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ExampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11
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A positive continuous integrand can have finite integral while unbounded on every tail

Example

There is a positive continuous f:[0,∞)→Rf:[0,\infty)\to\mathbb R for which ∫0∞f\int_0^\infty f converges although ff is unbounded on every tail.

Facts & Assumptions

Given: For each positive integer kk, put hk=2−k−2/kh_k=2^{-k-2}/k and let sks_k be the symmetric triangular function supported on [k−hk,k+hk][k-h_k,k+h_k], zero at the endpoints, and of height kk at its center. Define f(x)=1(1+x)2+∑k=1∞sk(x).f(x)=\frac1{(1+x)^2}+\sum_{k=1}^\infty s_k(x).

[L1]

The supports of the sks_k are pairwise disjoint, and every compact interval meets only finitely many of them.

[L2]

A triangle of height kk and half-width hkh_k has integral khk=2−k−2kh_k=2^{-k-2}.

[L3]

The geometric series ∑k=1∞2−k−2\sum_{k=1}^\infty2^{-k-2} converges (For ∣r∣<1|r| < 1, ∑k≥0rk=1/(1−r)\sum_{k \ge 0} r^k = 1/(1-r), and for ∣r∣≥1|r| \ge 1 the series diverges).

Verification

technique · construction
1.1

By [L1], local finiteness and matching zero endpoint values make the spike sum continuous; adding the positive continuous baseline preserves positivity and continuity. Direct differentiation gives primitive 1−1/(1+x)1-1/(1+x) for the baseline, whose improper integral is one.

L1
2.1

By [L2], additivity, and [L3], the total integral of all spikes is ∑k=1∞2−k−2<∞\sum_{k=1}^\infty2^{-k-2}<\infty. Given ε>0\varepsilon>0, choose KK so that both the geometric spike tail from KK and the baseline tail 1/(1+K)1/(1+K) are below ε/2\varepsilon/2. For K≤u<vK\le u<v, nonnegativity bounds the integral over every partial spike by the full spike area, so ∫uvf<ε\int_u^v f<\varepsilon. The Cauchy criterion therefore gives convergence of ∫0∞f\int_0^\infty f.

step 1.1L2L3
3.1

At every positive integer kk, f(k)≥sk(k)=kf(k)\ge s_k(k)=k. Integers occur arbitrarily far out, so ff is unbounded on every tail.

given∎

Depends on

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