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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.

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I think it should be $N=x^{3/2}$ and in the integral before this formula as well (and $x^{3/2}$ should be part of the exponent, too). –  Fabian Dec 31 '12 at 23:30
    
You found the asymptotic expansion of the Bi-function. The Ai-function is dominated by the saddle point at $u_0=-1$. The Ai-function should be defined such that it vanishes for $x\to \infty $ (I guess there is another typo in the post) which your solution does not do. –  Fabian Dec 31 '12 at 23:35
    
@Fabian: I've fixed the typo. –  Jonjo Dec 31 '12 at 23:37
    
@Fabian: Can you please point out the typo. did I define The Ai-function incorrectly? –  Jonjo Jan 2 '13 at 14:04
    
I guess Ai is defined such that it vanishes for $x\to\infty$ (and not fur $t\to\infty$ as you state in your post). –  Fabian Jan 2 '13 at 14:06
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