Integration Theorems

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Let $f$ be analytic in a simply connected domain $D$ , and let $z_0$ be a fixed point in $D$ . For each $z \in D$ , let $C_z$ be any contour in $D$ from $z_0$ to $z$ .

F(z) = \int_{C_z} f(\zeta)\,\text{d}\zeta

$F$ is analytic in $D$ and

F'(z) = f(z)

Definite Integrals

Let $f$ be analytic in a simply connected domain $D$ , and let $F$ be an antiderivative of $f$ in $D$ . If $C$ is any contour in $D$ from $z_1$ to $z_2$ ,

\int_C f(z)\,\text{d}z = F(z_2) - F(z_1)

Cauchy’s Integral Formula

Let $f$ be analytic in a simply connected domain $D$ , and let $C$ be a positively oriented simple closed contour lying in $D$ . If $z_0$ lies inside $C$ ,

f(z_0) = \frac{1}{2\pi i}\int_C \frac{f(z)}{z - z_0}\,\text{d}z

Equivalently,

\int_C \frac{f(z)}{z - z_0}\,\text{d}z = 2\pi i f(z_0)

Derivatives

Under the same conditions, for $n = 1, 2, 3, \ldots$ ,

f^{(n)}(z_0) = \frac{n!}{2\pi i}\int_C \frac{f(z)}{(z-z_0)^{n+1}}\,\text{d}z

Equivalently,

\int_C \frac{f(z)}{(z-z_0)^{n+1}}\,\text{d}z = \frac{2\pi i}{n!}f^{(n)}(z_0)

Written by

Sahithyan K.

October 8, 2026 1 min read

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