Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 1 / 27C / b

Near zero, write the amplitude as $t^{-1/4}(1+t{)}^{-1/4}={\sum }_{m\ge 0}(\genfrac{}{}{0ex}{}{-1/4}{m})t^{m-1/4}$. The large parameter localizes the integral at $t=0$. Termwise integration of the local expansion givesConsequentlyHere $(1/4{)}_m$ is the rising factorial. This is an asymptotic series, not an assertion that the full integrated Taylor series converges.

The needed result is Watson's lemma: if a locally integrable amplitude has an expansion $g(s)\sim \sum c_m{s}^{{\rho }_m-1}$ as $s\downarrow 0$, with ${\rho }_m>0$ increasing and suitable growth at infinity so its Laplace integral converges, then ${\int }_0^{\infty }{e}^{-ks}g(s)ds\sim \sum c_m\Gamma ({\rho }_m)k^{-{\rho }_m}$, with the corresponding finite-order remainder bounds. Apply it after $s=t^2$, when $g(s)=\frac{1}{2}{s}^{-5/8}(1+\sqrt{s}{)}^{-1/4}$ and ${\rho }_m=3/8+m/2$. The tail is exponentially small after any fixed positive cutoff.