What is the meaning of the lower limit in $V(r)=-\int_{\infty}^r E(r^{\prime})\,dr^{\prime}$? This is a follow up to my previous question I asked some time ago, I was doing some revision and realized I had no idea what eqn. $(1)$ actually meant (see below).
My question is regarding the solution to part b) of this homework-style question:


An infinite filled cylinder of radius $a$ contains a 3D charge density $\rho$. A thin-walled hollow cylinder of radius $b \gt a$ centered on the same axis surrounds it and contains a charge with the same charge per unit length, but with the opposite sign.
a) Compute the electric field $\vec E$ everywhere.
b) Compute the electrostatic potential $V$, defined by $\vec E = −\nabla V$, everywhere, subject to $$V(r \to \infty) = 0 \tag{A}$$


I only have a question regarding the solution to part b). But, unfortunately, for my question to make sense, I will have to typeset the full solutions to a) & b):

The arrangement is shown above and the solution to part a) is

By symmetry, the electric field is radial everywhere. For $r \lt a$, Gauss’s theorem in a cylinder of unit length (or use a length $L$ if preferred) gives $$\oint \vec E \cdot d\vec S =\frac{Q}{\varepsilon_0}\implies E 2 \pi r=\frac{\pi r^2 \rho}{\varepsilon_0}\implies E=\frac{\rho r}{2 \varepsilon_0}$$
For $a \lt r \lt b$ the charge enclosed is $\pi a^2 \rho$, so $$E 2 \pi r=\frac{\pi a^2 \rho}{2 \varepsilon_0}\implies E=\frac{\rho a^2}{2 r \varepsilon_0}$$
For $r \gt b$ the charge enclosed is zero, so $$E=0$$

The image below is just for clarity and it shows the cylinder as viewed from its cross-section:


The solution to part b) is

In cylindrical polars, the radial gradient is $\frac{\partial V}{\partial r}$, so $$V(r)=-\int_{\infty}^r E(r^{\prime})\,dr^{\prime}\tag{1}$$
Evidently $V=0$ for $r \gt b$.
For $a \lt r \lt b$,
$$V(r)=-\int_{b}^r \frac{\rho a^2}{2 r^{\prime} \varepsilon_0} \, dr^{\prime}=-\frac{\rho a^2}{2 \varepsilon_0}\ln\left(\frac{r}{b}\right)\tag{2}$$
For $r \lt a$,
$$V(r)=-\int_a^r \frac{\rho r^{\prime}}{2 \varepsilon_0} \,dr^{\prime}-\color{blue}{\frac{\rho a^2}{2 \varepsilon_0} \ln\left(\frac{a}{b}\right)} = \frac{\rho\left(a^2-r^2 \right)}{4 \varepsilon_0}-\frac{\rho a^2}{2 \varepsilon_0} \ln\left(\frac{a}{b}\right)\tag{3}$$


I have two questions:
1.I am questioning the presence of the blue term in $(3)$ above, for $r\lt a$. In order for $(3)$ to be true, then we must have $$V(r)=-\frac{\rho}{2\varepsilon_0}\int_a^r r^{\prime} \,dr^{\prime}-\frac{\rho a^2}{2 \varepsilon_0}\int_b^a \frac{1}{r^{\prime}}dr^{\prime}$$ Now the first integral makes sense, since $r \lt a$ but the second integral is in range $b \lt r \lt a$, but we already know that $r\lt a$, so why are we considering the contribution from eqn $(2)$? Or put in another way; why is the answer not just $$V(r)=-\int_a^r \frac{\rho r^{\prime}}{2 \varepsilon_0} \,dr^{\prime}=\frac{\rho\left(a^2-r^2 \right)}{4 \varepsilon_0}\color{red}{?}$$


*Why is there an $\infty$ in the lower limit of $(1)$?
I know that I can swap the limits and change sign, such that: $V(r)=-\int_{\infty}^r E(r^{\prime})\,dr^{\prime}=\int_{r}^\infty E(r^{\prime})\,dr^{\prime}$. But the form of electrostatic potential $\propto\,\frac{1}{r}$, and with the given boundary condition $V(r \to \infty) = 0$ $\big(\,\mathrm{from}\, \text{eqn}\,(A)\big)$. I understand eqn. $(A)$ because the potential is zero as the electric field, $\vec E$, tends to zero as $r \to \infty$. It was my understanding that closer objects (charges) have a larger potential (more negative, as the system is bound). But what I just can't figure out is why the solution is computed with the lower limit as infinity, this just seems nonsensical and looks like the integral is being carried out in the wrong order (from infinitely far away, to up-close at $r$). It seems to make more sense to integrate over $0 \lt r \lt \infty$. Why is it being done in this reverse order?

Images shown in this question were taken from this pdf by MIT

N.B
I wasn't sure whether this question belonged to M.S.E or P.S.E.

Update #1
I'm sorry for the confusion with my first question, I just noticed there were typo(s) in it, apologies for this - it's now fixed, thanks.

