How to obtain the asymptotic behavior of this integral? Consider the integral
$$ I(x) = \int_{0}^{\infty} \exp \bigl( - s^2/2 \bigr) \cos \bigl( xs + \lambda s^3 \bigr) \, \text{d} s,$$
where $\lambda >0$ is a constant. I would like to know the asymptotic behavior of this integral in the limit $x \rightarrow \infty$. Any ideas on how to proceed?
 A: This is an elaboration on Maxim's comment. Your integral is
$$
I(x) = \frac{1}{2}\int_{ - \infty }^{ + \infty } {\exp \left( { - \frac{1}{2}s^2  + i\lambda s^3  + ixs} \right)ds} .
$$
With the change of variables $s = \frac{t}{{(3\lambda )^{1/3} }} + \frac{1}{{6i\lambda }}$, we find
$$
I(x) \!=\! \frac{1}{2}\frac{1}{{(3\lambda )^{1/3} }}\exp \left( {\frac{x}{{6\lambda }} \!+\! \frac{1}{{108\lambda ^2 }}} \right)\int_{ - \infty  + i/(6\lambda )}^{ + \infty  + i/(6\lambda )} {\exp \left( {i\left( {\frac{{t^3 }}{3} \!+\! \frac{1}{{(3\lambda )^{1/3} }}\!\left( {x \!+\! \frac{1}{{12\lambda }}} \right)t} \right)} \right)dt}. 
$$
Pushing the contour downwards and using the know integral representation of the Airy function, we deduce
\begin{align*}
I(x) & = \frac{1}{2}\frac{1}{{(3\lambda )^{1/3} }}\exp \left( {\frac{x}{{6\lambda }} + \frac{1}{{108\lambda ^2 }}} \right)\int_{ - \infty }^{ + \infty } {\exp \left( {i\left( {\frac{{t^3 }}{3} + \frac{1}{{(3\lambda )^{1/3} }}\left( {x + \frac{1}{{12\lambda }}} \right)t} \right)} \right)dt} 
\\ &  = \frac{1}{{(3\lambda )^{1/3} }}\exp \left( {\frac{x}{{6\lambda }} + \frac{1}{{108\lambda ^2 }}} \right)\int_{ - \infty }^{ + \infty } {\cos \left( {\frac{{t^3 }}{3} + \frac{1}{{(3\lambda )^{1/3} }}\left( {x + \frac{1}{{12\lambda }}} \right)t} \right)dt} 
\\ & = \frac{\pi }{{(3\lambda )^{1/3} }}\exp \left( {\frac{x}{{6\lambda }} + \frac{1}{{108\lambda ^2 }}} \right)\operatorname{Ai}\left(\frac{1}{{(3\lambda )^{1/3} }}\left( {x + \frac{1}{{12\lambda }}} \right) \right).
\end{align*}
Now we know that
$$
\operatorname{Ai}(z) = \frac{1}{{2\sqrt \pi  z^{1/4} }}\exp \left( {-\frac{{2z^{3/2} }}{3}} \right)\left( {1 + \mathcal{O}\!\left( {\frac{1}{{z^{3/2} }}} \right)} \right)
$$
as $z\to \infty$ in the sector $|\arg z|\leq \pi-\delta$ ($<\pi$). Since
$$
\frac{2}{3}\frac{1}{{\sqrt {3\lambda } }}\left( {x + \frac{1}{{12\lambda }}} \right)^{3/2}  = \frac{{2x^{3/2} }}{{3\sqrt {3\lambda } }} + \frac{{x^{1/2} }}{{12\sqrt 3 \lambda ^{3/2} }} + \frac{1}{{576\sqrt 3 \lambda ^{5/2} x^{1/2} }} -  \cdots 
$$
and
$$
(3\lambda )^{1/12} \left( {x + \frac{1}{{12\lambda }}} \right)^{ - 1/4}  = (3\lambda )^{1/12} x^{ - 1/4} \left( {1 - \frac{1}{{48\lambda x}} +  \cdots } \right),
$$
we obtain
$$
I(x) = \frac{{\sqrt \pi  }}{{2(3\lambda x)^{1/4} }}\exp \left( {-\frac{{2x^{3/2} }}{{3\sqrt {3\lambda } }} + \frac{x}{{6\lambda }} - \frac{{x^{1/2} }}{{12\sqrt 3 \lambda ^{3/2} }} + \frac{1}{{108\lambda ^2 }}} \right)\left( {1 + \mathcal{O}\!\left( {\frac{1}{{x^{1/2} }}} \right)} \right)
$$
as $x\to \infty$ in the sector $|\arg x|\leq \pi-\delta$ ($<\pi$). More terms in the expansion can be obtained if necessary.
A: (I'll denote $\lambda$ as $t$) Call your integral $u_t(x)$. Then $u$ satisfies the differential equation
$$u''-\frac{1}{3t}u'-\frac{x}{3t}u=0.$$
This follows from using integration by parts on $u'$. We will look to expand $u$ around the point $\infty$ (equivalently $u(1/x)$ around $x=0$). To do so, we let $u=e^S$, and find that the ODE turns asymptotically into
$$(S')^2\sim\frac{x}{3t},$$
meaning that $S\sim - \frac{2}{3\sqrt{3}}\frac{x^{3/2}}{t^{1/2}}$ (we choose the sign of $S$ so that $u$ decays to zero for large $x$). Thus your asymptotic behavior as $x\to\infty$ should be dominated by
$$u_t(x)\sim \exp\left( -\frac{2}{3\sqrt{3}}\frac{x^{3/2}}{t^{1/2}}\right).$$
Here, I a not using $\sim$ in a technical sense, just abusing notation to say that this is the main contribution. You can substitute $S=-\frac{2}{3\sqrt{3}}\frac{x^{3/2}}{t^{1/2}}+h$ into the original differential equation, where $h=o(x^{3/2})$, and you find that $h(x)=x/6$ makes the ODE consistent. So the solution is dominated by a term of the form
$$u_t(x) \sim \exp\left( -\frac{2}{3\sqrt{3}}\frac{x^{3/2}}{t^{1/2}}+x/6\right).$$
Again, I am not using $\sim$ technically here. You can keep going to obtain lower order corrections.
Edit: a second round edits were made after reading Gary and Maxim's comments.
