$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\bZ}{\mathbb{Z}}$ $\newcommand{\bC}{\mathbb{C}}$ $\newcommand{\const}{\operatorname{const}}$ $\newcommand{\Hess}{\operatorname{Hess}}$ $\newcommand{\supp}{\operatorname{supp}}$ $\newcommand{\arg}{\operatorname{arg}}$

We consider \begin{equation} I(k)= \int_L e^{k \phi(z)}f(z)\,dz \label{eq-2.5.1} \end{equation} where now $L\subset \Omega$ is a contour from $z_0$ to $z_1$, $\Omega \subset \bC$ is a simple-connected domain and $\phi$, $f$ are holomorphic in $\Omega$. Recall from complex variables that in this framework $I(k)$ does not depend on the choice of $L$.

We are interested in the asymptotics of $I(k)$ as $k\to +\infty$.

Obviously we need to select a contour $L$ in such way that
$\max_{z\in L} \Re\phi(z)$ was a small as possible. Consider a landscape of $\Re \phi(z)$. From Complex Variables we know that both $\Re \phi(z)$ and
$\Im \phi(z)$ are harmonic functions. We also know that harmonic functions do not have local maxima or minima in the inner points, only *saddle* points.

**Lemma 1.**

- $|\nabla \Re\phi |=|\nabla \Im \phi|=|\phi'|$;
- $z_0$ is a stationary point of $\Re \phi$ if and only if $\phi'(z_0)=0$.
- $\partial^\alpha \Re\phi (z_0)=0$ for all $\alpha:\,|\alpha|\le m-1$ with $m\ge 2$ if and only if $\phi' (z_0)=\ldots =\phi^{(m-1)}(z_0)=0$.

*Proof.* From Complex Variables.

If $\phi' (z_0)=\ldots =\phi^{(m-1)}(z_0)=0$ and $\phi^{(m)}(z_0)\ne 0$ we call $(m-1)$ *multiplicity* of the saddle. If $m=2$ then the saddle is *simple*.

**Lemma 2.**
Consider lines along which $\Re \phi$ changes the fastest. Those are tangent to $\nabla \Re \phi$. Along these lines $\Im \phi=\const$.

**Lemma 3.**
Let $\phi' (z_0)=\ldots =\phi^{(m-1)}(z_0)=0$ and $A:=\phi^{(m)}(z_0)\ne 0$. Then in the vicinity of $z_0$

- $\{z:\,\Re \phi(z) =\Re \phi(z_0)\}$ consists of $2m$ "rays" (curved away from $z_0$) issued from $z_0$ in the directions \begin{equation} e^{i(\arg A + \pi k +\pi/2)/m}\qquad k=0,1,\ldots, 2m-1 \label{eq-2.5.2} \end{equation} dividing vicinity into $2m$ sectors in which $\Re \phi(z)\gtrless \Re\phi(z_0)$ alternatively. Here $\arg A$ is an argument of $A=|A|e^{i\arg A}$.
- In each of $m$ sectors in which $\Re \phi(z)< \Re\phi(z_0)$ there is a single ray of the steepest descent issued from $z_0$ in the direction \begin{equation} e^{i(\arg A + \pi k)/2m}\qquad k=1,3,\ldots, 2m-1 \label{eq-2.5.3} \end{equation} and in each of $m$ sectors in which $\Re \phi(z)> \Re\phi(z_0)$ there is a single ray of the steepest ascent issued from $z_0$ in the direction \begin{equation} e^{i(\arg A + \pi k +\pi/2)/m}\qquad k=0,2,\ldots, 2m-2. \label{eq-2.5.4} \end{equation}

*Proof.*
Sufficient to consider a toy-model $\phi(z)=Az^m$ with $z_0=0$.

$m=2$ | $m=3$ |

Orange lines are $\{z:\,\Re \phi(z) =\Re \phi(z_0)\}$, blue lines are of the steepest descent and red lines of the steepest ascent |

**Lemma 4.**
One can select contour from $z_0$ to $z_1$ in such a way that it can be broken into several contours $L_1,\ldots ,L_K$ such that

- In some contours $L_j$ equality $\Re\phi(z)=\max_{\zeta\in L} \Re\phi (\zeta)$ holds in the single point $z^*_j$ which is its end-point. Near this point $L_j$ is a line of the steepest descent.
- In the remaining contours $\Re\phi(z)<\max_{\zeta\in L} \Re\phi (\zeta)$ everywhere.

*Proof.* This lemma looks intuitively obvious (but the rigorous proof is a bit tedious).

Obviously we need to calculate only contributions of the contours of type (a). We consider just one contour $L$ of type (a) from $z^*$ to some point (does not matter).

**Theorem 1.**
Let $L$ be a contour from $z^*$ to some point (does not matter) and $\Re\phi(z)<\Re\phi(z^*)$ in each point of this contour $z\ne z^*$. Let
$\phi' (z^*)=\ldots =\phi^{(m-1)}(z^*)=0$ and $A:=\phi^{(m)}(z^*)\ne 0$, $m\ge 2$. Let $L$ be a contour of the steepest descent.

Then \begin{equation} I(k)\sim e^{ik\phi(z^*)}\sum_{n\ge 0}\kappa_n k^{-(n+1)/m} \label{eq-2.5.5} \end{equation} where \begin{equation} \kappa_0= \Gamma ((m+1)/m) |f^{(m)}(z^*)/m!|^{-1/m}e^{i\theta}f(z^*) \label{eq-2.5.6} \end{equation} where $e^{i\theta}$ is a direction of $L$ in $z^*$.

**Remark 1.**
If $m=1$ (\ref{eq-2.5.5}) holds with
\begin{equation}
\kappa_0= -(\phi'(z^*))^{-1}f(z^*).
\label{eq-2.5.7}
\end{equation}