 Solution

ID: past-exam-of-the-perturbation-methods-and-applied-complex-variables-course-of-the-university-of-oxford/2009/a/1/iii/solution

t = e^{iπ /6} r

r \geq 0

. Then

i t^{3} = - r^{3}

and

d t = e^{iπ /6} d r

. Its exact contribution is

I_{0} (x) = e^{iπ /6} \int_{0}^{\infty} e^{- x r^{3}} d r = \frac{e ^{iπ /6}}{3 x ^{1/3}} \int_{0}^{\infty} e^{- s} s^{- 2/3} d s .

Use the Gamma function integral and its recurrence

Γ (4/3) = Γ (1/3) /3

to obtain

I_{0} (x) = e^{iπ /6} Γ (4/3) x^{- 1/3} .

There is no remainder here: the complete infinite descent ray was integrated exactly.

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