Convergence of the Airy function from Integral by parts

Question

In the book Complex Analysis by Stein, it defines the Airy function as $$ \frac{1}{2\pi} \int_{-\infty}^\infty e^{i(x^3/3+sx)}\ dx \quad \text{for } s \in \mathbb{R} $$ It first says that

because of the rapid oscillations of the integrand as $|x| \to \infty$, the integral converges and represents a continuous function. of $s$.

Then, it perform the integral by part as we have $$ \frac{1}{i(x^2+s)}\frac{d}{dx} \left(e^{i(x^3/3+sx)}\right) = e^{i(x^3/3+sx)} $$ It says

If $a \geq 2|s|^{1/2}$, we can write the integral $$ \int_{a}^R e^{i(x^3/3+sx)}\ dx $$ as $$ \int_{a}^R e^{i(x^3/3+sx)}\ dx = \int_{a}^R \frac{1}{i(x^2+s)}\frac{d}{dx} \left(e^{i(x^3/3+sx)}\right)\ dx $$

I have two questions:

1. Can any help me elaborate the first quote? Why the rapid oscillations as $|x| \to \infty$ ensures the convergence of the integral?
2. In the second quote, why does the restriction $a \geq 2|s|^{1/2}$ matter?

Update: I did the integration by parts, and the second argument may have something to do with the result. Check below. \begin{align*} \int_{a}^R e^{i(x^3/3+sx)}\ dx &= \int_{a}^R \frac{1}{i(x^2+s)}\frac{d}{dx} \left(e^{i(x^3/3+sx)}\right)\ dx \\ &= \frac{1}{i(x^2+s)}e^{i(x^3/3+sx)}\Big|_a^R - \int_{a}^R e^{i(x^3/3+sx)}\left(\frac{d}{dx} \frac{1}{i(x^2+s)}\right)\ dx \\ &= \frac{1}{i(a^2+s)}e^{i(a^3/3+sa)}+\int_{a}^R ie^{i(x^3/3+sx)}\cdot \frac{-2x}{(s+x^2)^2}\ dx \end{align*} If $a \geq 2|s|^{1/2}$, then $a^2 \geq 4|s|$. But it actually cannot ensure $a^2+s$ to be non-zero.

Answer 1

With regard to the first question:

It's not so much the rapid oscillations themselves, but rather that the area associated with each oscillation decreases as $\lvert x\rvert \rightarrow \infty $.

Considering the real part of the integral:

$$ \Re(I) = \int_{-\infty}^{\infty} \cos{(x^3/3+sx)}dx = 2\int_{0}^{\infty} \cos{(x^3/3+sx)}dx$$

This integral can be realized as a summation of a sequence $(a_n)$, where $a_n$ is the area of each half-oscillation:

$$ \Re(I)=2\sum_{n=0}^{\infty} a_n$$

For this summation:

1. $a_n$ will alternate between positive and negative values
2. You can show that $a_n \rightarrow 0$
3. $\lvert a_n \rvert$ is decreasing

And hence by the alternating series test, the summation converges