If we have electric field as $$ \mathbf{E}\left(\mathbf{r}_{\mathrm{d}}, t\right)=\frac{1}{\varepsilon} \int_{\mathcal{V}} d \mathbf{r}^{\prime} \mathbf{K}\left(\mathbf{r}_{\mathrm{d}}-\mathbf{r}^{\prime}\right) \rho\left(\mathbf{r}^{\prime}, t\right) $$ where kernel is ${\bf K}\left(\mathbf{r}-\mathbf{r}^{\prime}\right)=\frac{1}{4\pi}\nabla\left(\frac{1}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}\right)$.
For Dirichlet boundary conditions for $\mathbf{r}_S$ on
- $\mathcal{S}=\{(x, y, z),-L / 2 \leq x, y \leq L / 2, z=0\}$,
- $\mathbf{r}_d$ in $\mathcal{V}=\{(x, y, z),-L / 2 \leq x, y \leq L / 2, z>0\}$,
- $n^{\prime}$ the coordinate normal to $\mathcal{S}$ and $\mathbf{r}_{\mathrm{d}}=(0,0,-d)$.
The kernel $\mathbf{K}(\mathbf{r})$ is expressed as a $2{\rm D}$ Fourier transform $$ \left(\mathcal{F}_{2 D}^{\mathbf{k}_{\|}}[\mathbf{K}(\mathbf{r})]=\int d \mathbf{r}_{\|} \mathbf{K}(\mathbf{r}) e^{-i \mathbf{k}_{\|} \cdot \mathbf{r}_{\|}}\right) $$ with respect to the surface coordinates ( yielding the Weyl representation of the Green's function ). $$\mathcal{F}_{2 D}^{\mathbf{k}_{\|}}\left[K_\nu(\mathbf{r})\right]=\frac{1}{2}\left[i\left(\delta_{\nu, x}+\delta_{\nu, y}\right) k_\nu /\left|\mathbf{k}_{\|}\right|+\delta_{\nu, z}\right] e^{-\left|\mathbf{k}_{\|}\right||z|}$$ Now my question is, How can we derive this 2D Fourier Transform of Kernel $\mathcal{F}_{2 D}^{\mathbf{k}_{\|}}[\mathbf{K}(\mathbf{r})]$ in a simple way? I have consulted this source, but I always get lost in the mathematics.
