 Solution

ID: 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} = arsinh x

, equal to

θ /2

in part (b). The hyperbolic function identity

(\frac{y}{1 + y})^{3/2} d y = 2 \frac{s i n h ^{4} u}{c o s h ^{2} u} d u = 2 (cosh^{2} u - 2 + sech^{2} u) d u

I (x) = [\frac{1}{2} sinh (2 u) - 3 u + 2 tanh u]_{0}^{u_{x}} = x (1 + x) + 2 \frac{x}{1 + x} - 3 arsinh x .

δ_{1}

therefore yields

δ_{2} (x) = 1 + \frac{3}{x} - 3 \frac{1 + x}{x ^{3}} arsinh x = 1 + \frac{3}{x} - 3 \frac{1 + x}{x ^{3}} ln (x + 1 + x) .

At small

x

, cancellation between the terms leaves the Taylor series

δ_{2} = 2 x /5 - 8 x^{2} /35 + 16 x^{3} /105 + O (x^{4})

. At large

x

arsinh x = \frac{1}{2} ln (4 x) + O (x^{- 1})

, so the correction to one is

[3 - \frac{3}{2} ln (4 x)] / 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.

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