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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/ii/paper-1/31c/c/solution

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Past exam of the mathematics course of the University of Cambridge / 2010 / ii / Paper 1 / 31C / c / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-14

The substitution

u = e^{- x}

changes the beta function to

B (s, t) = \int_{0}^{\infty} e^{- s x} (1 - e^{- x})^{t - 1} d x .

(1)

For fixed

t > 0

, its factor near zero expands as

(1 - e^{- x})^{t - 1} = x^{t - 1} [1 - \frac{t - 1}{2} x + O (x^{2})] .

(2)

Away from zero the factor is bounded, so the extension from the finite-interval Watson lemma to this exponentially damped tail is immediate. Therefore

B (s, t) = \frac{Γ ( t )}{s ^{t}} [1 - \frac{t ( t - 1 )}{2 s} + O (s^{- 2})] (s \to \infty) .

(3)

The change

u \mapsto 1 - u

gives

B (s, t) = B (t, s)

, so, with

s > 0

fixed,

B (s, t) = \frac{Γ ( s )}{t ^{s}} [1 - \frac{s ( s - 1 )}{2 t} + O (t^{- 2})] (t \to \infty) .

(4)

These expansions hold with the other parameter fixed; the first does not claim uniformity when

t

grows with

s

. The printed gamma-integral hint has inconsistent variable labels: the integration differential is

d x

and the positive parameter is

y

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