What does Griffith mean by adding a prime on integration variables? In the book "Introduction to Electrodynamics" by Griffith, the author mentions electric potential as a point function writes the equation for electric potential as

Then in a side note he write "To avoid any possible ambiguity, I should perhaps put a prime on the integration variable"

To what 'ambiguity' is he refering to and what how does adding the prime clarify it?
 A: The primed coordinate is what you are integrating over, whereas the unprimed coordinate is the point in space you are computing the potential for. If $r$ was left unprimed in $E(r)$ it could be seen as ambiguous whether we are talking about the variable of integration or not.
A: When talking about integrals, the variable of integration is "dummy", in the following sense. Suppose $f:\Bbb{R}\to\Bbb{R}$ is a function, then for any $x\in\Bbb{R}$,
\begin{align}
\int_0^xf(t)\,dt=\int_0^xf(s)\,ds=\int_0^xf(\xi)\,d\xi=\int_0^xf(\ddot{\smile})\,d \ddot{\smile}=\int_0^xf(@)\,d@=\int_0^xf(\sharp)\,d\sharp,
\end{align}
and so on. The actual symbol used does not matter except for $x$: what is completely nonsense notation is
\begin{align}
\int_0^xf(x)\,dx,
\end{align}
because the $x$ is being used in two places with different meanings, so it's just confusing and wrong. We can keep going: if you want, you can write
\begin{align}
\int_0^xf(y)\,dy= \int_0^xf(x')\,dx'=\int_0^{x}f(\tilde{x})\,d\tilde{x}.
\end{align}
Literally, any other symbol than $x$ can be used as the integration symbol. Same thing with line integrals
