TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-27

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Trigonometric integrals become contour integrals by the unit-circle substitution

Statement

Let $F (X, Y)$ be a rational expression in two variables, and assume that after the substitution

$X = \frac{z + z^{- 1}}{2}, Y = \frac{z - z^{- 1}}{2 i}$

the resulting rational function

$G (z) : = \frac{1}{i z} F ⁣ (\frac{z + z^{- 1}}{2}, \frac{z - z^{- 1}}{2 i})$

has no pole on $∣ z ∣ = 1$ . Then

$\int_{0}^{2 π} F (\cos θ, \sin θ) d θ = \int_{∣ z ∣ = 1} G (z) d z,$

where the right-hand side is taken on the positively oriented unit circle.

Facts & Assumptions

Given: A rational expression $F (X, Y)$ whose transformed integrand has no pole on the unit circle.

[L1]

Euler's formula gives $e^{i θ} = \cos θ + i \sin θ$ (Euler's formula: $\exp (i θ) = \cos θ + i \sin θ$ for every real $θ$ ).

Proof

technique · direct

1.1L1algebra

Put $z = e^{i θ}$ . By [L1], $\cos θ = \frac{z + z^{- 1}}{2}, \sin θ = \frac{z - z^{- 1}}{2 i} .$ Differentiating $z = e^{i θ}$ gives $d z = i z e^{0} d θ = i z d θ$ , so $d θ = d z / (i z)$ .

2.1step 1.1∎

As $θ$ runs from $0$ to $2 π$ , the variable $z$ traverses the unit circle once in the positive direction. Substituting the identities of step 1.1 into the real integral gives exactly the contour integral of $G (z)$ . The hypothesis that $G$ has no pole on $∣ z ∣ = 1$ is what makes the contour integral well defined.

Depends on

Used by

The integral of 1 / (a + cos theta) over [0, 2 pi] Example

Dependency tree · two levels

24 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

R. Howell and J. Mathews, Complex Analysis, Ch. 8 §8.2 (standard reference, not scraped)