I'm wondering if there is a general equation to apply Cauchy's Integral theorem for open curves. So far, I have been able to find that a curve over the bottom half of a circle works for Cauchy's integral theorem. In particular, what I have so far is that $$f^{\left(k\right)}\left(0\right)=k!\lim_{w\to\infty}\frac{\frac{\left(f\left(1\right)+\frac{f\left(-1\right)}{\left(-1\right)^{k}}\right)}{2}+\left(\sum_{m=2^{\left(w-1\right)}}^{2^{w}-1}\frac{f\left(e^{\left(\frac{2\pi im}{2^{w}}\right)}\right)}{\left(e^{\frac{2\pi im}{2^{w}}k}\right)}\right)}{2^{w-1}}=\frac{k!}{\pi}\int_{\pi}^{2\pi}\frac{f\left(e^{\theta i}\right)}{e^{k\theta i}}d\theta$$ With a change of variables, this becomes $$\frac{k!}{\pi}\int_{\pi}^{2\pi}\frac{f\left(e^{\theta i}\right)}{e^{k\theta i}}d\theta= -\frac{k!i}{\pi}\int_\gamma\frac{f\left(z\right)}{z^{k+1}}dz$$ Where $\gamma$ is the bottom half of a circle like shown below.
$\gamma$" />This equation only seems to come out cleanly because the points of the function nicely cancel out at this particular angle. Switching $\pi$ and $2\pi$ with different values doesn't work for random values, but I'm fairly certain that at least a few different angles should work. However, I'm most interested if there is a general equation for getting the derivatives of a function from an open curve, either just for parts of a unit circle, or for a general curve.
EDIT: I think I have discovered the formula for another angle, and it is indeed a fair bit less nice. I am still stuck on how to exactly to extend this equation onto the derivatives of $f$ at $0$, and right now this equation only works for power series, instead of working for Laurent series as well.
$$f(0)=\frac{1}{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}}f\left(e^{i\theta}\right)d\theta+\frac{1}{2\pi}\sum_{m=0}^{3}\left(\frac{1}{e^{m\pi i}}-\frac{1}{e^{2m\pi i}}\right)\int_{0}^{e^{\frac{m\pi i}{2}}}\frac{f\left(z\right)}{z}dz$$
EDIT 2: I've managed to find an equation for the derivative of $f$ at zero, though I'd like to eventually find a version that doesn't include $f(0)$ in the equation. $$f'\left(0\right)=-\frac{2}{\pi}f\left(0\right)+\frac{1}{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}}\frac{f\left(e^{i\theta}\right)}{e^{i\theta}}d\theta-\frac{1}{2\pi}\sum_{m=0}^{3}\frac{\int_{0}^{e^{\frac{m\pi i}{2}}}\frac{f\left(z\right)}{z^{2}}dz}{e^{2m\pi i}}+\frac{1}{2\pi}\sum_{m=0}^{3}\frac{\int_{0}^{e^{\frac{m\pi i}{2}}}\frac{f\left(z\right)-f\left(0\right)}{z^{2}}dz}{e^{m\pi i}}$$