Definition 4.2.1.
If \(F\) is holomorphic in the region \(G \subseteq \C\) and \(F'(z) = f(z)\) for all \(z \in G\text{,}\) then \(F\) is an antiderivative of \(f\) on \(G\), also known as a primitive of \(f\) on \(G\text{.}\)
Section 4.2 Antiderivatives
The central result about integration of a real function is the Fundamental Theorem of Calculus (Theorem A.0.3), and our next goal is to discuss complex versions of this theorem. The Fundamental Theorem of Calculus makes a number of important claims: that continuous functions are integrable, their antiderivatives are continuous and differentiable, and that antiderivatives provide easy ways to compute values of definite integrals. The difference between the real case and the complex case is that in the latter, we need to think about integrals over arbitrary paths in \(\C\text{.}\)
Example 4.2.2.
We have already seen that \(F(z) = z^2\) is entire and has derivative \(f(z) = 2z\text{.}\) Thus, \(F\) is an antiderivative of \(f\) on any region \(G \subseteq \C\text{.}\) The same goes for \(F(z) = z^2 + c\text{,}\) where \(c \in \C\) is any constant.
Example 4.2.3.
Since
\begin{equation*}
\frac{ d }{ \diff{z} } \left( \frac 1 {2i} (\exp(iz) -
\exp(-iz)) \right) \ = \ \frac 1 2 (\exp(iz) + \exp(-iz)) \, \text{,}
\end{equation*}
Example 4.2.4.
The function \(F(z) = \Log(z)\) is an antiderivative of \(f(z) = \frac 1 z\) on \(\C \setminus \R_{ \le 0 }\text{.}\) Note that \(f\) is holomorphic in the larger region \(\C
\setminus \{ 0 \}\text{;}\) however, we will see in Example 4.2.8 that \(f\) cannot have an antiderivative on that region.
Theorem 4.2.5.
Suppose \(G \subseteq \C\) is a region and \(\gg \subset G\) is a piecewise smooth path with parametrization \(\gg(t)\text{,}\) \(a \leq t \leq b\text{.}\) If \(f\) is continuous on \(G\) and \(F\) is any antiderivative of \(f\) on \(G\) then
\begin{equation*}
\int_\gg f \ = \ F \left( \gg(b) \right) - F \left( \gg(a)
\right) \, \text{.}
\end{equation*}
Proof.
This follows immediately from the definition of a complex integral and Theorem A.0.3(b), since \(\frac{ d }{ \diff{t} } F(\gg(t)) = f(\gg(t)) \,
\gg'(t)\text{:}\)
\begin{equation*}
\int_\gg f \ = \ \int_a^b f(\gg(t)) \, \gg'(t) \, \diff{t} \ = \ F
\left( \gg(b) \right) - F \left( \gg(a) \right) \, .
\end{equation*}
Example 4.2.6.
Since \(F(z) = \frac 1 2 \, z^2\) is an antiderivative of \(f(z) = z\) in \(\C\text{,}\)
\begin{equation*}
\int_\gg f \ = \ \frac 1 2 (1+i)^2 - \frac 1 2 \, 0^2 \ = \ i
\end{equation*}
for each of the three paths in Example 4.1.2.
There are several interesting consequences of Theorem 4.2.5. For starters, if \(\gg\) is closed (that is, \(\gg(a)=\gg(b)\)) we effortlessly obtain the following.
Corollary 4.2.7.
Suppose \(G \subseteq \C\) is open, \(\gg \subset G\) is a piecewise smooth closed path, and \(f\) is continuous on \(G\) and has an antiderivative on \(G\text{.}\) Then
\begin{equation*}
\int_\gg f \ = \ 0 \,\text{.}
\end{equation*}
This corollary is immediately useful as a test for existence of antiderivatives:
Example 4.2.8.
The function \(f: \C \setminus \{0\} \to \C\) given by \(f(z) = \frac 1 z\) satisfies \(\int_{\gg} f = 2\pi i\) for the unit circle \(\gg
\subset \C \setminus \{0\}\text{,}\) by Exercise 4.5.4. Since this integral is nonzero, \(f\) cannot have an antiderivative in \(\C \setminus
\{0\}\text{.}\)
Theorem 4.2.9.
Suppose \(G \subseteq \C\) is a region and \(z_0 \in G\text{.}\) Let \(f: G \to \C\) be a continuous function such that \(\int_\gg f = 0\) for any closed piecewise smooth path \(\gg \subset G\text{.}\) Then the function \(F: G \to \C\) defined by
\begin{equation*}
F(z) \ := \ \int_{\gg_z} f\text{,}
\end{equation*}
where \(\gg_z\) is any piecewise smooth path in \(G\) from \(z_0\) to \(z\text{,}\) is an antiderivative for \(f\) on \(G\text{.}\)
Proof.
