Steepest descent of integrand with a movable saddle? I want to apply the steepest descent method to the following integration:
$$
\int_0^\infty e^{-x^2 + i \sqrt{x^2 + 1} \cdot \lambda } dx
$$
It has movable saddle so I need to transform it into the standard form, something like
$$
\int_C g(z) e^{\lambda f(z) } dz
$$
I know for Gamma function:
$$ \Gamma(x+1)= \int_0^\infty e^{-t} t^{x} dt =\int_0^\infty e^{-t + x \ln t} dt $$
letting $t = x s$ transforms it into standard form. And for Airy function:
$$Ai(x) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} e^{i (t^3/3 + x t)} dt $$
letting $t = \sqrt{x} z$ does the trick. 
However, no change of variable seems to transform my integration into the standard form. For this reason I can not proceed at all. Any hint or suggestion ? Thanks!
 A: Let us here sketched a derivation. 


*

*OP's integral has a saddle point at $z=0$ with angular steepest descent  direction $\frac{\pi}{4}$. Therefore, inspired by the method of steepest descent, we rewrite OP's integral as
$$I(\lambda) ~:=~\int_{\mathbb{R}_+}\!\mathrm{d}z~ 
\exp\left\{-z^2+i\lambda\sqrt{1+z^2}\right\}
~=~ \frac{e^{i\lambda}}{2}J(\lambda),\qquad \lambda~>~0, \tag{1}$$
where 
$$\begin{align}J(\lambda) ~:=~&\int_{\mathbb{R}}\!\mathrm{d}z~ \exp\left\{-z^2+i\lambda\left(\sqrt{1+z^2}-1\right)\right\} \cr
~\stackrel{z=\frac{u}{\sqrt{\lambda}}\exp\left(\frac{i\pi}{4}\right)}{=}&
\frac{1}{\sqrt{\lambda}}\exp\left(\frac{i\pi}{4}\right)
K(\lambda),\end{align}\tag{2} $$ 
where
$$ K(\lambda)~:=~\int_{\mathbb{R}}\!\mathrm{d}u~ 
\exp\left\{-\frac{iu^2}{\lambda}-f_{\lambda}(u)\right\},\tag{3}$$
and where
$$ -f_{\lambda}(u)~:=~i\lambda\left(\sqrt{1+\frac{iu^2}{\lambda}}-1\right)
~=~-\frac{u^2}{2} + \frac{iu^4}{8\lambda}+ O(u^6\lambda^{-2}).\tag{4} $$ 

*Let us define the non-negative function
$$ f(u)~:=~\frac{u^2}{2}. \tag{5} $$
Clearly 
$$  f_{\lambda}(u)~\longrightarrow~ f(u)\quad\text{for}\quad\lambda~\to~ \infty \tag{6}$$
$u$-pointwise.

*We note that the real part is given by
$$ {\rm Re}~ f_{\lambda}(u)~=~\lambda\sqrt{\frac{\sqrt{1+\frac{u^4}{\lambda^2}}-1}{2}} .\tag{7}$$
Define a non-negative function 
$$ g(u)~:=~ \frac{|u|}{2}~\theta(|u|\!-\!\sqrt{2}) , \tag{8}$$
where $\theta$ denotes the Heaviside step function.
One may show that
$$\forall\lambda\geq\frac{1}{\sqrt{2}}\forall u\in\mathbb{R}:~~ 
{\rm Re}~ f_{\lambda}(u) ~\geq~ g(u),\tag{9}$$
because
$$\forall\lambda\geq\frac{1}{\sqrt{2}}\forall u\geq\sqrt{2}:~~ 
\sqrt{1+\frac{u^4}{\lambda^2}}-1  ~\geq~ \frac{u^2}{2\lambda^2}.\tag{10}$$

*It follows that $\exp\left\{-g(u)\right\}$ is a majorant function for the integrand (3). Lebesgue's dominated convergence theorem then shows that
the corresponding integral (3) satisfies
$$ K(\lambda)~\longrightarrow~ \int_{\mathbb{R}}\!\mathrm{d}u~ 
\exp\left\{-f(u)\right\}~=~\sqrt{2\pi} \quad \text{for} \quad\lambda~\to~ \infty. \tag{11} $$

*Next we can estimate corrections to any order in $1/\lambda$ we like. E.g.
$$\begin{align}K(\lambda)~\stackrel{(3)+(4)}{=}&\int_{\mathbb{R}}\!\mathrm{d}u~
\exp\left\{-\left(\frac{i}{\lambda}+\frac{1}{2}\right)u^2\right\}
\left(1 +\frac{iu^4}{8\lambda} + O(\lambda^{-2})\right)\cr
~=~&\sqrt{\frac{\pi}{\frac{i}{\lambda}+\frac{1}{2}}}+\frac{i}{8\lambda}\underbrace{\int_{\mathbb{R}}\!\mathrm{d}u~\exp\left\{-\frac{u^2}{2}\right\}u^4}_{=3\sqrt{2\pi}} + O(\lambda^{-2})\cr
~=~&\sqrt{2\pi}\left(1 -\frac{5i}{8\lambda} + O(\lambda^{-2})\right) ,\end{align}\tag{12}$$ 
and so forth. Hence OP's integral (1) has an asymptotic expansion

$$ I(\lambda)~\sim~ e^{i\lambda}\frac{1+i}{2}\sqrt{\frac{\pi}{\lambda}}\left(1 -\frac{5i}{8\lambda} + O(\lambda^{-2})\right)\quad \text{for}\quad \lambda~\to~\infty .\tag{13} $$

A: Not a full answer but too long for a comment.
It appears that the saddle point at $x=0$ is dominant. There are two more but they can be avoided (and they possibly lie on the branch cuts of the integrand anyway---that would be bad).
If we make the substitution $x = u/\sqrt{\lambda}$ we get
$$
\begin{align}
\int_0^\infty \exp\left\{-x^2+i\lambda\sqrt{x^2+1}\right\}dx &= \frac{1}{\sqrt{\lambda}} \int_0^\infty \exp\left\{-\frac{u^2}{\lambda}+i\lambda\sqrt{\frac{u^2}{\lambda}+1}\right\}du \\
&= \frac{e^{i\lambda}}{\sqrt{\lambda}} \int_0^\infty \exp\left\{-\frac{u^2}{\lambda}-i\lambda \left( 1-\sqrt{\frac{u^2}{\lambda}+1} \right) \right\}du.
\end{align}
$$
Numerically,
$$
\int_0^\infty \exp\left\{-\frac{u^2}{\lambda}-i\lambda \left( 1-\sqrt{\frac{u^2}{\lambda}+1} \right) \right\}du \to \int_0^\infty \exp\{iu^2/2\}\,du = \frac{1+i}{2}\sqrt{\pi}
$$
as $\lambda \to \infty$, but I'm not sure how to show it at the moment. This would imply that
$$
\int_0^\infty \exp\left\{-x^2+i\lambda\sqrt{x^2+1}\right\}dx \sim \frac{(1+i)e^{i\lambda}}{2} \sqrt{\frac{\pi}{\lambda}}
$$
as $\lambda \to \infty$.
