Mathematics.

contour integration

Cauchy's Integral Formula

Complex Analysis35 minDifficulty6 out of 10

You should know: cauchy integral theorem

Overview

Cauchy's integral formula expresses the value of an analytic function at any interior point of a simple closed contour C entirely in terms of the values of the function on the boundary C itself: f(a) = (1/2πi) ∮_C f(z)/(z−a) dz, for f analytic on and inside C and a inside C. This remarkable formula shows that a holomorphic function's interior behavior is completely determined by its boundary values — a phenomenon with no real-analysis analogue. Differentiating the formula under the integral sign (justified for analytic functions) produces a formula for every derivative f^(n)(a) as a contour integral, which is what proves that analytic functions are automatically infinitely differentiable.

Intuition

Cauchy's integral formula says that analytic functions have a kind of 'action at a distance' rigidity: knowing f only on a boundary loop C is enough to reconstruct f exactly at every point inside, via a weighted average around the loop with weight 1/(z−a). This is utterly unlike real functions, where knowing boundary values of a smooth real function tells you nothing about its interior values. The factor 1/(z−a) has a pole exactly at the point a where you're evaluating f, and the residue-like mechanics of that pole is precisely what picks out f(a) from the average over the whole loop.

Formal Definition

Definition

Let f be analytic on and inside a simple closed contour C (positively oriented), and let a be any point inside C:

f(a)=12πi∮Cf(z)z−a dzf(a) = \frac{1}{2\pi i}\oint_C \frac{f(z)}{z-a}\,dz
Cauchy's integral formula
f(n)(a)=n!2πi∮Cf(z)(z−a)n+1 dzf^{(n)}(a) = \frac{n!}{2\pi i}\oint_C \frac{f(z)}{(z-a)^{n+1}}\,dz
Generalized formula for derivatives

Worked Examples

  1. f(z)=e^z is entire, and a=1 lies inside |z|=2, so Cauchy's integral formula applies with f(a) = e^1 = e.

    f(z)=ez,a=1f(z) = e^z, \quad a = 1

  2. Cauchy's formula states f(a) = (1/2πi)∮_C f(z)/(z-a) dz, so rearrange to solve for the integral.

    ∮Cezz−1 dz=2πi⋅f(1)=2πi⋅e\oint_C \frac{e^z}{z-1}\,dz = 2\pi i \cdot f(1) = 2\pi i \cdot e

Answer: ∮_C e^z/(z-1) dz = 2πie.

Practice Problems

Difficulty 5/10

Evaluate ∮_C sin(z)/z dz where C is the unit circle |z|=1 (counterclockwise).

Difficulty 6/10

Evaluate ∮_C cos(z)/(z-π) dz where C is |z|=4 (counterclockwise).

Difficulty 7/10

Evaluate ∮_C e^{2z}/(z+1)^3 dz where C is |z|=2 (counterclockwise), using the generalized Cauchy integral formula.

Quiz

Cauchy's integral formula reconstructs f(a) for a inside a closed contour C using:
The generalized Cauchy integral formula for f^{(n)}(a) shows that:
∮_C e^z/(z-1) dz for C = |z|=2 (counterclockwise) equals:

Summary

  • Cauchy's integral formula: f(a) = (1/2πi)∮_C f(z)/(z-a) dz reconstructs an analytic function's interior values from its boundary values.
  • The generalized formula f^{(n)}(a) = (n!/2πi)∮_C f(z)/(z-a)^{n+1} dz shows analytic functions are automatically infinitely differentiable.
  • This formula is the computational engine behind evaluating many contour integrals directly, without needing residue calculus.

References

  1. WebsiteWikipedia — Cauchy's integral formula