I need help how to set up this integral
$$V(\mathbf r)=\frac{1}{4\pi\epsilon_0} \int_L \frac{\rho'_l}{\lvert \mathbf r – \mathbf{r'} \rvert}\mathrm{d}l'.
$$
I have a uniform line charge along the $z$-axis and want to calculate the electric potential between two points $A=(r_A,\phi_A,0)$ and $B=(r_B,\phi_B,0)$ (cylindrical coordinates).
Solution:
The line charge is aligned along the $z$-axis and the the source vector is
$\mathbf{r}=z\hat{\mathbf z}$ and the field vector is
$\mathbf{r'}=r'\hat{\mathbf r}+z'\hat{\mathbf z}$
so $$\lvert \mathbf{r}-\mathbf{r'} \rvert=\sqrt{(r')^2+(z-z')^2}.$$
I integrate along $z'$ from $-\infty$ to $\infty$
\begin{align}
V(\mathbf r) &=\frac{1}{4\pi\epsilon_0} \int_L \frac{\rho'_l}{\lvert \mathbf r – \mathbf{r'} \rvert}\mathrm{d}l' \\
&=\frac{\rho_l}{4\pi\epsilon_0} \int_{-\infty}^{\infty}\frac{1}{\sqrt{(r')^2+(z-z')^2}}\mathrm{d}z'\\
&=\frac{1}{4\pi\epsilon_0} \big [ \ln(z-z'+\sqrt{(r')^2+(z-z')^2})\big ]^{\infty}_{-\infty}\\
&= -\infty +\infty
\end{align}
The integral is indeterminate and I'm stuck here.
Are the vectors wrong, the limits? What have I missed?
Thanks!
If I calculate the line integral of the electric field I find the correct potential (however, I want to calculate with the integral above).
I integrate in the radial direction $\mathrm{d}\mathbf{r}$ from $r_A$ to $r_B$. The electric field is $\mathbf{E}(\mathbf{r})=\frac{\rho_l}{2\pi\epsilon_0 r}\hat{\mathbf{r}}$ so
\begin{align}V(\mathbf{r})&=-\int_L \mathbf{E}(\mathbf r) \cdot \mathrm{d}\mathbf{l}\\&=\frac{\rho_l}{2\pi\epsilon_0}\int_{r_A}^{r_B}\frac{1}{r}\mathrm{d}r=\frac{\rho_l}{2\pi\epsilon_0}\ln{\frac{\rho_B}{\rho_A}}\end{align}
Best Answer
Well actually, even if the general integral formula does not work since the charge is infinitely extended as correctly pointed out in the other answer, there is a way out using a suitable cutoff and a, say, renormalization procedure. Integrate from $-L$ to $L$ instead of from $-\infty$ to $+\infty$. \begin{align} V(\mathbf r) &=\frac{1}{4\pi\epsilon_0} \int_{[-L,+L]} \frac{\rho'_l}{\lvert \mathbf r - \mathbf{r'} \rvert}\mathrm{d}l' \\ &=\frac{\rho_l}{4\pi\epsilon_0} \int_{-L}^{+L}\frac{1}{\sqrt{(r')^2+(z-z')^2}}\mathrm{d}z'\\ &=\frac{1}{4\pi\epsilon_0} \left[ \ln \left|z-z'+\sqrt{(r')^2+(z-z')^2}\right|\right]^{+L}_{-L}\\ &= -\frac{\rho_l}{4\pi\epsilon_0} \ln \frac{r^2}{2L^2} + O(1/L) \end{align} The result, using $\ln(a^n) = n \ln a$ and $\ln(a/b)= \ln a - \ln b$ can be re-written as $$V(r)= -\frac{\rho_l}{2\pi\epsilon_0} \ln \frac{r}{\ell} + O(\ln (L/\ell))$$ for an arbitrarily fixed length unit $\ell$. We now renormalize the result just by dropping the final divergent term $ O(\ln(L/\ell))$ obtaining $$V(r)= -\frac{\rho_l}{2\pi\epsilon_0} \ln \frac{r}{\ell}$$ where you see that the so-called finite-renormalization ambiguities are all encapsulated in the arbitrary length scale $\ell$. However the gradient of the found function is not affected by the arbitrary choice of $\ell$.
If you compute (minus) the gradient of the found $V$ you have the correct electric field also obtained by Gauss law and symmetry arguments: $$\mathbf{E}(\mathbf{r})=\frac{\rho_l}{2\pi\epsilon_0 r}\hat{\mathbf{r}}\:.\tag{1}$$
There is a physical interpretation of the outlined procedure. You may consider a class of charged segments of length $L_n$ with $L_n \to + \infty$ as $n\to +\infty$ and compute $V_{L_n}({\bf r})$ for every $L_n$ where ${\bf r}$ is a fixed point in space. The crucial observation is that, since the potential is always defined up to an additive constant, at each step you can re-define $V_{L_n}$ subtracting a constant $C_n$ which logarithmically diverges as $L_n\to +\infty$. This limit procedure leads to the result found above.
There is no way to avoid the introduction of the arbitrary scale $\ell$ because the functional form of the electric field (1) says that any potential function producing it as gradient must have a logarithmic form. However the problem is that the argument of the logarithm is $r$ which has the dimension of a length. So, to make this argument dimensionless, one is forced to fix some scale $\ell$ arbitrarily which, however does not affect the electric field as it disappear when computing the derivative.