# Method of steepest descents for $\int_0^\infty\exp(iz(\frac{1}{3}t^3+t))dt\sim\frac{i}{z}$

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?

## 1 Answer

This is not perhaps explicitely about steepest descent: Let $h$ be the inverse of the function $t\mapsto \frac 13 t^3+t$. If we take $s=\frac 13 t^3+t$ as a new variable of integration, the integral becomes $\int_ 0^\infty \textrm{exp}(i zs) h'(s)\,ds$. An integration by parts gives (assuming $Im(z)\geq 0$) $\int_ 0^\infty \textrm{exp}(i zs) h'(s)\,ds = \Big /_0^\infty \frac 1{iz}\textrm{exp}(i zs) h'(s) - \int_0^\infty \frac 1{iz}\textrm{exp}(i zs) h''(s)\, ds\\= \frac iz + \frac i{z}\int_0^\infty \textrm{exp}(i zs) h''(s)\, ds$,
since $h'(0)=1$. Because $h''$ is integrable, the integral $\int_0^\infty \textrm{exp}(i zs) h''(s)\, ds$ approaches zero as $z\to \infty$ (assuming $Im(z)\geq 0$) and this should give the desired result.