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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/ii/paper-3/27c/solution

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Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 3 / 27C / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-06

Use the entire function

e^{ik z^{3}}

. The steepest descent contour starting at

1

is

z (u) = (1 + i u)^{1/3}, 0 \leq u < \infty,

with the cube-root branch starting at

1

. Along it

ik z^{3} = ik - k u

, so the real part decreases and the imaginary part stays constant. It approaches the ray

ar g z = π /6

, on which

z = e^{iπ /6} u^{1/3}

and

ik z^{3} = - k u

.

Close a contour using the interval

[0, 1]

, these two outgoing paths truncated at parameter

R

, and a short connecting arc. The two endpoints differ by

O (R^{- 2/3})

, and the integrand on the connector decays exponentially in

R

. Cauchy integral theorem therefore gives

\int_{0}^{1} e^{ik t^{3}} d t = I_{1} - I_{2}

.

The specified substitutions give

I_{1} = \frac{e ^{iπ /6}}{3} \int_{0}^{\infty} e^{- k u} u^{- 2/3} d u = \frac{e ^{iπ /6} Γ ( 1/3 )}{3 k ^{1/3}},

and

I_{2} = \frac{i e ^{ik}}{3} \int_{0}^{\infty} e^{- k u} (1 + i u)^{- 2/3} d u .

Expand

(1 + i u)^{- 2/3} = 1 - 2 i u /3 + O (u^{2})

near zero, and apply Watson lemma, or split the integral at a fixed small

u

and integrate the Taylor remainder. This gives the first two asymptotic terms

\int_{0}^{1} e^{ik t^{3}} d t = \frac{e ^{iπ /6} Γ ( 1/3 )}{3 k ^{1/3}} - \frac{i e ^{ik}}{3 k} + O (k^{- 2}) .

Keeping the next term gives

- 2 e^{ik} / (9 k^{2}) + O (k^{- 3})

. The fractional power comes from the degenerate stationary endpoint at zero; the oscillatory integer-power terms come from the ordinary endpoint at one.

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