The problem I'm working on is this:
Let $f(x) = x^4 −2x^2 +1 = (x^2 -1)^2 $, $x \in \mathbb{R}$. Prove that there exists an $ \alpha \in \mathbb{R}$ (and find its value) such that
$$ \frac{\lambda^{-\alpha}}{C} \leq \left| \int_{- \pi}^\pi e^{i \lambda (x^2 -1)^2} \sin(x)^4 \right| \leq C \lambda^{-\alpha}$$
as $\lambda \to \infty$, for some $C >0$.
I thought the stationary phase approximation would be the thing to do, but I actually don't really know that method at all.
Things I have tried:
- Chaning variables to $x^2 = u$ and then applying some contour integration. Did not seem to provide anything fruitful.
- Same change of variables, integrating by parts. Again, I didn't really find anything that interesting.
- Googling the stationary phase approximation, then being very unsure about how to apply it and what to do with the results.
If someone could work through this one with me, that would be much appreciated.
Here's one solution using the Riemann Lebesgue Lemma. Which states that if we have $f \in L^1([a,b])$ then the follwoing is true: $$\lim_{N \to \infty} \int_{[a,b]} f(x)e^{iNx}dx =0$$ So in your context use a change of variable maybe $u=(x^2-1)^2$ and transform your integral into a form where you can apply Riemann Lebesgue.
As always, I'm sure there are other ways. This was just the first I thought of. Hope this helps.