<s>[INST] Integrate Airy function from 0 to $\infty$ From the integral representation of Airy function
$$\mathrm{Ai}(x)=\int_{-\infty}^{\infty} \frac{\mathrm{d} \tau}{2\pi} \exp(-\mathrm{i}\tau x)\exp(-\mathrm{i}\frac{\tau^3}{3}),$$
It is easy to see that $\int_{-\infty}^{\infty} \mathrm{d} x\mathrm{Ai}(x) =1$. However, I am wondering how to find 
$$\int_{0}^{\infty} \mathrm{d}x \mathrm{Ai}(x).$$
From this website, the result of the above integral is $\frac{1}{3}.$ I could not follow the method in the reference given by that website. Could anyone give an alternative (more straightforward) derivation of  $\int_{0}^{\infty} \mathrm{d}x \mathrm{Ai}(x)=\frac{1}{3}$?
 [/INST] EDIT:
After posting my answer, I realized that the same approach can be used to find the Mellin transform of $\operatorname{Ai}(x)$. So I decided to modify my answer.

I don't know if this is the approach discussed in the reference, but the Airy function $\operatorname{Ai}(x)$ can be expressed in terms of the  modified Bessel function of the second kind of order $\frac{1}{3}$.
Specifically, $$\operatorname{Ai}(x)= \frac{1}{\pi} \sqrt{\frac{x}{3}} K_{1/3} \left(\frac{2}{3} x^{3/2} \right), \quad x>0 .$$
And an integral representation of the modified Bessel function of the second kind is $$K_{\nu}(x) = \frac{1}{2} \left(\frac{x}{2} \right)^{\nu} \int_{0}^{\infty}\exp\left(-t-\frac{x^{2}}{4t} \right) \, \frac{dt}{t^{\nu+1}}, \quad  x>0, $$
which can be derived from the integral representation$$K_{\nu}(x) =\int_{0}^{\infty} \exp(-x\cosh t) \cosh(\nu t) \, dt = \frac{1}{2} \int_{-\infty}^{\infty} \exp\left(-x \cosh t\right) e^{-\nu t} \, dt $$ by making the substitution $e^{t}= \frac{2}{x}u$.
Using this representation, and assuming that $a>0$, we get
$$ \begin{align}I(a) &= \int_{0}^{\infty} x^{a-1} \operatorname{Ai}(x) \, dx \\ &= \frac{1}{\pi \sqrt{3}}\int_{0}^{\infty} x^{a-1/2} \, K_{1/3}\left(\frac{2}{3}x^{3/2} \right) \, dx \\ &=\frac{1}{\pi} \, \frac{3^{2a/3-7/6}}{ 2^{2a/3-2/3}} \int_{0}^{\infty} u^{2a/3-2/3} K_{1/3}(u) \, du  \\  &= \frac{1}{\pi} \, \frac{3^{2a/3-7/6}}{ 2^{2a/3-2/3}}\int_{0}^{\infty} u^{2a/3-2/3} \, \frac{1}{2} \left(\frac{u}{2} \right)^{1/3} \int_{0}^{\infty} \exp \left(-t - \frac{u^{2}}{4t}\right) \, \frac{dt}{t^{4/3}} \, du \\ &= \frac{1}{\pi} \, \frac{3^{2a/3-7/6}}{ 2^{2a/3+2/3}}\int_{0}^{\infty}\frac{e^{-t}}{t^{4/3}} \int_{0}^{\infty}u^{2a/3-1/3} \exp \left(- \frac{u^{2}}{4t} \right) \, du \, dt  \tag{1}\\ &= \frac{3^{2a/3-7/6}}{2 \pi}\int_{0}^{\infty} t^{a/3-1} e^{-t} \int_{0}^{\infty} w^{a/3-2/3} e^{-w} \, dw  \, dt \\ &= \frac{3^{2a/3-7/6}}{2 \pi}\Gamma\left(\frac{a+1}{3}\right)  \int_{0}^{\infty} t^{a/3-1} e^{-t} \, \, dt \\&= \frac{3^{2a/3-7/6}}{2 \pi} \, \Gamma \left(\frac{a+1}{3} \right) \Gamma \left(\frac{a}{3} \right).  \end{align}$$

$(1)$ Since the integrand is nonnegative, Tonelli's theorem allows us to change the order of integration.

Therefore, $$\begin{align} \int_{0}^{\infty} \operatorname{Ai}(x) \, dx&= I(1) = \frac{1}{2\pi \sqrt{3}} \, \Gamma \left(\frac{2}{3} \right) \Gamma \left(\frac{1}{3} \right)\\ &= \frac{1}{2 \sqrt{3}} \, \pi \csc \left(\frac{\pi }{3} \right) \tag{2} \\ &=\frac{1}{2 \sqrt{3}} \left(\frac{2}{\sqrt{3}} \right) \\ &= \frac{1}{3}. \end{align}$$
$(2)$ Euler's reflection formula
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