Suppose $f$ is a smooth compactly supported function supported in $[-\pi, \pi]$. The problem I am working on is to show that $$\int_{\mathbb{R}}e^{ik(\sin x - x)}f(x)\, dx = \frac{f(0)}{k^{1/3}}\int_{\mathbb{R}}e^{-ix^{3}/6}\, dx + O(k^{-2/3})$$ as $k \rightarrow \infty$.
I noticed that $$\frac{f(0)}{k^{1/3}}\int_{\mathbb{R}}e^{-ix^{3}/6}\, dx = f(0)\int_{\mathbb{R}}e^{-ikx^{3}/3!}\, dx$$ and the $-x^{3}/3!$ term is precisely the next term in the power series expansion of $\sin x$ after I subtract off and $x$. Thus I need to show that $$\int_{\mathbb{R}}e^{-ikx^{3}/3!}(e^{ik(\sin x - x + x^{3}/3!)}f(x) - f(0))\, dx = O(k^{-2/3})$$ as $k \rightarrow \infty$. The $k^{-2/3}$ comes I think from the $x^{5}/5!$ term in the power series expansion of $\sin x$ and a change of variables. Does anyone know how to prove the above asymptotic?
As $k\to\infty$, $e^{ik\,g(x)}$ oscillates more and more quickly (higher and higher frequencies) so for any particular positive value that contributes to the integral there will be a corresponding negative value of the same magnitude to cancel that contribution, except...
... for values in the region around which the exponential is nearly stationary, i.e., where $g'(x)$ is nearly zero. Therefore, the largest contribution (in magnitude) to the integral comes from that region of stationarity. With $g(x) = \sin x - x$, stationarity happens at $x$ such that $$ (\cos x - 1) = 0 \qquad\implies\quad x=0 $$ So, expanding the integrand in a power series around $x=0$, you find $$ e^{ik\,(\sin x - x)}\,f(x) = \exp\left[-ik\,x^3/6 + O(x^5)\right]\left[f(0) + f'(0)\,x + O(x^2)\right] $$ and I think you can continue from there on your own.