Update #2
Looking at current answers so far, my second question is addressed very nicely, I'm still very confused about the integration limits: $$\int_a^r +\int_b^a$$ in my first question. I think it should just be $$\int_a^r$$ since $r \lt a$. Can anyone please help me to understand this better?
 A: 
But what I just can't figure out is why the solution is computed with the lower limit as infinity

Why not? It makes perfect sense if you just say it: I know the electric potential at radial infinity (zero) and I know how it changes when I move a little bit along the radial coordinate ($\frac{\partial V}{\partial r}=-E$), so I can find the potential at any point by starting with zero at infinity and summing up all the increments by integration ($V(r)=\int_\infty^r\frac{\partial V}{\partial r}\,dr'$). If getting from infinity to $r$ with tiny steps bothers you, then note that for any finite $\alpha$, we also have $V(r)=V(\alpha)-\int_\alpha^rE(r')\,dr'.$ Take $\lim_{\alpha\to\infty}.$ $V(r)$ doesn't change, and $V(\alpha)$ goes to zero, so you're left with the definition of an infinite integral bound $$\begin{aligned}V(r')&=\lim_{\alpha\to\infty}-\int_\alpha^rE(r')\,dr'\\&=-\int_\infty^rE(r')\,dr'.\end{aligned}$$
The issue with trying to start the integration at $r=0$ is that you don't know the voltage there. You could declare $r=0$ to be at $V=0,$ but then $V(r\to\infty)\neq0$ and so you haven't quite found $V$ according to the book's condition. Remember that the electric potential is not completely determined by the system. You have to make an arbitrary choice for its value somewhere, and then you have to be able to extend that choice to cover all of space. In this case the choice is made for you: make radial infinity go to zero potential. Then it's only natural to start from there, where you know the potential, and integrate inwards to find the potential everywhere else. If you're really dead set on it, note you can write $$V(r\to\infty)=V(r)-\int_r^\infty E(r')\,dr'=0$$ for all $r.$ Rearrange for $V(r)$ and we're back to where we started.
As to your first question, again, remember that $\int_a^bE(r)\,dr$ gives you $V(b)-V(a),$ not necessarily the actual value of $V(r)$ for any $r.$ Adding up a bunch of tiny differences still leaves you with difference. You have to choose one of the bounds to be a place where you already know $V$ and then add the correction produced by integration to that. So eqn (3) consists of an integral representing $V(r)-V(a)$ and adds $V(a)$ to find $V(r).$
A: Sirius, the main question remaining is about the intuition of this. And I believe the issue is with intuiting electric potential.
Electric potential $V$ is the potential energy stored in the field per unit charge. Another way to say it is that $V$ is the amount of energy that would be done in the form of work to bring a unit-charge to the location of said potential.
The maximum amount of work is to move a unit-charge from infinity. That way the force and displacement are in the same direction so that work $W= \int F(x)dx$ is maximized. And that work formula, divided by one coulomb and using a one-coulomb charge when calculating the force $F(x)= E(x) q$ ... equals electric potential. Remember $q=1$ and the negative in the potential equation is because $r = -x$ (the positive work, assuming $E>0$, to bring the particle in is the opposite direction). Hopefully this makes a lower limit of $\infty$ seem less weird.
Equations 2 and 3 above are bringing a charge in from infinity.
For Eqn 2, there’s no field or work from infinity down to $b$.
Equation 3 is bringing it from $\infty$ to a point $r<a$ as follows:
It takes zero work to go from $\infty$ to $b$, and the second term in Eqn 3 below is the work per unit-charge to go from $b$ to $a$ (or equivalently from $\infty$ to $a$), and the first term is to then to go from $a$ to $r<a$. The second term in Eqn 3 is just Eqn 2 for $r=a$, ie bringing the charge from $\infty$ to $a$. $$V(r)=-\int_a^r \frac{\rho r^{\prime}}{2 \varepsilon_0} \,dr^{\prime}-\frac{\rho a^2}{2 \varepsilon_0} \ln\left(\frac{a}{b}\right)$$
A: First, "the potential at $\vec r_p$" really means the potential difference between a reference point and the point $\vec r_p$.  If there is no charge at infinity, the reference point is by convention at infinity.  If there are charges at infinity, another reference point must be chosen.
I will illustrate the distinction with the example of a thin stick.  You can work out the details for the thick cylinder.
So consider a finite thin stick, placed along
$\hat{\mathbf{x}}$ between $-\ell$ and $\ell$ (total length $2\ell$), and carrying a linear charge density $\lambda$.  Let's compute "the potential" at a point $\vec r_p=z_p\hat{\mathbf{z}}$ due to this stick.
