For piecewise-C1 contours the Riemann–Stieltjes integral agrees with the parametric complex integral and the published real line integrals

On a nondegenerate smooth piece $[t_j,t_{j+1}]$ the integrands $u(\gamma (t))$ and $v(\gamma (t))$ are continuous, being composites of the continuous $f$ with the continuous $\gamma$, so [L5] makes each of the four component integrands Riemann integrable; the integrators $x_j,y_j$ are $C^1$ on the piece, hence continuous with continuously extending derivative. The hypotheses of [L1] therefore hold, and applying [L1] to the four component Stieltjes integrals and recombining gives $\int f(\gamma (t))(x_j^{\mathrm{\prime }}(t)+i{y}_j^{\mathrm{\prime }}(t))dt=\int f(\gamma (t)){\gamma }_j^{\mathrm{\prime }}(t)dt$.

On the same piece the arc-length integrator is $s_{{\gamma }_j}$, which by [L6] is differentiable with $s_{{\gamma }_j}^{\mathrm{\prime }}(t)=\mathrm{\mid }{\gamma }_j^{\mathrm{\prime }}(t)\mathrm{\mid }$, continuous because ${\gamma }_j$ is $C^1$; and $\mathrm{\mid }f(\gamma (t))\mathrm{\mid }$ is continuous, hence Riemann integrable by [L5]. So [L1] applies with $\alpha =s_{{\gamma }_j}$ and yields $\int \mathrm{\mid }f(\gamma )\mathrm{\mid }d{s}_{{\gamma }_j}=\int \mathrm{\mid }f(\gamma (t))\mathrm{\mid }\mathrm{\mid }{\gamma }_j^{\mathrm{\prime }}(t)\mathrm{\mid }dt$; summing over pieces and using [L3] to identify the total arc length gives the absolute-integral formula, which is the scalar line integral in [L2].

The real and imaginary parts in step 1.1 are exactly the vector line integrals of $(u,-v)$ and $(v,u)$ from [L2]. This uses the published real construction in a numbered step, with its piecewise-$C^1$ hypothesis unchanged.

Summing the identities over the pieces proves both displays. No equality of one-sided derivatives at corners is needed, and zero-speed pieces contribute $0$.