I wanted to compute the electric potential of an infinite charged wire, with uniform linear density $\lambda$.
I know that the potential can easily be calculated using Gauss law, but I wanted to check the result using the horrifying integral (assuming the wire is in the $z$ axis)
\begin{align}
&\phi({\bf r})=\int_{-\infty}^{+\infty}dz'
\frac{\lambda}{\sqrt{x^2+y^2+(z-z')^2}}
\end{align}
The antiderivative of the integrand is
\begin{equation}
g({\bf r},z')=-\lambda\log\left( \sqrt{x^2+y^2+(z-z')^2} + (z-z') \right)
\end{equation}
So the potential is:
\begin{equation}
\phi({\bf r})=\lim_{z' \rightarrow +\infty}g({\bf r},z')
-\lim_{z' \rightarrow -\infty}g({\bf r},z')
\end{equation}
The first limit converges:
\begin{equation}
\lim_{z' \rightarrow +\infty}g({\bf r},z')=
\lim_{z' \rightarrow +\infty}
-\lambda\log\left( \sqrt{x^2+y^2+(z-z')^2} + (z-z') \right)=
-\lambda\log(\sqrt{x^2+y^2})
\end{equation}
But the second limit diverges!
\begin{equation}
\lim_{z' \rightarrow -\infty}g({\bf r},z')=
\lim_{z' \rightarrow -\infty}
-\lambda\log\left( \sqrt{x^2+y^2+(z-z')^2} + (z-z') \right)=\infty
\end{equation}
Then $\phi=\infty$, which is absurd. So, why this calculation went wrong? Thanks
Best Answer
First, look at your integral for large $z'$. It goes as $1/z'$, which is divergent. So your math is fine. That's not a problem, however.
Remember that potentials are determined up to an additive constant. According to Gauss law, you should get that the field falls off as $1/\sqrt{x^2+y^2} = 1/r$, which means that the potential is indeed a logarithm, like what you have. That infinity is your "free constant" of the potential and is an artefact of the "infinitely long wire" assumption.