$I(x) = \int_a^b f(t) e^{x\phi(t)} dt$   as $x\rightarrow \infty$

For $\phi(t)$, $f(t)$ are generally complex, and the integral being along a complex contour in general.

See justification in notes..

Also: Handouts from lecture

Steepest descent contour refers to the contour of steepest descent of $u$, the real part of $\phi(t) = u(t) +iv(t)$. That is, the contour parallel to its gradient, $\nabla u(t)$. This is because, for $\phi(t)$ an analytic function, $\nabla u(t)$ is perpendicular to $\nabla v(t)$, so that the steepest descent contour is also a contour of constant imaginary part of $\phi(t)$. This later condition (together with others, depending on the problem) is often used to find the contour.

The other conditions may be:

• If the contour goes from one valley (valley in the $u$ landscape, which can only have saddle points, due to Cauchy–Riemann equations) to another valley, then the steepest descent contour must pass through a saddle point (in $u$ landscape). This is because in one valley $du/ds<0$ ($s$ being distance along path), and in the other $du/ds>0$. Therefore, at some point, $du/ds=0$. However, in a steepest descent contour $dv/ds=0$ everywhere. Therefore, at that point $du/ds=0$ and $dv/ds=0$, which implies that $\nabla u(t)$, so it is a saddle point. In this situation, the Laplace integral will get contribution from the saddle points (where $u$ is largest).
• If the contour starts at a certain point, not infinity, then that point has to be kept, and the Laplace integral may get a contribution from it.

Method of steepest descents

1. Deform the contour to be the steepest descent contour through the relevant saddle node(s).

2. Evaluate the local contribution from the saddle, exactly as in Laplace method.

3. Evaluate the local contribution from the end points, exactly as in Laplace method.

Remember that when deforming the contour we must include the contribution from any poles that we cross.

Example: Steepest descents on the gamma function

Example: Steepest descents on the Airy function

Both of these are moveable saddle problems, so we first need to rescale variables, so that the saddle is fixed.

Revise Watson lemma, and examples