Past exam of the perturbation methods and applied complex variables course of the University of Oxford / 2009 / A / 1 / iv

The endpoint branch has

t = (1 + i s)^{1/3}

and

d t = (i /3) (1 + i s)^{- 2/3} d s

. Its reversed orientation in the deformed contour gives

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

s = 0

(1 + i s)^{- 2/3} = 1 - \frac{2 i}{3} s + O (s^{2}) .

I_{1} (x) = - \frac{i e ^{i x}}{3 x} - \frac{2 e ^{i x}}{9 x ^{2}} + O (x^{- 3}) .

Adding the exact cubic-saddle contribution gives the two leading terms

I (x) = e^{iπ /6} Γ (4/3) x^{- 1/3} - \frac{i e ^{i x}}{3 x} + O (x^{- 2}) .

If the next term is retained, it is

- 2 e^{i x} / (9 x^{2})

. The different powers reflect a degenerate saddle at zero and an ordinary endpoint at one.

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