<s>[INST] Integral involving Airy function I met the following integral when I was reading a paper:
$$\int_0^\infty Ai(y)dy=\frac{1}{3},$$
where
$$Ai(y)=\frac{1}{\pi}\int_0^\infty \cos(\alpha y+\frac{\alpha^3}{3})d\alpha.$$
The paper adopted one asymptotic result of the Airy function,
$$\int_0^x Ai(y)dy \sim \frac{1}{3}-\frac{1}{2}\pi^{-\frac{1}{2}}x^{-\frac{3}{4}}\exp{\left(-\frac{2}{3}x^\frac{3}{2}\right)}, \quad \text{for large x}.$$
You may find this result on Eq.(10.4.82), Handbook of Mathematical Functions, Abramowitz and Stegun. A similar result is given in Eq.(10.4.83),
$$\int_0^x Ai(-y)dy \sim \frac{2}{3}-\frac{1}{2}\pi^{-\frac{1}{2}}x^{-\frac{3}{4}}\cos{\left(\frac{2}{3}x^\frac{3}{2}+\frac{\pi}{4}\right)}, \quad \text{for large x}.$$
However, I was wondering whether we can see the first result by calculating the integrals directly, and
$$\int_0^\infty Ai(-y)dy=\frac{2}{3}$$
as well.
Any advice or references are appreciated.
 [/INST] You can obtain it using the integral representation
$$
\operatorname{Ai}(y) = \frac{{\sqrt 3 }}{{2\pi }}\int_0^{ + \infty } {\exp \left( { - \frac{{t^3 }}{3} - \frac{{y^3 }}{{3t^3 }}} \right)dt} ,\quad |\arg y|<\tfrac{\pi}{6}
$$
(cf. http://dlmf.nist.gov/9.5.E6). Indeed, by the Fubini theorem and the Gauss multiplication theorem for the gamma function, it is found that your integral is
\begin{align*}
& \frac{{\sqrt 3 }}{{2\pi }}\int_0^{ + \infty } {\exp \left( { - \frac{{t^3 }}{3}} \right)\int_0^{ + \infty } {\exp \left( { - \frac{{y^3 }}{{3t^3 }}} \right)dy} dt} \\ &\mathop  = \limits^{x = y^3 /(3t^3 )} \frac{1}{{2\pi }}\frac{1}{{3^{1/6} }} \int_0^{ + \infty } {\exp \left( { - \frac{{t^3 }}{3}} \right)t \int_0^{ + \infty } e^{ - x} x^{1/3 - 1} dx dt} \\ & = \frac{1}{{2\pi }}\frac{1}{{3^{1/6} }}\Gamma \left( {\frac{1}{3}} \right)\int_0^{ + \infty } {\exp \left( { - \frac{{t^3 }}{3}} \right)tdt} 
\\ &\mathop  = \limits^{s = t^3 /3} \frac{1}{{2\pi }}\frac{1}{{\sqrt 3 }}\Gamma \left( {\frac{1}{3}} \right)\int_0^{ + \infty } {e^{ - s} s^{2/3 - 1} ds} \\ & = \frac{1}{{2\pi }}\frac{1}{{\sqrt 3 }}\Gamma \left( {\frac{1}{3}} \right)\Gamma \left( {\frac{2}{3}} \right) = \frac{1}{3}.
\end{align*}
For a more general result, see http://dlmf.nist.gov/9.10.E17. It can be derived the same way.
</s>