How to derive the macroscopic dielectric function? I'm following Matteo Gatti's slides to repeat the derivation of macroscopic dielectric function $\epsilon_M$:
$$\epsilon_M=\dfrac{1}{\epsilon^{-1}_{\vec{G}=0,\vec{G}'=0}(\vec{q},\omega)}.$$
On page 41 one needs to perform the following derivations:

Here are my derivations:

Note:


*

*$V_{tot}$ and $V_{ext}$ is a periodic potential in real space.


One can see to obtain the final result I have introduced a delta function, is it reasonable? Or do I miss something to arrive at the final result?
 A: Perhaps it's missing a prefactor? If you split the sum as
$$
\sum_{\textbf{q}'} \sum_{\textbf{G}'} V_{ext}(\textbf{q}' + \textbf{G}', \omega) \int d\textbf{r}_1 d\textbf{r}_2 e^{-i(\textbf{q}+\textbf{G})\cdot \textbf{r}_1} \epsilon^{-1}(\textbf{r}_1, \textbf{r}_2, \omega) e^{i(\textbf{q}'+\textbf{G}')\cdot \textbf{r}_2},
$$
then you can shift $\textbf{G'} \rightarrow \textbf{G}' - \textbf{q}' + \textbf{q}$, which yields
$$
\sum_{\textbf{q}'} \sum_{\textbf{G}'} V_{ext}(\textbf{q} + \textbf{G}', \omega) \int d\textbf{r}_1 d\textbf{r}_2 e^{-i(\textbf{q}+\textbf{G})\cdot \textbf{r}_1} \epsilon^{-1}(\textbf{r}_1, \textbf{r}_2, \omega) e^{i(\textbf{q}+\textbf{G}')\cdot \textbf{r}_2} \\
= \sum_{\textbf{q}'} \sum_{\textbf{G}'} V_{ext}(\textbf{q} + \textbf{G}', \omega) \epsilon^{-1}_{\textbf{G}\textbf{G}'}(\textbf{q}, \omega).
$$
The sum over $\textbf{q}'$ is simply the number of $\textbf{q}$-points, that's why I think it's missing a prefactor.
A: I found the problem can be solved if one looks at the definition as following:
\begin{align}
f(\vec{q}+\vec{G},\vec{q}'+\vec{G}';\omega)
& =
\int d\vec{r}d\vec{r}' 
e^{-i(\vec{q}+\vec{G})\cdot\vec{r}}
f(\vec{r},\vec{r}';\omega)
e^{-i(\vec{q}'+\vec{G}')\cdot\vec{r}}
= f_{\vec{G},\vec{G}'}(\vec{q},\vec{q}';\omega) \\
& \Rightarrow f_{\vec{G},\vec{G}'}(\vec{q},\omega) \equiv f_{\vec{G},\vec{G}'}(\vec{q},\vec{q}';\omega) \delta_{\vec{q},\vec{q}'}.
\end{align}
