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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2013/ib/paper-2/13d/iii/solution

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Past exam of the mathematics course of the University of Cambridge / 2013 / ib / Paper 2 / 13D / iii / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-07

For

z = R + i y

,

0 \leq y \leq π

, the numerator modulus is

e^{- 2 R y / π}

and the denominator modulus is at least

1 - e^{- 2 R}

. Hence the magnitude of the right-side contour integral is at most

\frac{1}{1 - e ^{- 2 R}} \int_{0}^{π} e^{- 2 R y / π} d y = \frac{π}{2 R} .

On

z = - R + i y

the corresponding bounds are

e^{2 R y / π}

and

e^{2 R} - 1

, giving exactly the same bound

π / (2 R)

. Thus the total vertical contribution is $O (R^{- 1})$ and vanishes. This integral estimate handles the endpoints where a pointwise uniform-decay argument would fail.

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