Cauchy's Integral Formula

Equivalent forms

f^{(n)}(z_0)=\frac{n!}{2\pi i}\oint_C\frac{f(z)}{(z-z_0)^{n+1}}dz

Boundary data dictates the entire interior — one contour integral reconstructs a function and all its derivatives at once.

Symbol	Name	SI unit	Dimension	Range
$f(z_0)$	Interior valueoutput Function value at an interior point	1	1	—
$z_0$	Interior point Point enclosed by the contour	1	1	-3 – 3
$z$	Contour variable Point running around the loop C	1	1	—
$n$	Derivative order Order of derivative recovered	1	1	0 – 5

Where it holds

f holomorphic on and inside a simple closed contour C (counterclockwise),

z_{0}

interior. Singularities inside require the residue theorem instead.

Dimensional analysis

f(z_0) \to

value of f

(1/2\pi i)\oint f/(z-z_{0}) dz \to f\cdot (1

/length

)\cdot

length

The dz and

1/(z-z_{0})

dimensions cancel, returning the dimension of f.

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Discovery

Augustin-Louis Cauchy · 1831

Cauchy developed the formula through the 1820s–30s as he built rigorous complex analysis. It followed from his integral theorem (that loop integrals of holomorphic functions vanish) plus a small circle around the singularity. It underpins the residue theorem and much of 19th-century mathematical physics.

Research status: stable

Real-world applications

Kramers–Kronig relations linking dispersion and absorption
Evaluating real integrals via contour methods
Analytic continuation and Green's-function techniques

Common misconceptions

It requires f holomorphic everywhere inside — a pole breaks it (use residues)
The contour must enclose $z_{0}$ ; if $z_{0} is$ outside, the integral is zero
Orientation matters: counterclockwise gives + $2\pi i$

Experimental verification

A theorem of analysis; its physical consequences (dispersion relations, Kramers–Kronig) are confirmed in optics and electronics — causality forcing a link between a material's absorption and refraction.

Derivation

Since

f(z)/(z-z_{0})

is holomorphic except at

z_{0}

, Cauchy's theorem lets you shrink the contour C to a tiny circle of radius

\varepsilon

around

z_{0}

without changing the integral.

Parametrize

z = z_{0} + \varepsilon e^{i\theta }

:

dz = i\varepsilon e^{i\theta }d\theta

, and

(z-z_{0}) = \varepsilon e^{i\theta }

, so the integrand becomes

f(z_{0}+\varepsilon e^{i\theta })\cdot i d\theta

.

As \varepsilon \to 0

,

f\to f(z_{0})

, giving

\oint = 2\pi i f(z_{0})

.

Differentiating under the integral in

z_{0}

yields the formula for all derivatives.

What if…

What if

z_{0}

lies outside the contour?

The integrand is holomorphic inside, so by Cauchy's theorem the integral is exactly zero.

What if there's a pole of order n+1?

The derivative formula applies: you pick up $(2\pi i/n!)$ times the n-th derivative of f at $z_{0}$ .

1

A simple contour integral

Given ·

\oint e^z/(z-1) dz

over |

z|=2

Find · the value

Steps

$f(z)=e^z$ is holomorphic; $z_{0}=1$ is inside | $z|=2$
Cauchy: $\oint = 2\pi i f(1)$
$= 2\pi i\cdot e$

Result ·

2\pi i\cdot e^{1} = 2\pi ie

Cauchy's Integral Theorem→residue theoremAnalytic (Holomorphic) Functions→