 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2016/iii/paper-336/1/b/i/solution

Codex  0 Created 2026-10-03 Updated 2026-10-06

d = 1/ 2

. At the positive real endpoint the contour is real and crosses the saddle point at zero. The Gaussian integral gives

f (0, λ) = \frac{π}{λ} - \int_{d}^{\infty} e^{- λ x^{2}} d x - \int_{1}^{\infty} e^{- λ x^{2}} d x \sim \frac{π}{λ} .

Both omitted tails are exponentially small. At the negative real endpoint the contour is oriented leftward and does not cross the saddle point:

f (π, λ) = - \int_{d}^{1} e^{- λ x^{2}} d x .

\int_{d}^{\infty} e^{- λ x^{2}} d x \sim e^{- λ d^{2}} / (2 λ d)

. The upper-end correction is

O (e^{- λ} / λ)

and is much smaller. Therefore

f (π, λ) \sim - \frac{e ^{- λ /2}}{2 λ} .

The sign comes from contour orientation, not from a different Gaussian convention.

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