Simplification of the exchange term in the Hartree-Fock treatment of an interacting paramagnetic electron gas In the Hartree-Fock treatment of the interacting electron gas it is assumed that solutions are planewaves of the form
\begin{equation}
\phi_{i}^{HF} = \phi_{{\bf{k}}\lambda} = \frac{1}{\sqrt{\Omega}}e^{i{\bf{k}}{\bf{r}}}\chi_{\lambda}
\end{equation}
where $\chi_{\lambda}$ is the spin function for spin up $\begin{pmatrix}1\\ 0 \end{pmatrix}$ and spin down $\begin{pmatrix}0\\ 1 \end{pmatrix}$. I am trying to understand the proof that the planewave functions are solutions to the self-consistent Hartree-Fock equations, which for the jellium model are reduced to
\begin{equation}
-\frac{\hbar}{2m}\nabla^{2}\phi({\bf{r}}, \sigma) + \sum_{\sigma'}\int V_{x}({\bf{r}}, {\bf{r'}}, \sigma, \sigma')\phi_{i}({\bf{r}'}, \sigma') d{\bf{r'}} = \epsilon_{i}\phi_{i}({\bf{r}}, \sigma)
\end{equation}
where
\begin{equation}
V_{x}({\bf{r}}, {\bf{r'}}, \sigma, \sigma') = - \sum_{j}^{occ} \frac{e^{2}}{|{\bf{r}}- {\bf{r'}}|}\phi_{j}^{*}({\bf{r'}}, \sigma')\phi_{j}({\bf{r}}, \sigma)
\end{equation}
Substituting the planewave solutions to the equation
\begin{equation}
\frac{\hbar^{2}k^{2}}{2m} \phi_{{\bf{k}}\lambda}({\bf{r}}) -  \frac{1}{\Omega^{3/2}}\sum_{k'< k_{F}}\sum_{\lambda'}\int d{\bf{r'}}\frac{e^2}{|{\bf{r}}- {\bf{r'}}|}e^{-i{\bf{k'r'}}}e^{-i{\bf{k'r}}}e^{-i{\bf{kr'}}}\delta_{\lambda' \lambda}\chi_{\lambda} = \epsilon_{\lambda}({\bf{k}})\phi_{{\bf{k}}\lambda}({\bf{r}})
\end{equation}
Assuming a paramagnetic electron gas, i.e. equal occupation of ${\bf{k}}$ spin up and ${\bf{k}}$ spin down orbitals the exchange term (second term on the left-hand side) can be simplified
\begin{equation}
-\frac{1}{\Omega}\sum_{k' < k_{F}} \int d{\bf{r'}}\frac{e^2}{|{\bf{r}}- {\bf{r'}}|} e^{-i({\bf{k}}-{\bf{k'}})({\bf{r}}- {\bf{r'}})}\phi_{{\bf{k}}\lambda}({\bf{r}})=\\
= \bigg( -\frac{1}{\Omega} \sum_{k' < k_{F}} \frac{4\pi e^{2}}{|{\bf{k}} - {\bf{k'}}|^{2}}\bigg)\phi_{{\bf{k}}\lambda}({\bf{r}})
\end{equation}
I am confused about the final step, as I am trying to integrate the expression with respect to $\bf{r'}$, but getting $\Gamma(0, i[({\bf{k}} - {\bf{k'}})({\bf{r}} - {\bf{r'}})])$ as a result. Could you please help me understand how the derivation in the last step is done? Thanks in advance.
 A: I don't know whether this is (completely) mathematical rigorous, but intuitively I'd say that the last equality can be derived by noting that
$$ -\Delta_\mathbf{r} \, \frac{1}{|\mathbf{r}-\mathbf{r}^\prime|}  = 4\pi\,\delta(\mathbf{r}-\mathbf{r}^\prime) \quad $$
and performing a Fourier transformation on both sides.
A: My best guess is that since for,
$$\int d{\bf{r'}}\frac{e^2}{|{\bf{r}}- {\bf{r'}}|} e^{-i({\bf{k}}-{\bf{k'}})\cdot ({\bf{r}}- {\bf{r'}})}$$
the integral is carried out over $\bf{r'}$ everything else is essentially fixed when we integrate, so lets make the substitution $\bf{w} = {\bf{r'}}- {\bf{r}}$ and $\bf{z} = {\bf{k'}}- {\bf{k}}$,
$$\int d{\bf{w}}\frac{e^2}{|\bf{w}|} e^{-i\bf{z}\cdot\bf{w}}$$
this is now simply the Fourier transform of the coulomb potential. The full derivation of that Fourier transform is quite long but here is a good example I found https://blog.cupcakephysics.com/electromagnetism/math%20methods/2014/10/04/the-fourier-transform-of-the-coulomb-potential.html, anyway we get.
$$\int d{\bf{w}}\frac{e^2}{|\bf{w}|} e^{-i\bf{z}\cdot\bf{w}} = \frac{4\pi e^{2}}{\vert \bf{z} \vert^{2}} = \frac{4\pi e^{2}}{\vert {\bf{k}}- {\bf{k'}} \vert^{2}}$$
and therefore.
$$\int d{\bf{r'}}\frac{e^2}{|{\bf{r}}- {\bf{r'}}|} e^{-i({\bf{k}}-{\bf{k'}})\cdot ({\bf{r}}- {\bf{r'}})} = \frac{4\pi e^{2}}{\vert {\bf{k}}- {\bf{k'}} \vert^{2}}$$
