Past exam of the mathematics course of the University of Cambridge 2002 iii Paper 48 2 a Solution Created 2026-10-03 Updated 2026-10-14
For real , the multiplier has modulus one andThe triangle inequality applied to the integral therefore givesThe hypotheses and ensure endpoint convergence. Differentiation in is justified on compact parameter intervals by integrable bounds with an additional factor . Differentiating both the exponential multiplier and the power of makes the logarithmic denominator cancel:Here the scaled Gamma function integral is obtained by . Integrating this derivative from zero to yieldsThe same modulus bound also gives the useful parameter-shift remainder bound for a logarithmic Laplace integralbecause the denominator magnitude is at least .
Past exam of the mathematics course of the University of Cambridge 2002 iii Paper 48 2 b Solution Created 2026-10-03 Updated 2026-10-14
Fix a small , for example , independently of . Taking imaginary parts of the identity in part (a) givesThe parameter-shift remainder bound for a logarithmic Laplace integral makes the last term , smaller than for every fixed .
The endpoint function is analytic near zero. For any fixed , its Taylor expansion isIntegrating this expansion against useswhile the error is and the extension from to infinity contributes an exponentially small tail. This is Watson's lemma, and givesEquivalently the coefficients have the stated exponential generating function . In particular,Thus the apparent term vanishes: the first nonzero terms are . The derivatives of the gamma function supply the successive logarithmic coefficients.