Past exam of the cosmology course of the University of Oxford / 2025 / 1 / e / Solution

In the integral from part (d), set $y={\sinh }^{2}u$. Its upper endpoint is $u_x=\mathrm{arsinh}\sqrt{x}$, equal to $\theta /2$ in part (b). The hyperbolic function identitygives the antiderivativeMultiplication by ${\delta }_1$ therefore yieldsAt small $x$, cancellation between the terms leaves the Taylor series ${\delta }_2=2x/5-8x^2/35+16x^3/105+O(x^4)$. At large $x$, $\mathrm{arsinh}\sqrt{x}=\frac{1}{2}\ln (4x)+O(x^{-1})$, so the correction to one is $[3-\frac{3}{2}\ln (4x)]/x+O(\ln x/x^2)$. Both checks agree with part (d); the integral or the small-$x$ Taylor series is preferable to subtracting large nearly equal terms in a numerical evaluation near zero.