False statementConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-27

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FALSE: pointwise decay alone makes every large semicircle integral vanish

Statement

False claim: if $f (z) \to 0$ pointwise on the upper semicircle $∣ z ∣ = R$ as $R \to \infty$ , then $\int_{γ_{R}^{+}} f (z) d z \to 0.$

Facts & Assumptions

Given: The function $f (z) = 1 / z$ and the upper semicircles $γ_{R}^{+} (t) = R e^{i t}$ .

[L1]

The correct large-arc lemma assumes control of $z f (z)$ , not only of $f (z)$ itself (A rational large-semicircle integral vanishes under the zR(z) to 0 condition).

Refutation

technique · direct

1.1

Along the upper semicircle, $f (R e^{i t}) = e^{- i t} / R \to 0$ pointwise. But $\int_{γ_{R}^{+}} \frac{d z}{z} = \int_{0}^{π} \frac{i R e^{i t}}{R e^{i t}} d t = i π .$

givenalgebra

2.1

So the arc integral does not tend to $0$ even though the pointwise values do. [step 1.1, L1] ∎ This is exactly why requires the stronger hypothesis on $z f (z)$ .

Depends on

A rational large-semicircle integral vanishes under the zR(z) to 0 condition

Used by

The function 1 / z makes the large-semicircle shortcut fail Counterexample

Dependency tree · two levels

2 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

L. V. Ahlfors, Complex Analysis, 3rd ed., Ch. 4 §5.3 (standard reference, not scraped)