Line integral of a point charge I am trying to teach myself Electrodynamics through self-study of Griffiths' Introduction to Electrodynamics, and I am having difficulty with a calculation that involves a line integral of a point charge.  I posted this question to Math Stack Exchange yesterday, but I was unable to get an answer. I'm not really sure that my question was understood. I have attempted to clarify it below.
I have been stuck on this for quite some time, and would appreciate any help that could be offered!
Cheers!

This question concerns a calculation in Section 2.2.4 of Griffiths' book on an Introduction to Electrodynamics (4th Ed.), in which he shows that the field of a point charge is curl-free.
The field of a point charge at the origin is given in spherical coordinates by $$\mathbf{E}=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\ \hat{\mathbf{r}}.$$
Here $\epsilon_0$ is the permittivity of free space, $q$ is the magnitude of the charge, $r$ is the radius from the origin, and $\hat{\mathbf{r}}$ is radial spherical basis vector.  He shows that this field is curl-free by demonstrating that its line integral around any closed loop is zero.  So he starts by considering the integral $$\int_{\mathbf{a}}^{\mathbf{b}}\mathbf{E}\cdot d\mathbf{l}$$ over an arbitrary path in $\mathbb{R}^3$ (that presumably doesn't include the origin).  This integral is calculated in spherical coordinates, so observing that the infinitesimal displacement vector is given by $$d\mathbf{l}=dr\ \hat{\mathbf{r}}+r\ d\theta\ \hat{\boldsymbol\theta}+r\ \sin\theta\ d\phi\ \hat{\boldsymbol\phi}$$ leads to the conclusion $$\mathbf{E}\cdot d\mathbf{l}=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\ dr$$
(in Griffiths' notation $\phi$ is the azimuthal angle, and $\theta$ is the polar angle).  Griffiths proceeds to evaluate the integral $$\int_\mathbf{a}^\mathbf{b}\mathbf{E}\cdot d\mathbf{l}=\frac{1}{4\pi\epsilon_0}\int_\mathbf{a}^\mathbf{b}\frac{q}{r^2}\ dr=\frac{-1}{4\pi\epsilon_0}\frac{q}{r}\bigg|^{r_b}_{r_a}=\frac{1}{4\pi\epsilon_0}\bigg(\frac{q}{r_a}-\frac{q}{r_b}\bigg).$$
Here $r_a$ and $r_b$ are the radii associated with $\mathbf{a}$ and $\mathbf{b}$.  It is then argued that the integral around any closed path is zero, so that $\mathbf{E}$ is curl-free by Stokes theorem.  I cannot understand how he got from $$\frac{1}{4\pi\epsilon_0}\int_\mathbf{a}^\mathbf{b}\frac{q}{r^2}\ dr$$ to $$\frac{-1}{4\pi\epsilon_0}\frac{q}{r}\bigg|^{r_b}_{r_a}.$$ This step seems to assume path independence, which is equivalent to what he is trying to prove.  Thus this logic seems a bit circular.  For suppose we do not use the fact that the integral is path-independent.  Then it must be the case that we need to parameterize $r$ as a function of $\phi$ and $\theta$, so that $r=r(\phi,\theta)$.  This parameterization would in turn change the infinitesimal displacement.  This would make the integral considerably more difficult to evaluate.
How does this step not assume path-independence?
And also, this seems to assume that the path does not cross the origin.  Does mean that our field $\mathbf{E}$ is curl-free everywhere but the origin, at which it is undefined?
I would be very grateful to anyone who could help clear up my confusion!
 A: Maybe it would have helped you if Griffiths had left in the other two terms in the path integral?
$$\int {\bf E} \cdot d{\bf l} = \frac{1}{4\pi \epsilon_0} \int \frac{q}{r^2}\ dr + \int E_{\theta} r\ d\theta + \int E_{\phi} r \sin \theta\ d\phi, $$
but since $E_{\theta} = E_\phi=0$ in this case, then the last two terms disappear. The remaining integral is indeed path independent in the sense that it does not matter what excursions you take in $\theta$ and $\phi$ because the integrand does not depend on $\theta$ or $\phi$, all that matters is the starting and ending values of $r$.
That the line integral of a closed path is zero does indicate that the field is curl free (unless you hve chosen a pathological path where the E-field is always perpendicular to the path.
A: This part $$\int_\mathbf{a}^\mathbf{b}\mathbf{E}\cdot d\mathbf{l}=\frac{1}{4\pi\epsilon_0}\int_\mathbf{a}^\mathbf{b}\frac{q}{r^2}\ dr=\frac{-1}{4\pi\epsilon_0}\frac{q}{r}\bigg|^{r_b}_{r_a}$$ doesn't assume path independence. It follows from $$\mathbf{E}\cdot d\mathbf{l}=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\ dr$$
The integrand only depends on the $r$ coordinate, so the integral just becomes an integral over $r$. This doesn't assume path independence, it actually shows that the integral only depends on the starting and ending $r$ coordinates. 
Therefore, if you understand why $$\mathbf{E}\cdot d\mathbf{l}=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\ dr
$$ then you're good to go.
A: Let $r$ be a function of $\theta$ and $\phi$ (I'll write $r$ as $r$ itself and not $r(\theta,\phi)$ just to make things less cluttered). Now, we know that
$$\mathrm d r=\frac{\partial r}{\partial \theta}\mathrm d \theta+\frac{\partial r}{\partial \phi}\mathrm d\phi$$
Let's represent $\displaystyle \frac{\partial r}{\partial \theta}=\dot r_{\theta}$ and $\displaystyle \frac{\partial r}{\partial \phi}=\dot r_{\phi}$. Then, your integral becomes (I am ignoring the constants for now):
$$\int_\mathrm{a}^\mathrm{b}\frac{1}{r^2}\ dr= \int_\mathrm{a}^\mathrm{b} \left(\frac{\dot r_{\theta}\mathrm d \theta}{r^2}+ \frac{\dot r_{\phi}\mathrm d \phi}{r^2} \right)\tag{1}$$
However, we know that
$$\mathrm d\left(-\frac{1}{r}\right)=\frac{\dot r_{\theta}\mathrm d \theta}{r^2}+ \frac{\dot r_{\phi}\mathrm d \phi}{r^2}$$
Therefore, the integral in equation $(1)$ is converted to
$$\int_\mathrm{a}^\mathrm{b} \mathrm d\left(-\frac{1}{r}\right)=\left|-\frac{1}{r}\right|_{\mathrm a}^{\mathrm b}=\frac 1 a -\frac 1 b$$
Thus even if you explicitly involve path dependence, you will get the same result.
