$\lim_{n\to\infty} \int_0^1e^{2\pi inp(x)}=0$

Question

Let $p$ non constant real polynomial, I want to prove that $\lim_n \int_0^1e^{2\pi inp(x)}=0$

I am not sure even how to begin with this, whether to use measure theory, fourier, complex analysis. It reminds me somewhat of Riemann-Lebesgue lemma from Fourier, but I am not really sure how to proceed with that.

My intuition would be to try and use measure theory/convergence theorems to prove $\lim_n\int_0^1cos(2\pi n x^m)=0$ for every m and similarly for sine and then somehow generalize for general polynomials. Take the simple case for $m=2$ for example. Taking $u=2\pi nx^2$ we get $\int_0^1cos(2\pi n x^m)=\int_0^{2 \pi n} \frac{\sqrt{2\pi n}}{4\pi n}\frac{1}{\sqrt{u}}cos(u)=\int_{\mathbb R} \frac{\sqrt{2\pi n}}{4\pi n}\frac{1}{\sqrt{u}}cos(u)\chi_{[0,2\pi n]}$ but this doesn't seem to help much as I can't use any of the convergence theorems.

Any hints as to at least what would be a good strategy to approach this problem would be appreciated.

Answer 1

Split $[0,1]=\bigcup_j [a_j,b_j]$ where $p$ is injective on $(a_j,b_j)$ (ie. $p'$ doesn't vanish) then $$\int_{a_j}^{b_j} e^{2i\pi np(x)}dx=\int_{p(a_j)}^{p(b_j)} \frac{e^{2i\pi n y}}{p'(p^{-1}(y))}dy$$

Since $$\int_{p(a_j)}^{p(b_j)} |\frac{1}{p'(p^{-1}(y))}|dy= |b_j-a_j|$$ then$\frac1{p'(p^{-1}(y))}$ is $L^1[p(a_j),p(b_j)]$ and we can apply Riemann-Lebesgue lemma.

Answer 2

By the stationary phase method, the integral is approximately given by $$ \int_0^1 e^{inp(x)}dx= \sum_{x_0} \sqrt{\frac{2\pi}{np''(x_0)}}e^{inp(x_0)+\mathrm{sign}(p''(x_0))i\pi/4}+o(1/\sqrt{n}),$$ where the sum is over the stationary points of $p(x)$ in the integration interval (roots of $p'(x)$).

So it decays like $\sim 1/\sqrt{n}$ for large $n$, provided there are no double roots of $p'(x)$. If there are double roots, it will decay slower, but will still decay.