# proving the saddle point method in a specific case

$1:$ the problem

Let $f : U \to \Bbb{C}$ be analytic on some open set $U$ that includes the closed unit ball. Define a path $\gamma$ by:

$\gamma(t) = e^{i t}$, -$\pi < t \leq \pi$

I want to show that the following holds:

$\int_\gamma\ e^{\lambda\ f(z)}\ dz =\ (\frac{2\pi}{-\lambda\ f''(1)})^{1/2}\ e^{\lambda\ f(1)}\ + O(e^{\lambda\ f(1)}/\lambda^{3/2}) \hspace {10 mm}(\lambda \to +\infty)$

under the following assumptions on $f$:

$f(1)$ is real, $f''(1)$ is real, and $\forall$ $\left|z\right|=1$ s.t. $z\neq 1$, $\Re(f(z)) < f(1)$.

I'm not sure that this is true. Can someone tell me if additional assumptions are required on $f$ and/or help me out with the proof?

$2:$ my attempt at proof

$\int_\gamma e^{\lambda\ f(z)}dz\ =\ \int_{-\pi}^\pi\ i\ e^{it}\ e^{\lambda\ f(e^{i\ t})}\ =\ i\ \int_{-r}^r\ e^{\lambda\ [\ f(1)\ -\ \frac{1}{2}f''(1)\ t^2\ +\ T(t)\ ]}\ +\ ...$

where $r\ \in\ (0,\ \pi)$ and $T$ is a function satisfying

$\left|T(t)\right|\ \leq\ M\ t^4\$ for some constant $M$. I have omitted terms that I already know how to deal with (ie, I can show they get sufficiently small as $\lambda$ gets large).

If $T$ were not there, I could use the Laplace method (I have included this below for reference). With $T$ there, though, the Laplace method cannot be immediately applied, since the function in the exponent is not real. If I expand the exponent, I get terms like

$\frac{1}{n!}\ \int_{-r}^r\ (M\ \lambda\ t^4)^n\ e^{\lambda\ [\ f(1)\ -\ \frac{1}{2}\ f''(1)\ t^2\ ]}$

The Laplace method only says that this term $=\ O(e^{\lambda\ f(1)}\ \lambda^{n\ -\ \frac{3}{2}})\ \hspace {10mm}\ (\lambda \to +\infty)$, which is not very helpful.

$3:$ Laplace method

Let $g:[a,b] \to \Bbb{C},\ h:[a,b] \to R$ be smooth (the assumption of smoothness can be weakened, but this is good enough for my purposes). Suppose that h has an absolute minimum at $x_0 \in (a,b)$ and that $\frac{dh}{dx}\ \neq\ 0\ on\ (a,b)/{x_0}.$

Then we have

$\int_a^b\ g(x)\ e^{-\lambda\ h(x)}dx\ =\ (\frac{2\pi}{\lambda\ h''(x_0)})^{1/2}\ g(x_0)\ e^{-\lambda\ h(x_0)}\ +\ O(e^{-\lambda\ h(x_0)}\ /\ \lambda^{3/2}) \hspace {10 mm} (\lambda \to +\infty)$

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