Integrability of Airy function I work with well known Airy function given for evey $x\in R$ by 
$$Ai(x) := \frac{1}{2\pi}\int_R e^{i(x\xi + \xi^3)} \, d\xi = \lim_{K\to+\infty} \frac{1}{2\pi}\int_{-K}^K e^{i(x\xi + \xi^3)} \, d\xi.$$ 
My question is, how to prove that the intergral $\int_R Ai(x) \, dx$ exists, or in the other words, how to prove the existence of the limit
$$\lim_{K\to+\infty} \frac{1}{2\pi}\int_{-K}^K Ai(x) \, dx$$
?
 A: Asymptotics for the Airy function are known (see the first formula here). These asymptotics immediately prove that the function is integrable. (For that matter, so does the existence of its Fourier transform, further down that page.)
A: The existence $\int_1^\infty A_i (x) dx$ is straightforward via the following asymptotics 
$$\text{Ai}(x) \sim \frac{\exp(-\frac{2}{3} x^{3/2})}{x^{1/4}}, \quad x \to +\infty$$ 
The existence of $\int_{-\infty}^{-1} \text{Ai}(x) dx$ is less obvious, despite we have the asymptotics $x \to -\infty$
\begin{align*}
\text{Ai}(x) &\sim \frac{\cos\Big(\frac{\pi}{4} + \frac{2}{3}x^{3/2}\Big)}{x^{1/4}}\\
\text{Ai}'(x) &\sim -\frac{1}{4}\frac{\cos\Big(\frac{\pi}{4} + \frac{2}{3}x^{3/2}\Big)}{x^{5/4}} - x^{1/4}\sin\Big(\frac{\pi}{4} + \frac{2}{3} x^{3/2}\Big)
\end{align*}
The integral does not converge absolutely. To prove the convergence, one can use the property
$$\text{Ai}''(x) = x \text{Ai}(x).$$
One has via integration by part
\begin{align*}
\int_{-\infty}^{-1} \text{Ai}(x) dx &= \int_{-\infty} ^{-1} \frac{\text{Ai}''(x)}{x} dx\\
&= -\text{Ai}'(-1) + \int_{-\infty}^{-1} \frac{\text{Ai}'(x)}{x^2} dx
\end{align*} 
The second integral now converges absolutely.
