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I want to calculate the electric potential of a uniformly charged wire with infinite length $\rho(\vec{r}') = \lambda \delta(x') \delta(y')$ with Green function $G(\vec{r}, \vec{r}') = \frac{1}{4\pi \epsilon_0}\frac{1}{|\vec{r} - \vec{r}'|}$.

$$\nabla^2 \phi(\vec{r}) = - \frac{\rho(\vec{r})}{\epsilon_0}$$ $$\phi(\vec{r}) = \int_V G(\vec{r}, \vec{r}') \rho(\vec{r}') d\vec{r}'$$

I get $$\phi(\vec{r}) = \frac{\lambda}{4\pi \epsilon_0} \int_{-\infty}^{\infty} \frac{1}{\sqrt{x^2 + y^2 + (z-z')^2}} dz'$$ which yields $$\phi{(\vec{r})} = \frac{\lambda}{4 \pi \epsilon_0} \left[ \ln|(z-z') + \sqrt{x^2 + y^2 + (z-z')^2}| \right]_{-\infty}^{\infty}$$ and not $$\phi(\vec{r}) = \frac{\lambda}{2\pi \epsilon_0} \ln r$$ what I should get if we assume that potential at $\phi(r=1) = 0$.

How do I get the known potential whit green function?

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These types of source distributions fail to give the correct answer via Green's function method as you have just shown. This can be appreciated if you consider how you find the green's function solution to begin with. You assume a point source and the calculate the response in the potential field using the differential equation. Then you add up all the point sources to get your original source distribution. When you don't have a regular source distribution, integrating over many dirac delta function's give you weird unphysical infinities as you have it here. However, I don't have enough pure math knowledge to explain rigorously why this doesn't work.

Here, you actually need to find the Green's function in 2-d by solving the Laplace's equation and then imposing the necessary boundary conditions at $x = 0$ and $y = 0$. This should give you the green's function in 2-D and the solution to your problem!

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