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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ii/paper-3/7a/a/solution

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Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 3 / 7A / a / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-03

Differentiating under the contour integration sign gives

z w^{''} + w = \int_{γ} e^{z t} (z t^{2} f (t) + f (t)) d t .

Since

\partial_{t} e^{z t} = z e^{z t}

, integration by parts turns this into

[e^{z t} t^{2} f (t)]_{\partial γ} + \int_{γ} e^{z t} [f (t) - \frac{d}{d t} (t^{2} f (t))] d t .

The integral vanishes when

(t^{2} f)^{'} = f, \frac{f ^{'}}{f} = \frac{1}{t ^{2}} - \frac{2}{t} .

Hence, up to an irrelevant multiplicative constant,

f (t) = t^{- 2} e^{- 1/ t} .

The remaining endpoint condition is

[e^{z t - 1/ t}]_{\partial γ} = 0.

Thus each open end of $γ$ must lie where $Re (z t - 1/ t) \to - \infty$ ; a closed contour also makes the boundary term vanish when the integrand is single-valued along it.

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