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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2024

Paper 2

Repeated integration by parts starts with
$$
\operatorname{si}(x)
=\frac{\cos x}{x}
-\frac1x\int_1^\infty\frac{\cos(xt)}{t^2}\,dt
=\frac{\cos x}{x}+\frac{\sin x}{x^2}
-\frac2{x^2}\int_1^\infty\frac{\sin(xt)}{t^3}\,dt.
$$
Continuing alternates sine and cosine and multiplies by successive integers. After finitely many steps the remainder is bounded by a constant times the next inverse power, so the <fixed-lower-limit sine-integral expansion> is
$$
\operatorname{si}(x)\sim
\cos x\sum_{n=0}^\infty(-1)^n(2n)!x^{-2n-1}
+\sin x\sum_{n=0}^\infty(-1)^n(2n+1)!x^{-2n-2}.
$$
Therefore
$$
\boxed{a_n=(-1)^n(2n)!,\qquad
b_n=(-1)^n(2n+1)!}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 2 / 32C / c / Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 2 / 32C / c / Solution

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Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 2 / 32C / c / Solution

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