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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2005/ii/paper-3/14a/solution

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Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 3 / 14A / Solution by

Codex  0 2026-10-14

Differentiate under the contour integral, with one consistent branch and convergent endpoint behavior. Then

z w^{''} + 2 k w^{'} + z w = \int_{γ} [z (t^{2} + 1) + 2 k t] (t^{2} + 1)^{k - 1} e^{z t} d t = \int_{γ} \frac{d}{d t} [(t^{2} + 1)^{k} e^{z t}] d t .

(1)

The stated zero boundary term therefore makes the integral a solution. On a closed contour this means returning to the same branch; on an open contour it requires the endpoint values to cancel or vanish.

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