Eqs. (\[eq17\]) and (\[eq18\]) can be rewritten in dimensionless forms as $$I^{(R)}_{t,q} = \left ( \frac{mn_R V_0}{2\pi \hbar^3 k_F^2} \right )  \int_0^1 dx \frac{g_{t,q}(x) e^{-2 x a_R}}{\sqrt{1-x^2}(x+s)^2}
\label{eq19}$$ and $$I^{(B)}_{t,q}  = \left ( \frac{mn_B V_0}{2\pi \hbar^3 k_F^2} \right ) 
 \int_0^1 dx \frac{g_{t,q}(x)}{\sqrt{1-x^2}(x+s)^2}
\label{eq20}$$ where $a_R \equiv 2k_F d_R$ and $s=q_{TF}/2k_F$, and $g_t(x) = 2x^2$, $g_q(x) = 1$.

To proceed further analytically, we now make the assumption that, by definition, the remote dopants are far enough that the condition $a_R = 2k_F d_R \gg 1$ is satisfied. We note that for any arbitrarily large value of $d_R$, this “remote" impurity condition breaks down at a low enough carrier density such that $n \alt 1/ 8\pi d_R^2$. Thus, the distinction between ‘far’ and ‘near’ impurities in our 2-impurity model starts disappearing at very low carrier density. Putting $a_R \gg 1$ as well as $s=q_{TF}/2k_F \gg 1$ we can obtain the following asymptotic expressions for $I_{t,q}^{(R)}$ $$I^{(R)}_{t} = \left ( \frac{m n_R}{2\pi \hbar^3 k_F^2} \right ) \left ( \frac{2\pi e^2}{\kappa} \right )^2
\left ( \frac{2}{3s^2} \right ) \frac{1}{a_R^3},
\label{eq24}$$ $$I^{(R)}_{q} = \left ( \frac{m n_R}{2\pi \hbar^3 k_F^2} \right ) \left ( \frac{2\pi e^2}{\kappa} \right )^2
\frac{1}{a_R s^2},
\label{eq25}$$ $$I^{(B)}_{t} = \left ( \frac{m 