I am working on the following problem:
Let $D \subset \mathbb C$ be a domain, $f: D \to \mathbb C$ a continuous function and $\gamma : [\alpha, \beta] \to D$ a contour. Assume that $\int_\gamma f$ only depends on the initial and final points of $\gamma$. Fix a point $a \in D$ and define a function $F_a : D \to \mathbb C$ by \begin{align} F_a(z) := \int_{\gamma_{a,z}} f, \end{align} where $\gamma_{a,z} : [\alpha, \beta] \to D$ is a smooth contour with initial point $a$ and final point $z$. Then $F_a'(z) = f(z)$ in $D$, that is $F_a$ is an antiderivative of $f$ in $D$.
I tried to prove this:
Proof: We have that \begin{align*} F_a'(z) &= \lim_{h \to 0} \frac{F_a(z+h) - F_a(z)}{h} = \lim_{h \to 0} \frac{1}{h} \left(\int_{\gamma_{a,z+h}} f(w) \, dw - \int_{\gamma_{a,z}} f(w) \, dw\right) \\ &= \lim_{h \to 0} \frac{1}{h} \left(\int_{\gamma_{a,z}} f(w) \, dw + \int_{[z,z+h]} f(w) \, dw - \int_{\gamma_{a,z}} f(w) \, dw\right) = \lim_{h \to 0} \frac{1}{h} \int_{[z,z+h]} f(w) \, dw \\ &= \lim_{h \to 0} \frac{1}{h} \int_0^1 f((1-t)z + t(z+h)) \cdot ((1-t)z + t(z+h))'\, dt \\ &= \lim_{h \to 0} \frac{1}{h} \int_0^1 f(z-tz+tz+th) \cdot h\, dt = \lim_{h \to 0} \int_0^1 f(z+th)\, dt. \end{align*} Hence for every $\epsilon > 0$ there exists $\delta > 0$ such that for all $h \in D$ with $|h| < \delta$ \begin{align*} \left|\int_0^1 f(z+th)\, dt - F_a'(z)\right| < \frac{\epsilon}{2}. \end{align*} Since $f$ is continuous in $z \in D$, there is $\delta' > 0$ such that for any $h \in D$ with \begin{align*} t|h| = |th| = |(z+th)-z| < \delta' \Rightarrow |h| < \frac{\delta'}{t}, \quad (t > 0) \end{align*} we have \begin{align*} |f(z+th)-f(z)| < \frac{\epsilon}{2}. \end{align*} Therefore for all $h \in D$ with $|h| < \min\left\{\delta, \frac{\delta'}{t}\right\}$ \begin{align*} |F'_a(z) - f(z)| &\le \left|F'_a(z) - \int_0^1 f(z+th) \, dt\right| + \left|\int_0^1 f(z+th) \, dt - f(z)\right| \\ &< \frac{\epsilon}{2} + \left|\int_0^1 f(z+th) - f(z) \, dt\right| \le \frac{\epsilon}{2} + \int_0^1 |f(z+th) - f(z)| \, dt \\ &= \frac{\epsilon}{2}+ \int_0^1 \frac{\epsilon}{2} \, dt = \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon. \end{align*}
My problems: 1. I don't know if \begin{align*} \int_{\gamma_{a,z+h}} f(w) \, dw = \int_{\gamma_{a,z}} f(w) \, dw + \int_{[z,z+h]} f(w) \, dw \end{align*} is possible? 2. I don't know how to go on from $\lim_{h \to 0} \int_0^1 f(z+th)\, dt$.
Edit: Is this technical $\epsilon$-$\delta$-argument correct? Maybe some modifications are necessary.
No, it's exactly right. You can go different ways, but that is the nicest I know.
Concerning question 1., this is technically possible, the premise
gives you the possibility to choose whatever path you find convenient. You should however explicitly state that $h$ be small, e.g. $\lvert h\rvert < r$, where $r > 0$ is such that the disk with radius $r$ and centre $z$ is contained in $D$.
Concerning question 2., use the continuity of $f$ in $z$ to conclude that for all small enough $h$, you have
$$\left\lvert \int_0^1 f(z+th)\,dt - f(z)\right\rvert < \varepsilon.$$
Regarding the update, part of the task is to show that the limit
$$\lim_{h\to 0} \frac{F_a(z+h)-F_a(z)}{h}$$
exists at all, so you can't split into
$$\left\lvert F_a'(z) - \int_0^1 f(z+th)\,dt\right\rvert + \left\lvert \int_0^1 f(z+th)\,dt - f(z)\right\rvert.$$
However, as you have
$$\frac{F_a(z+h) - F_a(z)}{h} = \int_0^1 f(z+th)\,dt,$$
you show a) the existence of the limit and b) $F_a'(z) = f(z)$ when you show
$$\left\lvert \int_0^1 f(z+th)\,dt - f(z)\right\rvert \to 0.$$
The last part shows that, although there is a small problem with your use of $\frac{\delta'}{t}$, since your bound has to be independent of $t$. But you know that when $\lvert h\rvert < \delta'$, then for every $t \in [0,1]$ you have $\lvert th\rvert = t\lvert h\rvert < \delta'$, so all you need is $\lvert h\rvert < \delta'$ to conclude
$$\left\lvert \int_0^1 f(z+th)\,dt - f(z)\right\rvert = \left\lvert \int_0^1 f(z+th)-f(z)\,dt\right\rvert \leqslant \int_0^1 \lvert f(z+th)-f(z)\rvert\,dt < \int_0^1 \varepsilon\,dt = \varepsilon.$$