Past exam of the mathematics course of the University of Cambridge / 2010 / ii / Paper 1 / 31C (Asymptotic Methods)

Repeated integration by parts would require derivatives of $f$ and finite endpoint values of those derivatives. The hypothesis gives only continuity away from zero, and even the leading behavior $x^{\alpha }$ can diverge at zero or have singular derivatives when $\alpha$ is nonintegral. Thus those integrations need not be defined. The appropriate method integrates the asymptotic powers themselves, rather than differentiating an insufficiently regular function.

Fix an integer $N\ge 0$ and writeFor each power, substitute $y=\lambda x$ and extend the upper limit to infinity; the discarded tail is exponentially small. Since $\alpha +j\beta >-1$, the gamma integral givesTo control the remainder, choose $\delta >0$ such that $|r_N(x)|\le ϵx^{\alpha +N\beta }$ for $0<x<\delta$. Its integral there is at most $ϵ\Gamma (\alpha +N\beta +1){\lambda }^{-\alpha -N\beta -1}$. On $[\delta ,b]$ the remainder is bounded and its integral is $O(e^{-\lambda \delta })$. Since $ϵ$ is arbitrary, we have the full Watson lemma expansionMore precisely, after the term $j=N$, the remainder is $o({\lambda }^{-\alpha -N\beta -1})$. This proof uses neither differentiability of $f$ nor convergence of its asymptotic series.

The substitution $u=e^{-x}$ changes the beta function toFor fixed $t>0$, its factor near zero expands asAway from zero the factor is bounded, so the extension from the finite-interval Watson lemma to this exponentially damped tail is immediate. ThereforeThe change $u\mapsto 1-u$ gives $B(s,t)=B(t,s)$, so, with $s>0$ fixed,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 $dx$ and the positive parameter is $y$.