TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-16

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Complex and absolute line integrals are invariant under increasing continuous reparametrization

Statement

Let $ϕ : [c, d] \to [a, b]$ be a strictly increasing continuous bijection, let $γ : [a, b] \to C$ be rectifiable, and let $f$ be continuous on the trace of $γ$ . Then $\int_{γ \circ ϕ} f d z = \int_{γ} f d z, \int_{γ \circ ϕ} ∣ f ∣ ∣ d z ∣ = \int_{γ} ∣ f ∣ ∣ d z ∣ .$ For singleton source and target intervals the same identities hold by the zero-integral convention.

Facts & Assumptions

Given: A rectifiable contour, a continuous integrand, and a reparametrization $ϕ$ as in the Statement.

[L1]

Under a strictly increasing continuous bijection between nondegenerate compact intervals, the real Riemann–Stieltjes change-of-variable formula holds (Change of variable for the Riemann–Stieltjes integral).

[L2]

Arc length is invariant under continuous surjective monotone reparametrization, including the stated singleton cases (Arc length is invariant under every continuous surjective monotone reparametrization, including pauses and reversal).

[L3]

Published piecewise- $C^{1}$ line integrals are invariant under orientation-preserving reparametrization and change sign under orientation reversal (Scalar line integrals are parametrization-independent; vector line integrals retain orientation and change sign when it reverses).

[L4]

The complex integral is the combination of four component Riemann–Stieltjes integrals, and the absolute integral is the Riemann–Stieltjes integral against arc length (The complex line integral over a rectifiable path as a componentwise Riemann–Stieltjes integral, The absolute line integral over a rectifiable path using its arc-length function).

Proof

technique · cases

1.1assume-case nondegenerateL1L4

Assume first that both intervals are nondegenerate. Apply [L1] to each of the four component integrals in [L4]; their recombination is unchanged.

1.2assume-case singletonalgebra

If both intervals are singletons, both complex and absolute integrals are $0$ by definition.

2.1step 1.1L1L2L4

For the absolute integral in [L4], [L2] identifies the reparametrized arc-length integrator, and [L1] gives the same Stieltjes integral.

3.1step 1.1step 2.1step 1.2L3cases-exhaustive∎

The cases exhaust the Statement and prove both identities. On piecewise- $C^{1}$ contours this is exactly the increasing half of [L3]; decreasing reparametrization is excluded and instead changes the complex integral's sign.

Depends on

Used by

Dependency tree · two levels

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Sources

E. Stein and R. Shakarchi, Complex Analysis, Ch. 1, §3 (standard reference, not scraped)