Total curvature of a piecewise smooth planar curve I am trying to show that the total curvature of a piecewise smooth planar curve $\alpha$ has the property that 
$\displaystyle \int_{C}\kappa $ $\text{d}s + \displaystyle \sum_{j=1}^{l} \epsilon_{j} = \pm 2\pi$,
where $\epsilon_{j}$  at $\alpha(s_{j})$ is the angle between $\alpha_{-}^{'}(s_{j}) = \displaystyle \lim_{s \to s_{j}^{-}}\alpha^{'}(s)$ and $\alpha_{+}^{'}(s_{j}) = \displaystyle \lim_{s \to s_{j}^{+}}\alpha^{'}(s)$, such that $|\epsilon_{j}| \leq \pi$ for all $j$.
Intuitively this is obvious, though I'm experiencing difficulty proving it technically. Note this is a generalization of Hopf Umlaufsatz (where $C$ is a simple closed plane curve), and so I have the proof of that. However, the generalization calls for exterior angles which now trouble me.
My gut instinct tells me to prove that $\displaystyle\sum_{j=1}^{l}\epsilon_{j} = 2\pi - \int_{C}\kappa$ $\text{d}s$, where we traverse $C$ in the counterclockwise direction.
$\textbf{Progress:}$ Here is a rough sketch of my intuition:
Suppose we have a closed, simple, piecewise smooth planar curve $C$. WLOG assume $\alpha$ is positively oriented in the counterclockwise direction. Let $\epsilon_{1}, \dots, \epsilon_{l}$ be the exterior angles of $\alpha$ at the "vertices" (corners) $\alpha(s_{1}), \dots, \alpha(s_{l})$ respectively.
So we consider $\displaystyle \sum_{j=1}^{l} \int_{s_{j}}^{s_{j+1}} \kappa$ d$s$ with the convention that $s_{l+1}=s_{1}$. This sum is the total curvature of the curve not accounting for the exterior angles.
Somewhere I would like to use the fact that for any polygon (with at least $2$ vertices) the sum of the exterior angles adds to $2\pi$, but I'm realizing after sketching some counterexamples that this doesn't need to be true for nonpolygonal curves.
I'm also aware that $\text{cos}\epsilon_{j} = T(s_{j}^{-})\cdot T(s_{j}^{+})$ and $\text{sin}\epsilon_{j} = N_{\pm}(s_{j}^{-})\cdot T(s_{j}^{+})$. From here, however, is where I begin to slump.
$\textbf{Progress II:}$ Another way I was thinking about it is to take our closed, planar, piecewise smooth curve $C$ with, say, $l$ corners, and polygonalize it. Partition the interval $I=[0,s_{m}]$ on which $\alpha : I \longrightarrow \mathbb{R}^3$ is defined. So $\mathcal{P}(I) = [s_{1}, s_{2}, \dots, s_{t}, \dots, s_{m}]$, where $s_{1}=0$, and $l$ of those points map injectively to corners on $C$. Now we have a rigid polygon. Thus, we know that $\displaystyle \int_{s_{j}}^{s_{j+1}} \kappa $ d$s = 0$ for all $j$, and the exterior angles at each $\alpha(s_{j})$ all add to $2\pi$.
Thus, $\displaystyle \sum_{j=1}^{m} \int_{s_{j}}^{s_{j+1}} \kappa$ d$s + \displaystyle \sum_{j=1}^{m} \epsilon_{j} = 0 \pm 2\pi$.
Now, consider $\displaystyle \lim_{m \to \infty} \mathcal{P}(I)$; that is, polygonalize $C$ even further. Thus each successive edge $[\alpha(s_{j}),\alpha(s_{j+1})]$ becomes smaller, obtaining a better approximation to the curve $C$. Since we've retained the $l$ corners, and $\displaystyle \lim_{m \to \infty} \displaystyle \sum_{j=1}^{m} \int_{s_{j}}^{s_{j+1}} \kappa$ d$s = \displaystyle \int_{C} \kappa$ d$s$, we obtain the result. 
Again, these are just rough sketches of intuition. Any feedback, hints, etc. would be greatly appreciated! What a fun problem! =]
 A: It seems this is often considered a bit tedious to do rigorously. One approach proceeds via smoothing the corners. This is advocated in, for example, "Differential Geometry of Curves and Surfaces" by Kristopher Tapp. While I find that book very nice, this is one of the places where it gives up on being 100% rigorous. I believe Kreyszig's book also omits some details (see the section on Gauss-Bonnet). I have written up a note with a proof, with a detailed description of a corner (and cusp) smoothing procedure. It is a bit long for stackexchange, but you can read it at here. All references to pages etc. are to Tapp's book.
On the other hand, I think your second approach is potentially promising. One needs to show that for any arc the sum of exterior angles of a sufficiently fine polygonization approaches the curvature integral, which could follow from the definition of the Riemann integral applied to  $\int \kappa ds$ (works more directly if one replaces a curve not with arcs but with pieces of tangents; one has to think through the details, which I have not). One thing to be careful about, however, is to either make sure the curve obtained by attaching together the polygonzations is a boundary of a polygon (i.e. non-self-intersecting), or to show it has the "sum of exterior angles" equal to $2\pi$ even if it is not. This may require some care.
