k\cos\theta)^2-q^2} \right]\\ & = \frac{4\pi\alpha}{(2\pi)^3}\int_0^{k_F}\,dk\left[\frac{k\left(\tanh^{-1}\left(\frac{2k}{q}\right)- \tanh^{-1}\left(\frac{-2k}{q}\right)\right)}{2q} \right] \end{align} Since $\tanh^{-1}(x) = \frac{1}{2}[\ln(1+x)-\ln(1-x)]$ \begin{align} \epsilon(\mathbf{q}, 0) -1 &= \frac{8\pi^2 e^2}{\frac{\hbar^2q^2}{2m}}\frac{1}{(2\pi)^3}\int_0^{k_F}\,dk\frac{k\left[\ln\left(\frac{q+2k}{q-2k}\right)- \ln\left(\frac{q-2k}{q+2k}\right)\right]}{q}\\ & = \frac{2\pi\alpha}{(2\pi)^3}\int_0^{k_F}\,dk\frac{k\ln\left(\frac{q+2k}{q-2k}\right)^2}{q} \\ &= \frac{\pi\alpha}{(2\pi)^3}\int_0^{k_F}\,dk\frac{k\ln\left(\frac{q+2k}{q-2k}\right)}{q} \\ \epsilon(\mathbf{q}, 0)&=1 + \frac{4\pi^2 e^2}{\frac{\hbar^2q^2}{2m}}\frac{1}{(2\pi)^3}\frac{2k_Fq+(4k_F^2-q^2)\ln\left|\frac{q+2k}{q-2k}\right|}{8q} \end{align} Thanks to @Meng Cheng for the constant, but problems remain.
