Tricky steepest descent applied to an inverse Fourier transform

The saddle points of $\phi$ are $$k_{1,2} = \frac {i(a \pm \sqrt {a^2 - 3 b})} {3 b}.$$ Let $a > 0, b > 0$ for simplicity. Consider the case $a^2 - 3 b < 0$. The contour lines $\operatorname{Re} \phi = \text{const}$ will look like this, dividing the plane around a saddle point into four regions:

The contour lines go through the saddle points in the directions $1$ and $i$. $\operatorname{Re} \phi$ will have maxima at the saddle points if the integration contour (the green line) goes through $k_1$ in the direction $(-1)^{1/4}$ and through $k_2$ in the direction $(-1)^{-1/4}$. With the parametrizations $k = k_{1, 2} + (-1)^{\pm 1/4} \xi$, expanding $\phi(\xi)$ around $\xi = 0$ to the second order and extending the integration range to infinity, the leading term is given by $$\int_{-\infty}^\infty f(k) e^{t \phi(k)} dk \sim \\ \left( (-1)^{1/4} f(k_1) e^{\phi(k_1) t} + (-1)^{-1/4} f(k_2) e^{\phi(k_2) t} \right) \int_{-\infty}^\infty e^{-t \xi^2 \sqrt {3 b - a^2}} d\xi = \\ \frac {\sqrt {2 \pi} e^{-a x/(3 b)}} {(3 b - a^2)^{1/4}} \exp\left( -\frac {a (9 b - 2 a^2) t} {27 b^2} \right) \frac 1 {\sqrt t} (\cos \omega(t) + \sin \omega(t)), \\ \omega(t) = \frac {\sqrt {3 b - a^2} ((6 b - 2 a^2) t + 9 b x)} {27 b^2}.$$

Now consider $a^2 - 3 b > 0$. The contours of $\operatorname{Re} \phi$ will look like this:

The directions of the contour lines are $(-1)^{\pm 1/4}$. Thus we need an integration contour going only through the saddle point $k_2$ in the direction $1$. With the parametrization $k = k_2 + \xi$, the leading term is given by $$\int_{-\infty}^\infty f(k) e^{t \phi(k)} dk \sim f(k_2) e^{\phi(k_2) t} \int_{-\infty}^\infty e^{-t \xi^2 \sqrt {a^2 - 3b}} d\xi = \\ \frac {\sqrt {\pi} e^{(\!\sqrt {a^2 - 3 b} - a)x / (3b)}} {(a^2 - 3 b)^{1/4}} \exp\left( -\frac {(2 (a^2 - 3 b)^{3/2} - a (2 a^2 - 9 b)) t} {27 b^2} \right) \frac 1 {\sqrt t}.$$ Finally, let $b = a^2/3$. There is one saddle point of the second order (the third derivative is non-zero):

Taking $k = k_1 + (-1)^{1/6} \xi$ for the left part of the contour and $k = k_1 + (-1)^{-1/6} \xi$ for the right part, we obtain $$\int_{-\infty}^\infty f(k) e^{t \phi(k)} dk \sim \\ f(k_1) e^{\phi(k_1) t} \left( (-1)^{1/6} \int_{-\infty}^0 e^{a^2 t \xi^3/3} d\xi + (-1)^{-1/6} \int_0^\infty e^{-a^2 t \xi^3/3} d\xi \right) = \\ \frac {\Gamma \left( \frac 1 3 \right) e^{-x/a}} {3^{1/6} a^{2/3}} \frac {e^{-t/(3 a)}} {t^{1/3}}.$$