Method of Steepest descent integral I am looking to evaluate the following asymptotic integral: 
Find the leading term of asymptotics as $\lambda\to\infty$
$$I(\lambda)=\int_0^1\cos(\lambda x^3)dx$$
Using method of steepest descent along a certain contour. I am having trouble approaching this problem as I don't understand it well. Any help would be appreciated.
 A: To start, recognize that
$$
I(\lambda) = \operatorname{Re} \int_0^1 e^{i\lambda x^3}\,dx.
$$
Now there are several questions that you can ask to get yourself going:


*

*Where is the saddle point?

*What are the paths of steepest descent away from the saddle point?

*How can I deform my contour so that it follows this path of steepest descent?
The last one is a bit tricky since the endpoints of the contour are finite.  You only need to follow a portion of the path of steepest descent though; you can have the contour return to its start/endpoint afterwards.
Let me know if you get stuck on any of these.
A: A general blueprint to solve this type of problems is explained in my Math.SE answer here. Applied to OP's example, one derives that
$$ \int_0^1  \! \mathrm{d}z~e^{i\lambda z^3}~=~J(0)-e^{i\lambda}J(1), \tag{1}$$
where
$$ J(a)~=~\frac{i}{3} \int_0^{\infty} \! \mathrm{d}u~   \frac{e^{-\lambda u}}{(iu+a^3)^{\frac{2}{3}}} .\tag{2} $$
The integral associated with the upper endpoint $a=1$ yields
$$ J(a\!=\!1)~\stackrel{(2)}{=}~\frac{i}{3} \int_0^{\infty} \! \mathrm{d}u~ e^{-\lambda u}\left( 1  - \frac{2i}{3}u + O(u^2) \right)
~=~ \frac{i}{3\lambda}   + \frac{2}{9\lambda^2} + O(\lambda^{-3}) .\tag{3} $$
The integral associated with the lower endpoint $a=0$ yields
$$ J(a\!=\!0)~\stackrel{(2)}{=}~ \frac{e^{\frac{i\pi}{6}}}{3} \int_0^{\infty} \! \mathrm{d}u~   \frac{e^{-\lambda u}}{u^{\frac{2}{3}}}~=~ \frac{e^{\frac{i\pi}{6}}\Gamma\left(\frac{1}{3}\right)}{3\lambda^{\frac{1}{3}}}.\tag{4} $$
This leads to the OP's sought-for expansion

$${\rm Re}  \int_0^1  \! \mathrm{d}z~e^{i\lambda z^3}~\stackrel{(1)+(3)+(4)}{=}~ \frac{\Gamma\left(\frac{1}{3}\right)}{2\sqrt{3}\lambda^{\frac{1}{3}}} + \frac{\sin(\lambda)}{3\lambda}  + O(\lambda^{-2}).\tag{5} $$

