Stationary phase method for $\int_{-\infty}^{\infty}f(t)\exp(ix(t^3-t))dt$

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.