 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-4/7e/solution

Codex  0 Created 2026-09-23 Updated 2026-09-25

y (z) = \int_{C} e^{z t} f (t) d t .

Differentiating under the integral and substituting into

z y^{''} + y^{'} - zy = 0

gives

\int_{C} e^{z t} [z (t^{2} - 1) f (t) + t f (t)] d t = 0.

Since

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

[e^{z t} (t^{2} - 1) f (t)]_{\partial C} - \int_{C} e^{z t} [(t^{2} - 1) f^{'} (t) + t f (t)] d t = 0.

It is therefore enough to choose

f

and

C

so that

(t^{2} - 1) f^{'} (t) + t f (t) = 0

and the boundary term vanishes. Solving the first-order equation on

- 1 < t < 1

gives

f (t) = \frac{K}{1 - t ^{2}} .

Take

C

to be the interval

[- 1, 1]

. Although

f

has integrable endpoint singularities,

(t^{2} - 1) f (t) = - K 1 - t^{2}

vanishes at both endpoints, so the boundary term is zero. Thus

y (z) = K \int_{- 1}^{1} \frac{e ^{z t}}{1 - t ^{2}} d t .

z = 0

, substituting

t = sin θ

gives

\int_{- 1}^{1} \frac{d t}{1 - t ^{2}} = π .

The normalization

y (0) = 1

therefore sets

K = 1/ π

I_{0} (z) = \frac{1}{π} \int_{- 1}^{1} \frac{e ^{zs}}{1 - s ^{2}} d s .

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