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TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-21
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A jointly continuous finite-interval parameter integral of holomorphic functions is holomorphic

Statement

Let ab be real, let ΩC be open, and let φ:[a,b]×ΩC be jointly continuous. Suppose φ(t,) is holomorphic on Ω for every t[a,b]. Then the componentwise Riemann integral

F(z):=abφ(t,z)dt

is holomorphic on Ω.

If, in addition, zφ(t,z) exists everywhere and is jointly continuous on [a,b]×Ω, then

F(z)=abzφ(t,z)dt.

If φ:[a,b]×ΩC is jointly continuous and φ(t,) is holomorphic for every t, then F(z)=abφ(t,z)dt is holomorphic on Ω.

Facts & Assumptions

Given: Real numbers ab, an open set ΩC, and a jointly continuous function φ:[a,b]×ΩC whose z-slice is holomorphic for every parameter; the identification C=R2 from C is the real coordinate plane, with coordinate arithmetic.

[L1]

Complex-valued Riemann integration is the componentwise vector integral in R2, with zero integral when the limits agree and with linearity on every nondegenerate interval (The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral).

[L2]

On a piecewise-C1 contour, the complex contour integral equals the sum of the parameter integrals of f(γ(s))γ(s) over its smooth pieces (For piecewise-C1 contours the Riemann–Stieltjes integral agrees with the parametric complex integral and the published real line integrals).

[L3]

A Riemann-integrable real function on a product of nondegenerate closed rectangles has equal iterated integrals in either order when all sections are integrable (Riemann--Fubini on product rectangles, with lower and upper section integrals and content-zero exceptional sections).

[L4]

A holomorphic function has zero integral around every contained filled triangle (Goursat's triangle theorem: a holomorphic function integrates to zero around every triangle contained in its domain).

[L5]

A continuous function with zero integral around every contained filled triangle is holomorphic (Morera's theorem: vanishing triangle integrals characterize holomorphy among continuous functions).

[L6]

If H is a primitive of a continuous function h on a neighbourhood of a rectifiable contour γ, then γh=H(γ(b))H(γ(a)) (The line integral of a continuous function admitting a primitive is that primitive's endpoint increment along every rectifiable path).

[L7]

A continuous map on a compact metric space is uniformly continuous (Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous).

[L9]

Every continuous real function on a closed nondegenerate rectangle is Riemann integrable (Every continuous function on a closed nondegenerate rectangle in Rm is Riemann integrable).

[L10]

Let ab. If a<b and h:[a,b]Rm is integrable, then th(t)2 is integrable, and in all cases abh2abh2 (For ab and f:[a,b]Rm integrable when a<b, abf2abf2; for a<b, f2 is integrable).

Proof

technique · direct
1.1

If a=b, [L1] makes F=0, so all conclusions are immediate. Suppose a<b. For each z, continuity of tφ(t,z) and [L9] make the componentwise integral exist; near any fixed z0, choose a compactly contained closed disc K, so [L8] makes [a,b]×K compact, [L7] makes φ uniformly continuous there, and [L10] with [L11] gives F(z)F(z0)2(ba)supt[a,b]φ(t,z)φ(t,z0)0, hence F is continuous.

givenL1L7L8L9L10L11
1.2

For a filled triangle ΔΩ, parametrize each directed edge by its affine map on [0,1]; [L2] rewrites the edge contribution to ΔF(z)dz as an iterated parameter integral on [a,b]×[0,1].

givenL2
1.3

For the differentiation-under-the-integral conclusion, now assume ψ:=zφ exists and is jointly continuous. Fix zΩ and a closed disc about z contained in Ω; for sufficiently small nonzero h, [L6] and the parametrization in [L2] on the segment from z to z+h give (φ(t,z+h)φ(t,z))/h=01ψ(t,z+sh)ds, and [L7] on the compact parameter-disc product from [L8] makes this quotient converge to ψ(t,z) uniformly in t.

givenL2L6L7L8
2.1

For the basic holomorphy conclusion, each real and imaginary component of the edge integrand in step 1.2 is continuous, hence Riemann integrable by [L9]; [L3] interchanges its parameter and edge integrals, and [L4] makes the resulting inner contour integral Δφ(t,z)dz zero for every fixed t, so ΔF(z)dz=0.

step 1.2L3L4L9
3.1

For the basic holomorphy conclusion, the continuity from step 1.1 and the vanishing triangle integrals from step 2.1 satisfy [L5], so F is holomorphic on Ω.

step 1.1step 2.1L5
4.1

For the differentiation-under-the-integral conclusion, by linearity in [L1], the difference quotient of F minus abψ(t,z)dt is the integral over t of the error in step 1.3; [L10] and [L11] bound its modulus by (ba) times the uniform error, which tends to zero, so F(z)=abzφ(t,z)dt, including the already settled case a=b.

step 1.3L1L10L11

Depends on

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