Saddle point and stationary point approximation of the Airy equation Happy New Year to you all.  
Let $$\tag 1 J(N)=\int_a^b e^{Nf(x)}dx$$ where $N\in\mathbb R$ and $N>>1$ and $f(x)$ has a global maximum at $x=x_0$ with  Taylor expansion $$f(x) \approx f(x_0)-|f''(x_0)|\frac{(x-x_0)^2}{2}.$$ Then by the saddle point approximation, we get 
$$ \tag 2 J(N)\approx \frac{e^{Nf(x_0)}}{\sqrt{N}}\sqrt{\frac{2\pi}{|f''(x_0)|}}\left(1+O\left(\frac{1}{N}\right)\right).$$ 
Let $$\tag 3  Ai(x)= \int_\gamma \exp{\left(tx-\frac{t^3}{3}\right)}dt$$ where $\gamma$ is a contour in $\mathbb{C}$ chosen such that the integral converges and the integrand vanishes as $t\to\infty$.   
First question: For $x>0$, I want to find an approximation to (3). So, by letting $t=u \sqrt{x}$ , I get $$ Ai(x) = \sqrt{x} \int_{\mathbb{R}_+}\exp{\left(\left(u -\frac{u^3}{3}\right)x^{3/2}\right)}$$ Doing so allows me to identify the following:$$ N= x^{3/2}~~~f(u)=u-u^3/3~~~$$ from which I get $$ Ai(x>0) = \sqrt{2\pi}e^{{2/3}z^{3/2}}x^{-1/4}\left(1+O\left(z^{-3/2}\right)\right).$$ However, the solution is given as $$ Ai(x>0) = \sqrt{2\pi}e^{{-2/3}z^{3/2}}x^{-1/4}\left(1+\left(x^{-3/2}\right)\right).$$ Obviously, the my trouble stems from the fact that using  (2), I keep getting 2/3 and not -2/3. I'd be very happy if someone could point me in the right direction.   
Second question: For $x<0$, can I get help in using the stationary phase approximation to approximate (2)? 
Please, if what I've done above is complete hokum, let me know as well. Thanks.
 A: I) Judging from OP's eq. (2), it seems OP is conflating Laplace's method in real analysis, and the method of steepest descent in complex analysis. The latter has extra complex phase factors, which undoubtedly resolves OP's sign problem.
II) In more detail, the Airy function of first kind is defined as
$$ Ai(z)~:=~\int_{i\mathbb{R}+0^{+}} \! \frac{du}{2\pi i} \exp\left(\frac{u^3}{3}-uz \right)~\stackrel{u=\sqrt{r} t}{=}~\frac{\sqrt{r}}{2\pi i} \int_{i\mathbb{R}+0^{+}} \! dt~\exp\left(r^{\frac{3}{2}} S(t,\theta) \right) , \qquad z~=~re^{i\theta},$$ 
$$ S(t,\theta)~:=~\frac{t^3}{3}-te^{i\theta}, \qquad S^{\prime}(t,\theta)~:=~t^2-e^{i\theta},\qquad S^{\prime\prime}(t,\theta)~:=~2t. \tag{1}$$  
Here $r\gg 1$. There are two critical points:
$$ t_{\sigma} ~=~\sigma e^{\frac{i\theta}{2}}, \qquad S_{\sigma}~=~ - \frac{2\sigma}{3}e^{\frac{3i\theta}{2}} , \qquad S^{\prime\prime}_{\sigma}~=~  2\sigma e^{\frac{i\theta}{2}}, \qquad \sigma ~\in~\{\pm 1\} .\tag{2}$$ 
The steepest decent contribution from each critical point comes from a Gaussian integral:
$$ I_{\sigma}~=~\frac{\sqrt{r}}{2\pi i}\sqrt{\frac{2\pi}{-r^{\frac{3}{2}} S^{\prime\prime}_{\sigma}} } \exp\left(r^{\frac{3}{2}} S_{\sigma} \right)$$
$$ ~=~\frac{1}{2\sqrt{\pi}i \sqrt[4]{r}}  \exp\left(-\frac{2\sigma}{3}r^{\frac{3}{2}} e^{\frac{3i\theta}{2}} -\frac{i\theta}{4}\right)\left\{\begin{array}{rcl}  \pm i & \text{for}& \sigma~=~+1 \cr\cr 1 & \text{for}& \sigma~=~-1\end{array}  \right\} .\tag{3}$$
Case $z<0 ~\Leftrightarrow ~\theta=\pi$: Critical points: $t_{\sigma}=\sigma i$. The $Ai$-contour along the imaginary axis is reproduced by a combination of both the two steepest decent contours:  
$$ Ai(-r)~\sim~\pm I_{+1}-I_{-1}~=~\frac{1}{\sqrt{\pi} \sqrt[4]{r}}\sin\left( \frac{2}{3}r^{\frac{3}{2}}+\frac{\pi}{4}\right)$$
$$~=~\frac{1}{\sqrt{\pi} \sqrt[4]{r}}\cos\left( \frac{2}{3}r^{\frac{3}{2}}-\frac{\pi}{4}\right)\quad \text{for} \quad r\to \infty.  \tag{4}$$
Case $z>0 ~\Leftrightarrow~ \theta=0$: Critical points: $t_{\sigma}=\sigma$. The $Ai$-contour  along the imaginary axis is reproduced by the steepest decent contour corresponding to the critical point $t_{+1}$:  
$$ Ai(r)~\sim~\pm I_{+1}~=~\frac{1}{2\sqrt{\pi} \sqrt[4]{r}}\exp\left(- \frac{2}{3}r^{\frac{3}{2}}\right)\quad \text{for} \quad r\to \infty. \tag{5}$$
The $\pm$ in eqs. (3), (4) & (5) are correlated. See also the Wikipedia page for the Stokes phenomenon.
