Boundary integral method to solve Poisson equation

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Suggest how to solve Poisson equation \begin{equation} σ ∇^2 V = - I δ(x-x_s) δ(y-y_s) δ(z-z_s) \nonumber \end{equation} by using the boundary integration method to calculate the potential $V(r,z)$ with the help of changing the Poisson equation into cylindrical polar co ordinates? Where $V$ is electric Potential (scalar)[volts]. Solution depends only on $r$ and $σ$ is constant. $r$ is horizontal coordinate(direction) $0\leq r \leq \infty $ and $z$ is vertical co ordinate(direction) $-\infty\leq r \leq \infty $. What boundary conditions are appropriate?

=> Poisson equation \begin{equation} σ ∇^2 V = - I δ(x-x_s) δ(y-y_s) δ(z-z_s) \nonumber \end{equation} can be written as \begin{equation} σ ∇^2 V = - I δ(X-X_0) \nonumber \end{equation}

where $I$ is current or source intensity and we take $I$ as $1$.

$δ(x-x_s) δ(y-y_s) δ(z-z_s)$ is source term.

$ δ$ is Dirac delta function.

The Laplacian in cylindrical coordinates is:

\begin{equation} - ∇^2 V= \frac{1}{r} \frac{\partial}{\partial r}[ r \frac{\partial v}{\partial r}] + \frac{1}{r^2}\frac{\partial^2 V}{\partial \theta^2}+\frac{\partial^2 V}{\partial z^2} \end{equation} $\theta$ is not relevant, so it goes to zero.

\begin{equation} - ∇^2 V= \frac{1}{r} \frac{\partial}{\partial r}[ r \frac{\partial v}{\partial r}] +\frac{\partial^2 V}{\partial z^2} \end{equation} cylindrical polar co ordinates is
\begin{equation} \int ^{r= ∞}_ {r=0} \int^ {z=∞} _{ z= - ∞} {δ(r) δ(z-z_s)} dz dr. \end{equation} where $r_s$ is $0$.

( green function) $G_{3D}= \frac{- 1}{4πσ r} = \frac{∂G}{∂r} + \frac{1}{r} ≈ 0$ on boundary $\Gamma_\infty$

where $r$ is $|X-X_0|$. $\sigma$ is constant with respect to one of the variables $r$ or $z$ but solution depend upon $r$ only.

No current flux on a surface with normal $n$: $ σ \frac{∂V}{∂n} =0$ [ Neumann boundary condition ]

now my question is how do I tackle the problem by using the boundary integral method with the help above equations.

can anyone please help me.