An Exact Formula for a Zero of a Continuous Function Under Minimal Conditions

Abstract

This paper is a self-contained companion to the author's first deposit (v1, zenodo.org/records/20085431), which established an exact double integral formula for the unique zero x* of a continuous function f:[a,b]→R under minimal conditions (continuity, f(a)>0, f(b)<0, uniqueness). The v1 formula involves integration over the unbounded domain [a,b]×(0,+∞). The present work introduces the elementary change of variables u = t/(1+t), which maps (0,+∞) bijectively onto (0,1) and transforms the formula into a double integral over the compact square [a,b]×[0,1]: x* = (a+b)/2 + (1/π) ∫₀¹ ∫ₐᵇ f(x)/[(1-u)²+u²f(x)²] dx du = (a+b)/2 + (1/π) ∫ₐᵇ ∫₀¹ f(x)/[(1-u)²+u²f(x)²] du dx A single rational kernel K(x,u) = f(x)/[(1-u)²+u²f(x)²] appears on a bounded domain. We prove: (i) K ∈ L¹([a,b]×[0,1]) with exact norm π(b-a)/2(ii) Both integration orders are valid (Fubini-Tonelli)(iii) The singularity at (x*,1) is integrable and harmless(iv) The sign function sgn(f(x)) is identified as the inner integral of K in u — a consequence, not an axiom The formula is validated on f(x) = -x + cos(x) on [0,π/2], whose unique zero is the Dottie number x*≈0.739085133215161. All proofs are elementary and self-contained. No knowledge beyond standard real analysis is required.