 Solution

Put $t=\sqrt{z}\tau$ and $L=z^{3/2}$. The steepest descent phase is $F(\tau )=-{\tau }^3/3+\tau$, with saddles at $\pm 1$. The specified contour can be deformed through the negative saddle, where $F(-1)=-2/3$; the positive saddle would produce the growing exponential and is not the saddle for this contour.

An explicit descending contour is $\tau =u+iv$, where $u=-\sqrt{1+v^2/3}$ and $v$ increases from negative to positive infinity. Its imaginary phase is $v(1-u^2+v^2/3)=0$, and its ends approach the prescribed rays $4\pi /3$ and $2\pi /3$. The real part of the phase has a strict maximum at $v=0$, with quadratic term $-v^2$. Since the integrand is entire and decays in the connecting sectors, contour deformation introduces no residue.

Near the saddle, write $\tau =-1+is+O(s^2)$. Then $F(\tau )=-2/3-s^2+O(s^3)$ and $d\tau =i(1+O(s))ds$. The Gaussian saddle contribution is thereforeThe remainder of the descending contour has smaller real phase, so does not alter this leading term.