Fundamental theorem of contour integration

Statement

Let $U \subseteq \mathbb{C}$ be an open set and let $f : U \to \mathbb{C}$ be continuous. Let $F : U \to \mathbb{C}$ be holomorphic with $F'(z) = f(z)$ for each $z \in U$. Let $\gamma : [a, b] \to U$ be a continuous and piecewise continuously differentiable path. Then the contour integral of $f$ along $\gamma$ is $$\int_\gamma f(z) \, dz = F(\gamma(b)) - F(\gamma(a)).$$ In particular, $\int_\gamma f(z) \, dz = 0$ if $\gamma$ is closed.

Remarks

The function $F$ is a primitive of $f$ on $U$. The integral depends only on the end points of $\gamma$. For each integer $n \neq -1$, $z^n$ has the primitive $z^{n+1}/(n+1)$ on $\mathbb{C} \setminus \{0\}$. So $\oint_\gamma z^n \, dz = 0$ for each closed path $\gamma$ that does not pass through $0$. The function $1/z$ has no primitive on $\mathbb{C} \setminus \{0\}$, because its integral over the unit circle is $2 \pi i \neq 0$.

Explanations

L1 human

This is the complex version of $\int_a^b F'(t) \, dt = F(b) - F(a)$. If you know a function whose derivative is $f$, you get the integral along any path from its values at the two ends.