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

Use the entire function $e^{ik{z}^3}$. The steepest descent contour starting at $1$ iswith the cube-root branch starting at $1$. Along it $ik{z}^3=ik-ku$, so the real part decreases and the imaginary part stays constant. It approaches the ray $\arg z=\pi /6$, on which $z=e^{i\pi /6}{u}^{1/3}$ and $ik{z}^3=-ku$.

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 ${\int }_0^1{e}^{ik{t}^3}dt=I_1-I_2$.

The specified substitutions giveandExpand $(1+iu{)}^{-2/3}=1-2iu/3+O(u^2)$ 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 termsKeeping the next term gives $-2e^{ik}/(9k^2)+O(k^{-3})$. The fractional power comes from the degenerate stationary endpoint at zero; the oscillatory integer-power terms come from the ordinary endpoint at one.