Through some calculations I ended up with the integral $\int_0^\infty\exp(iz(\frac{1}{3}t^3+t))dt$. I would like to obtain the result that the behaviour of the integral is $\sim\frac{i}{z}$ as $z\rightarrow\infty$

I am not very good in obtaining a nice asymptotic approximation but my guess here is to use the method of steepest descents. Calculating the derivative of $\frac{1}{3}t^3+t$ and setting it to zero gives me $t_1=i,t_2=-i$ but how can we gain the directions of steepest descent now?