From $\int_Vd^3x \rho(\vec x)\mathrel{\mathop{=}\limits^!}Q_V$ and by use of Dirac's delta distribution one finds that the charge density for the uniformly charged infinite plane is $\sigma\cdot\delta(z)$ and that the charge density for the uniformly charged infite wire is $\kappa/\pi\cdot\delta(r)/r$, where $\sigma/\kappa$ are some constants with dimension charge per area/lenght and $r$ denotes the orthogonal distance from the wire.

I want to calculate the electric potential associated to both densities, the general ansatz from coulomb superposition is:
$\qquad\varphi(\vec x) = \frac1{4\pi\varepsilon_0}\int_{R^3}d^3r \frac{\rho(\vec r)}{\|\vec r-\vec x\|}.$

Both times I try to do the integral and end up with $\forall\vec x:\varphi(\vec x)=+\infty$.

Components: $\vec x = (a,b,h)$.
$4\pi\varepsilon_0/\sigma\cdot\varphi(\vec x)= \int_{-\infty}^\infty dx\int_{-\infty}^\infty dy\int_{-\infty}^\infty dz \frac{\delta(z)}{ \sqrt{(x-a)^2+(y-b)^2+(z-h)^2}} $
$\quad= \int_{-\infty}^\infty dx\int_{-\infty}^\infty dy \sqrt{(x-a)^2+(y-b)^2+h^2}^{-1}$
which can be reparametrised using polar coordinates and choosing new variables $(x-a)/h$ and $(y-b)/h$:
$\quad=2\pi h\int_0^\infty dr\;r\sqrt{1+r^2}^{-1} = 2\pi h\left.\sqrt{1+r^2}\right|_{r=0}^\infty = \infty$ while this result is independent of $a,b,h$.

Components: $\vec x = s\hat\rho(\alpha) +h\hat z$.
$4\pi^2\varepsilon_0/\kappa\cdot\varphi(\vec x)= \int_0^\infty r\;dr\int_0^{2\pi}d\phi\int_{-\infty}^\infty dz\; \delta(r)/\big(r\sqrt{\|\vec x-\vec r(r,\phi,z)\|}\big) $
after the integral over $r$ the integrand is not anymore dependent on $\phi$ and note that $\forall a>0,f:\int_0^adx\,f(x)\delta(x)=f(0)/2$:
$\quad = \pi\int_{-\infty}^\infty dz\; \sqrt{s^2+(z-h)^2}^{-1}=\left.\pi\, \mathrm{asinh}(w)\right|_{-\infty}^\infty=\infty$,
again transforming $w=(z-h)/s$.

Literature results:
Plane: $\varphi(\vec x) = \varphi(h) = C-|h|\sigma/(2\varepsilon_0)$
Wire: $\varphi(\vec x)=\varphi(s)=C-\kappa\ln(s)/(2\pi\varepsilon_0)$


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