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ID: past-exam-of-the-perturbation-methods-and-applied-complex-variables-course-of-the-university-of-oxford/2008/a/3/i/solution

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Past exam of the perturbation methods and applied complex variables course of the University of Oxford / 2008 / A / 3 / i / Solution by

Codex  0 2026-10-31

Let

b = π /4

and introduce

J (x) = \int_{0}^{b} e^{i x t^{2}} tan^{2} t d t, I (x) = Re J (x) .

(1)

Deform the contour upward to the steepest descent contour

t = e^{iπ /4} r

from zero and the endpoint path

t = (b^{2} + i s)^{1/2}

from

b

. More precisely the original integral is the first path traversed forward minus the endpoint path traversed forward. Connect the truncated paths where

Im (t^{2}) = R

and let

R \to \infty

. That connector has exponentially small factor

e^{- x R}

, and no pole of

tan t

is crossed: all its poles are real and outside the original segment.

On the first path

tan^{2} t = t^{2} + O (t^{4})

, so its leading contribution is

J_{0} (x) = e^{3 iπ /4} \int_{0}^{\infty} r^{2} e^{- x r^{2}} d r + O (x^{- 5/2}) = e^{3 iπ /4} \frac{π}{4 x ^{3/2}} + O (x^{- 5/2}) .

(2)

The last equality follows by

u = x r^{2}

Γ (3/2) = π /2

On the endpoint path

d t = i d s / (2 b^{2} + i s)

, and

tan^{2} b = 1

. Its negative orientation gives the leading endpoint term

J_{b} (x) = - i e^{i x b^{2}} \int_{0}^{\infty} e^{- x s} \frac{t a n ^{2} b ^{2} + i s}{2 b ^{2} + i s} d s = - \frac{i e ^{i x b^{2}}}{2 b x} + O (x^{- 2}) .

(3)

Taking real parts and inserting

b = π /4

gives

I (x) = \frac{2}{π x} sin (\frac{π ^{2} x}{16}) - \frac{2 π}{8 x ^{3/2}} + O (x^{- 2}) .

(4)

The oscillatory first coefficient can vanish at particular values of

x

; the stated remainder still holds as an additive asymptotic expansion.

Figure 1.
The quadratic phase descends from the stationary endpoint and from the nonstationary upper endpoint
. **Details**.

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