Deriving asymptotic expansion of $\int_{-1}^{1} e^{i\lambda(\frac{x^3}{3}+x)} dx$

Question

I'm attempting to use integration by parts to find the expansion for the above integral. I'm able to derive the correct sequence, however I'm not sure how to demonstrate that the error bound is $O(\lambda ^{-n})$. My working is as follows:

\begin{align*}\int \frac{1}{(x^2+1) \cdot i\lambda} \cdot \frac{d}{dx}e^{i\lambda(\frac{x^3}{3}+x)} dx &= \frac{e^{i\lambda(\frac{x^3}{3}+x)}}{(x^2+1) \cdot i\lambda}\biggr{|}_{-1}^{1} +\frac{1}{i\lambda} \int \frac{2x}{(x^2+1)^2} \cdot e^{i\lambda(\frac{x^3}{3}+x)} dx\\ &=\frac{1}{\lambda}\sin\left(\frac{4\lambda}{3}\right)+R(\lambda)\end{align*}

I'd like to show that $R(\lambda)$ is of order $O(\lambda^{-2})$. I thought to just take the supremum of both functions over the range, but noting that $|\frac{2x}{(x^2+1)^2}|<1$ on the domain, this gives the less useful $$|R(\lambda)|\leq \frac{2}{\lambda} $$

Answer 1

A trick is to integrate by parts once more $$R(\lambda)=\frac{1}{i \lambda} \int_{-1}^1 \frac{2x}{(x^2+1)^2} e^{i \lambda (x^3/3+x)}dx=\frac{1}{i \lambda} \int_{-1}^1 \frac{1}{i \lambda}\frac{2x}{(x^2+1)^3} \frac{d}{dx}e^{i \lambda (x^3/3+x)}dx$$ so $$R(\lambda)= -\frac{1}{\lambda^2} \left.\frac{2x}{(x^2+1)^3} e^{i \lambda (x^3/3+x)} \right|_{-1}^1+\frac{1}{\lambda}^2 \int_{-1}^1 \frac{d}{dx} \frac{2x}{(x^2+1)^3} e^{i \lambda (x^3/3+x)}dx=O(\lambda^{-2}).$$

Answer 2

Hint: This integral is equal to $$\int_{-1}^{1} \cos{(\frac{kx^3}{3}+kx)} dx$$

Answer 3

rewrite your integral to $\int_{-1}^{1}[cos\lambda\int{(x^2+1)dx+isin\lambda\int{(x^2+1)dx}]dx}\sim{sin\lambda\int(x^2+1)dx\vert_{-1}^{1}}-isin\lambda{\int(x^2+1)dx\vert_{-1}^{1}}$

therefore, the error's order is $2\lambda$, then use the condition of asymptotic exapnasion to assume $O(f(x)/g(x))=1/x^2\sim1$. now you can multiple $O(2\lambda)$ with $O(\lambda^{-2})$ to get $|R(\lambda)|\leq \frac{2}{\lambda} $. thank you!