We break up the stick is small portions of size $dx_s$.  The portion of this size located at $x_s$ contains a small amount of charge $dq_s=\lambda dx_s$ so "the potential" due to this small amount of charge is just
\begin{align}
dV=\frac{dq_s}{4\pi\epsilon_0 \sqrt{{x_s}^2 +{z_p}^2}}\, , \tag{1}
\end{align}
where $\sqrt{x_s^2+z_p^2}$ is the distance between the little piece of stick at $x_s$ and your point of interest $\vec r_p=z_p\hat{\textbf{z}}$.
Here, I'm using the expression for "the potential" of a point charge of magnitude $dq_s$.  I can do this because clearly there is no charge at infinity here (the stick is of finite size).  I might as well get the total potential by adding the small potentials from the small pieces of stick.
The net potential due to all the small portions of the stick is just
\begin{align}
V(\vec r_p)=\int_{-\ell}^{\ell} dV= \int_{-\ell}^{\ell} \frac{\lambda dx_s}{4\pi\epsilon_0 \sqrt{{x_s}^2 +{z_p}^2}}\, . 
\end{align}
The integration is done by trig substitution and yields
\begin{align}
V(\vec r_p)=\frac{\lambda}{4\pi\epsilon_0}\log\left(1+\frac{2\ell\left(\ell+\sqrt{\ell^2+z_p^2}\right)}{z_p^2}\right)\, .
\end{align}
The $\log$ is typical of problems with cylindrical symmetry.  You can see immediately there are problems with this expression as in increase the length $2\ell$ of the stick: in the limit of an infinitely long stick, you would get basically $\log(1+\infty)$, which does not evaluate.  This is because (1) was used under the assumption there was no charge at infinity, something that is clearly no longer true if you make the charged stick infinitely long: there are then charges at $\pm \infty\hat{\bf{x}}$.
The same reasoning holds if you start with an infinite thin stick and use Gauss' law to get first the field and then try to get "the potential".   The field of the infinitely long stick is then
\begin{align}
\vec E= \frac{1}{2\pi \epsilon_0}\frac{\lambda}{R}\hat{\bf{R}}
\end{align}
where $R$ is the distance from the wire and $\hat{\bf{R}}$ is the radial unit vector in cylindrical coordinates.  Choosing a path along $\hat{\bf{z}}$ if $\vec r_p$ is on the $\hat{\bf{z}}$ axis,  and using the naive definition
\begin{align}
V(\vec r_p)=-\int_{-\infty}^{z_p} \vec E\cdot dz \hat{\bf{z}}=\frac{\lambda}{2\pi \epsilon_0}\log(z)\bigl\vert_{\infty}^{z_p}
\end{align}
which again diverges, for the same reason as before.
In case like the infinitely long wire, or the thick cylinder as you have in your example, one always runs into this problem.  Hence, it is dangerous to define "the potential" using $V(\vec r_p)=-\int_{\infty}^{z_p}  E\cdot dz 
\hat{\textbf{z}}$ although it makes perfect sense to compute a potential difference
\begin{align}
\Delta V(a,b)=-\int_{a}^{b} \vec E\cdot dz  
\hat{\textbf{z}}\, ,\tag{2}
\end{align}
provided that both $a$ and $b$ are finite.  Note that in such calculations, it is technically incorrect to think of
$\Delta V(a,b)$ as the difference between "the potential" $V(a)$ at $a$ and "the potential" $V(b)$ at $b$ since neither $V(a)$ nor $V(b)$ make sense.
In your specific problem, you are asked to evaluate the potential difference between $a$ and some other point inside your thick cylinder, so it makes sense to use (2) with $b=r$ in your specific case.
I do not understand why $V=0$ for $r>b$.  It might be that $V$ is constant but there's no reason to suggest this constant is $0$: for all we know if could be any constant $V_0$.  Now maybe the reference potential is set at $0$ outside the arrangement, but it's not clear this is so in what you have posted.  Of course if the potential is constant ($0$ or $V_0$), the field will be $0$ in that region (as it is).
Now, assuming you set $V=0$ at $b$, then you will do work in going from $b$ to $r$ when $r$ is in between $b$ and $a$.  For this you use the expression for $\vec E$ in between the cylinders. You will also do additional work once you get to inside the inner thick cylinder.
The only reason I see to have one limit of the integral to be $\infty$ is by setting $V=0$ outside the arrangement.  The problem is that $\infty$ along the $\hat{\bf{z}}$ axis is the same $\infty$ as along the $\hat{\textbf{x}}$ axis, and you can go from one to the other by a circle of infinite radius. Nevertheless, assuming the reference is chosen in this way, for any point outside the arrangement with $r>b$, there is no field to $V(r)=V(b)=0$ for any $r>b$.  In this case, $\int_b^R \vec E\cdot d\vec \ell=0$ everywhere $R>b$, and you might as well use  $\int_\infty^R \vec E\cdot d\vec \ell$: this will not mess up your declaration that $V=0$ at $r=b$.
For the reasons given in the first part of my answer, I would never set up a problem like this, where there are charges at infinity and  declare $V=0$ at $\infty$.
