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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2009/iii/paper-7/3/a/solution

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Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 7 / 3 / a / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-14

The improper integral is assumed to have a finite limit. Fix

λ > 1

and put

I (U, V) = \int_{U}^{V} (β (x) - x) x^{- 2} d x

. Convergence implies

I (X, λ X) \to 0

and

I (X / λ, X) \to 0

X \to \infty

. By monotonicity,

I (X, λ X) \geq β (X) \int_{X}^{λ X} x^{- 2} d x - lo g λ = (1 - λ^{- 1}) \frac{β ( X )}{X} - lo g λ,

and

I (X / λ, X) \leq (λ - 1) \frac{β ( X )}{X} - lo g λ .

Thus

\frac{l o g λ}{λ - 1} \leq lim inf_{X \to \infty} \frac{β ( X )}{X} \leq lim sup_{X \to \infty} \frac{β ( X )}{X} \leq \frac{l o g λ}{1 - λ ^{- 1}} .

Both outer bounds tend to one as

λ ↓ 1

β (X) / X \to 1

. This monotone integral Tauberian lemma uses monotonicity to control the value at a point from integral averages on adjacent multiplicative intervals; convergence of the integral alone would not control arbitrary local spikes.

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