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Our main method for calculating the Riemann integral \[\int\limits_r^s g(t) \intd t\] is to find $G : [r,s] \to \R$ differentiable with $G' = g$ and apply the fundamental theorem of calculus to get \[\int\limits_r^s g(t) \intd t = \bigg[ G(t) \bigg]_r^s = G(s) - G(r)\] easily. The difficult part is finding such a $G$.
In the previous section we defined the contour integral \[\int\limits_\gamma f\] for $\gamma$ a contour. In principle such contour integrals can be calculated as Riemann integrals, but finding antiderivatives may be too challenging in practice. Moreover, this course is not about calculating contour integrals using the tools of Riemann integration. It is about the interactions between contour integration, holomorphic functions and domains. As a first result in this direction, we introduce here antiderivatives in the complex setting and will see how the intricacies of domains in the complex plane make their existence relatively rare, in stark comparison with the real case.
Fix a domain $D \subset \C$ and $f : D \to \C$ continuous. A holomorphic function $F : D \to \C$ is an antiderivative of $f$ is $F'(z) = f(z)$ for all $z \in D$.
Fix a domain $D \subset \C$ and $f : D \to \C$ continuous. If $F : D \to \C$ is an antiderivative of $f$ then \[\int\limits_\gamma f = F(\gamma(b)) - F(\gamma(a))\] for every smooth path $\gamma : [a,b] \to \C$ in $D$.
Write $f(\gamma(t)) \gamma'(t) = u(t) + iv(t)$ and $F(\gamma(t)) = U(t) + i V(t)$ as real and imaginary parts. From \[(F(\gamma(t))' = F'(\gamma(t)) \gamma'(t) = f(\gamma(t)) \gamma'(t)\] we obtain $U'(t) = u(t)$ and $V'(t) = v(t)$. Now \[ \begin{align*} \int\limits_\gamma f & = \int\limits_a^b f(\gamma(t)) \gamma'(t) \intd t \\ & = \int\limits_a^b u(t) \intd t + i \int\limits_a^b v(t) \intd t \\ & = \bigg[ U(t) \bigg]_a^b + i \bigg[ V(t) \bigg]_a^b = F(\gamma(b)) - F(\gamma(a)) \end{align*} \] as claimed. $\square$
Define $f : \C \to \C$ by $f(z) = z^2$ and let $\gamma$ be any contour from $0$ to $1+i$. If we can find $F : \C \to \C$ holomorphic with $F'(z) = z^2$ then \[\int\limits_\gamma f = F(1+i) - F(0)\] by the theorem. We can take $F(z) = \frac{1}{3} z^3$ so that the integral is $\frac{1}{3} (1 + 3i - 3 - i) = - \frac{2}{3} + i \frac{2}{3}$.
Notice that the right-hand side only involves $\gamma(a)$ and $\gamma(b)$ and is therefore the same no matter the route from start to finish! If $f$ has an antiderivative then its contour integrals only depend on the start and end points. This is called path independence. In particular, the following is true.
Fix a domain $D \subset \C$ and $f : D \to \C$ continuous. If $F : D \to \C$ is an antiderivative of $f$ and $\gamma$ is any closed contour in $D$ then $\displaystyle\int\limits_\gamma f = 0$.
If $\gamma$ starts at $w \in \C$ then \[\int\limits_\gamma f(z) \intd z = F(w) - F(w) = 0\] by the theorem. $\square$
The hypothesis that $f$ has an antiderivative is crucial, as the following example shows.
Define $f : \C \setminus \{0\} \to \C$ by $f(z) = 1/z$ and $\gamma : [0,2\pi] \to \C$ by $\gamma(t) = e^{it}$. Then \[\int\limits_\gamma f = \int\limits_0^{2\pi} f(e^{it}) i e^{it} \intd t = \int\limits_0^{2\pi} \frac{1}{e^{it}} i e^{it} \intd t = 2 \pi i\] even though $\gamma$ is a closed contour! It must therefore be the case that there is no holomorphic function $F : \C \setminus \{0\} \to \C$ with $F'(z) = 1/z$!