It seems that many Posts have solved the Fourier transform of the Coulomb interaction $V(r)=1/r$ which is $v(k) = 4\pi / k^2$. This is not my question.
I have come across the Fourier transform (Hansen & McDonald "Theory of Simple Liquids" Ch 10) $$\sum_i \sum_{j < i} Z_i Z_j e^2 V(|r_i - r_j |) = \frac{1}{2 L^3} \sum_k v(k) \left( \rho_k \rho_{-k} - \sum_i Z_i^2 e^2 \right)$$ where $L^3$ is the volume of the periodic box and $\rho_k$ is the $k$ Fourier mode of the charge density. I do not know how to manipulate the LHS to produce the RHS of this equation. I have given my attempt below, but I have used two guesses (marked in bold). If someone could confirm the guesses or deny them I would be appreciative.
Starting from the LHS we let our $j$ index run from $1,..., N$, $$ \frac{1}{2} \sum_i \sum_{j \neq i} Z_i Z_j e^2 V(|r_i - r_j |) $$ Let $\rho(r) = \sum_i Z_i e \delta(r-r_i)$ (and follow roughly the same steps as this stack post) $$ \frac{1}{2} \int_0^L dr_1 \int_0^L dr_2 V(|r_1 - r_2 |) \left( \rho(r_1) \rho(r_2) - \sum_i Z_i^2 e^2 \delta(r_1 - r_i) \delta(r_1 - r_2) \right) $$ This is the same format as the RHS, but I need to take the Fourier transform.
Consider the first term in the parenthesis and insert the Fourier expansion for $\rho(r) = \frac{1}{V} \sum_k \rho_k e^{i k \cdot r }$ $$ \frac{1}{2 L^2} \int_0^L dr_1 \int_0^L dr_2 V(| r_1 - r_2 |) \sum_k \rho_k e^{i k \cdot r_1 } \sum_{k'} \rho_{k'} e^{i k' \cdot r_2 }$$ Grouping terms $$ \frac{1}{2L^6} \sum_{k,k'} \rho_k \rho_{k'} \int_0^L dr_1 \int_0^L dr_2 V(| r_1 - r_2 |) e^{i k \cdot r_1 + k' \cdot r_2 } $$ There is no contribution when the plane waves are out of phases, so $k'=-k$ that way $k \cdot r_1 + k' \cdot r_2 = k \cdot (r_1 - r_2)$. This gives $$ \frac{1}{2L^6} \sum_{k} \rho_k \rho_{-k} \int_0^L dr_1 \int_0^L dr_2 V(| r_1 - r_2 |) e^{i k \cdot ( r_1 - r_2 ) } $$ Guessing $\int_0^L dr_1 \int_0^{L} dr_2 V(| r_1 - r_2 |) e^{i k \cdot ( r_1 - r_2 ) } = L^3 v(k)$, then $$ \frac{1}{2L^3} \sum_{k} \rho_k \rho_{-k} v(k) $$
Consider the second term in the parenthesis $$ \frac{1}{2} \int_0^L dr_1 \int_0^L dr_2 V(| r_1 - r_2 |) \sum_i Z_i^2 e^2 \delta(r_1 - r_i) \delta(r_1 - r_2)$$ Recognize that $\delta(r_1 - r_2) = \frac{1}{V} \sum_k e^{i k \cdot (r1 - r2)}$ $$ \frac{1}{2L^3} \sum_{k} \sum_i Z_i^2 e^2 \int_0^L dr_1 \int_0^L dr_2 V(| r_1 - r_2 |) \delta(r_1 - r_i) e^{i k \cdot ( r_1 - r_2 ) } $$ Guessing $\int_0^L dr_1 \int_0^L dr_2 V(| r_1 - r_2 |) \delta(r_1 - r_i) e^{i k \cdot ( r_1 - r_2 ) } = v(k)$ then $$ \frac{1}{2L^3} \sum_{k} v(k) \sum_i Z_i^2 e^2 $$
This expressions completes the derivation. But I am not confident that my guesses are right. Please confirm or deny!