There are two statements that we have to prove: first, that our definition of \(F\) is sound—that is, the integral defining \(F\) does not depend on which path we take from \(z_0\) to \(z\)—and second, that \(F'(z) = f(z)\) for all \(z \in G\text{.}\)
Suppose \(G \subseteq \C\) is a region, \(z_0 \in G\text{,}\) and \(f: G \to \C\) is a continuous function such that \(\int_\gg f = 0\) for any closed piecewise smooth path \(\gg \subset G\text{.}\) Then \(\int_\sigma f\) evaluates to the same number for any piecewise smooth path \(\sigma \subset G\) from \(z_0\) to \(z \in G\text{,}\) because any two such paths \(\sigma_1\) and \(\sigma_2\) can be concatenated to a closed path first tracing through \(\sigma_1\) and then through \(\sigma_2\) backwards, which by assumption yields a zero integral:
\begin{equation*}
\int_{ \sigma_1 } f \ - \ \int_{ \sigma_2 } f \ = \ \int_{
\sigma_1 - \sigma_2 } f \ = \ 0 \, \text{.}
\end{equation*}
This means that
\begin{equation*}
F(z) \ := \ \int_{\gg_z} f
\end{equation*}
is well defined. By the same argument,
\begin{equation*}
F(z+h) - F(z) \ = \ \int_{\gg_{z+h}} f \ - \ \int_{\gg_z} f \ = \ \int_\gg f
\end{equation*}
for any path \(\gg \subset G\) from \(z\) to \(z+h\text{.}\) The constant function \(1\) has the antiderivative \(z\) on \(\C\text{,}\) and so \(\int_\gg 1 = h\text{,}\) by Theorem 4.2.5. Thus
\begin{align*}
\frac{ F(z+h) - F(z) }{ h } - f(z) \amp \ = \ \frac 1 h
\int_\gg f(w) \, \diff{w} \ - \ \frac{ f(z) }{ h } \int_\gg
\diff{w}\\
\amp \ = \ \frac 1 h \int_\gg (f(w)-f(z)) \, \diff{w} \, \text{.}
\end{align*}
If \(|h|\) is sufficiently small then the line segment \(\lambda\) from \(z\) to \(z+h\) will be contained in \(G\text{,}\) and so, by applying the assumptions of our theorem for the third time,
\begin{align}
\frac{ F(z+h) - F(z) }{ h } - f(z) \amp \ = \ \frac 1 h
\int_\gg (f(w)-f(z)) \, \diff{w}
\nonumber\notag\\
\amp \ = \ \frac 1 h \int_\lambda
(f(w)-f(z)) \, \diff{w} \,\text{.}\tag{4.2}
\end{align}
We will show that the right-hand side goes to zero as \(h \to
0\text{,}\) which will conclude the theorem. Given \(\epsilon > 0\text{,}\) we can choose \(\delta > 0\) such that
\begin{equation*}
\abs{ w-z } \lt \delta \qquad \Longrightarrow \qquad \abs{
f(w) - f(z) } \lt \epsilon
\end{equation*}
because \(f\) is continuous at \(z\text{.}\) (We also choose \(\delta\) small enough so that (4.2) holds.) Thus if \(|h| \lt \delta\text{,}\) we can estimate with Proposition 4.1.8(d)
\begin{align*}
\abs{ \frac 1 h \int_\lambda (f(w)-f(z)) \, \diff{w} }
\amp \ \le
\ \frac 1 {|h|} \max_{ w \in \lambda } \abs{ f(w) - f(z) }
\length(\lambda)\\
\amp \ = \ \max_{ w \in \lambda } |f(w) - f(z)|\\
\amp \lt \epsilon \,\text{.}
\end{align*}
There are several variations of Theorem 4.2.9, as we can play with the assumptions about paths in the statement of the theorem. We give one such variation, namely, for polygonal paths, i.e., paths that are composed as unions of line segments. You should convince yourself that the proof of the following result is identical to that of Theorem 4.2.9.
Corollary 4.2.10.
Suppose \(G \subseteq \C\) is a region and \(z_0 \in G\text{.}\) Let \(f: G \to \C\) be a continuous function such that \(\int_\gg f = 0\) for any closed polygonal path \(\gg
\subset G\text{.}\) Then the function \(F: G \to \C\) defined by
\begin{equation*}
F(z) \ := \ \int_{\gg_z} f \,\text{,}
\end{equation*}
where \(\gg_z\) is any polygonal path in \(G\) from \(z_0\) to \(z\text{,}\) is an antiderivative for \(f\) on \(G\text{.}\)
If you compare our proof of Theorem 4.2.9 to its analogue in \(\R\text{,}\) you will see similarities, as well as some complications due to the fact that we now have to operate in the plane as opposed to the real line. Still, so far we have essentially been “doing calculus” when computing integrals. We will now take a radical departure from this philosophy by studying complex integrals that stay invariant under certain transformations of the paths we are integrating over.
