Stationary phase method for $\int_{-\infty}^{\infty}f(t)\exp(ix(t^3-t))dt$ I am currenty struggling with the integral $\int_{-\infty}^{\infty}f(t)\exp(ix(t^3-t))dt$ where $f(t)$ is smooth and $f\rightarrow 0$ as $t\rightarrow +-\infty$. I want to obtain the leading asymptotic beahviour as $x\rightarrow \infty$
I would not have a problem if the boundaries of the integral are finite, as stated here http://www.math.unl.edu/~scohn1/8423/intasym4.pdf  (Formula (2))
$g(t)=t^3-t$ and $g'(t)=0$ at $+-\sqrt{\frac{1}{3}}$
What can I do to use the formula stated in the link above? 
 A: If we split the interval of integration up into four parts,
$$
\int_{-\infty}^{\infty} = \int_{-\infty}^{-1} + \int_{-1}^0 + \int_0^1 + \int_1^\infty,
$$
the inner two integrals are of the type considered in the PDF you linked, so it just remains to show that the first and last integrals are asymptotically smaller than them as $x \to \infty$ (and hence that they do not contribute to the leading-order asymptotic).
I'll just consider the last integral,
$$
I(x) = \int_1^\infty f(t) \exp\left[ix(t^3-t)\right]\,dt,
$$
since the process for the first, $\int_{-\infty}^{-1}$, should be similar.
The substitution $s = t^3-t$ defines an increasing, concave bijection $t(s) : [0,\infty) \to [1,\infty)$.  For large $s$ we have
$$
t \sim s^{1/3}
$$
and for small $s$ we have
$$
t = 1 + \frac{s}{2} + O(s^2).
$$
We'll then write
$$
f(t)\,dt = f(t(s))t'(s)\,ds,
$$
so that
$$
I(x) = \int_0^\infty f(t(s))t'(s)e^{ixs}\,ds.
$$
Note that, since $t \sim s^{1/3}$ for large $s$, we have
$$
t'(s) \sim \frac{1}{3}s^{-2/3}
$$
for large $s$.  Integrating by parts thus yields
$$
\begin{align}
I(x) &= \frac{1}{ix}\left[f(t(s))t'(s)e^{ixs}\right]_0^\infty - \frac{1}{ix} \int_0^\infty \frac{d}{ds} \Bigl[f(t(s))t'(s)\Bigr]e^{ixs}\,ds \\
&= -\frac{f(1)}{2ix} - \frac{1}{ix} \int_0^\infty \frac{d}{ds} \Bigl[f(t(s))t'(s)\Bigr]e^{ixs}\,ds, \tag{1}
\end{align}
$$
since $t(0) = 1$ and $t'(0) = \frac{1}{2}$.  Now
$$
\frac{d}{ds} \Bigl[f(t(s))t'(s)\Bigr] = f'(t(s))t'(s)^2 + f(t(s))t''(s),
$$
and for large $s$ we have
$$
f'(t(s))t'(s)^2 \sim f'(t(s)) \left( \frac{1}{3} s^{-2/3} \right)^2 \tag{2}
$$
and
$$
f(t(s))t''(s) \sim f(t(s)) \left( -\frac{2}{9} s^{-5/3} \right). \tag{3}
$$
Since $f(t(s)) \to 0$ as $s \to \infty$ the expression in $(3)$ is integrable, and if the expression in $(2)$ is integrable as well (for instance if $f'(r)$ is bounded) then the integral in $(1)$ exists and is bounded.  Thus
$$
I(x) = O\left(\frac{1}{x}\right).
$$
This is smaller than the estimates you would get for the integrals over finite intervals, which would be something on the order of $1/\sqrt{x}$.  So, you can throw the tails of the integral out if all you're interested in is its leading-order approximation.
A: I) OP's integrand reads
$$\begin{align} g(z)~:=~&f(z)e^{ixS(z)}, \qquad S(z)~:=~z^3-z, \cr S^{\prime}(z)~=~&3z^2-1,\qquad S^{\prime\prime}(z)~=~6z,\end{align} \tag{1}$$
with 2 critical points at
$$z_{\pm}~=~ \pm 3^{-\frac{1}{2}}\tag{2}$$
with angular steepest decent direction $\pm\frac{\pi}{4}$, respectively.

$\uparrow$ Fig. 1. The complex $z$-plane with OP's original integration contours $-{\cal C}_1$, and two other contours ${\cal C}_2\equiv {\cal C}_-$ and  ${\cal C}_3\equiv{\cal C}_+$. The shaded regions denote exponentially decaying sectors. (Figure taken from Ref. [W].)
II) OP seems mostly interested in the stationary phase approximation for real smooth functions $f$ on the real line $\mathbb{R}$. In this answer we will just mention that if $f$ is also holomorphic in appropriate regions of the complex plane $\mathbb{C}$, then one may apply the method of steepest descent. OP's integral
$$ I ~=~ \int_{\mathbb{R}+i0^+} \mathrm{d}z~g(z)~=~I_- + I_+ \tag{3} $$
can in the latter case be deformed to a sum of 2 steepest descent contours
$$\begin{align} I_{\pm}~:=~& \int_{{\cal C}_{\pm}} \mathrm{d}z~g(z) \cr 
    ~\sim~& \sqrt{\frac{2\pi}{-ix S^{\prime\prime}(z_{\pm})}}g(z_{\pm}) \cr
~=~&e^{\pm\frac{i\pi}{4}} \sqrt{\frac{\pi}{x \sqrt{3}}} f(z_{\pm}) e^{ix(3^{-\frac{3}{2}}-3^{-\frac{1}{2}})} \qquad\text{for}\qquad x~\to~\infty \end{align} \tag{4}$$
through the 2 critical points $z_{\pm}= \pm 3^{-\frac{1}{2}}$, respectively.
Note that the final formula (4) for $I=I_- + I_+$ only refers to objects on the real line. Therefore we expect that a real analysis would lead to the same result for an appropriate class of real smooth functions $f$.
References:

*

*[W] E. Witten, Analytic Continuation Of Chern-Simons Theory, arXiv:1001.2933; p. 23-29, 48-49. A related 2015 KITP lecture by Witten, A New Look At The Path Integral Of Quantum Mechanics, can be found on YouTube.

