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  $\int_{0}^{\infty} \mathrm{d}x \mathrm{Ai}(x)=\frac{1}{3}$?
 A: $$\begin{align}
\frac1{2\pi}\int_0^L \int_{-\infty}^\infty e^{-itx}e^{-it^3/3}\,dt\,dx&=\frac1{2\pi}\int_{-\infty}^\infty \left(\int_0^L e^{-itx}\,dx \right)\,e^{-it^3/3}\,dt\\\\
&=\frac1{2\pi i}\int_{-\infty}^\infty \left(\frac{1-e^{-iLt}}{t} \right)\,e^{-it^3/3}\,dt\\\\
&=\frac1{\pi }\int_{0}^\infty \frac{\sin(Lt)}{t}\,\cos(t^3/3)\,dt-\frac1{\pi }\int_{0}^\infty \frac{1-\cos(Lt)}{t}\,\sin(t^3/3)\,dt
\end{align}$$
It is straightforward (See the analysis in the OP of this question ) to show that 
$$\lim_{L\to \infty}\int_{0}^\infty \frac{\sin(Lt)}{t}\,\cos(t^3/3)\,dt=\frac{\pi}{2}$$
and that 
$$\lim_{L\to \infty}\int_{0}^\infty \frac{1-\cos(Lt)}{t}\,\sin(t^3/3)\,dt=\frac{\pi}{6}$$
Putting everything together reveals
$$\int_0^\infty \text{Ai}(x)\,dx=\frac13$$
as was to be shown!

Alternatively, we can use distributions and write
$$\begin{align}
\frac1{2\pi}\int_0^\infty \int_{-\infty}^\infty e^{-itx}e^{-it^3/3}\,dt\,dx&=\frac1{2\pi}\int_{-\infty}^\infty \left(\int_0^\infty e^{-itx}\,dx \right)\,e^{-it^3/3}\,dt\\\\
&=\frac1{2\pi }\text{PV}\left(\int_{-\infty}^\infty \left(\pi \delta(t)+\frac{1}{it} \right)\,e^{-it^3/3}\,dt\right)\\\\
&=\frac12-\frac1{2\pi }\int_{-\infty}^\infty \frac{\sin(t^3/3)}{t} \,dt\\\\
&=\frac12-\frac16\\\\
&=\frac13
\end{align}$$
as expected!
A: 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
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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$\ds{\mrm{Ai}\pars{x} =
\int_{-\infty}^{\infty}{\dd\tau \over 2\pi}\,\exp\pars{-\ic\tau x}
\exp\pars{-\ic\,{\tau^{3} \over 3}}\,,
\qquad\qquad\int_{0}^{\infty}\mrm{Ai}\pars{x}\,\dd x:\ {\large ?}}$.

\begin{align}
&\int_{0}^{\infty}\mrm{Ai}\pars{x}\,\dd x =
\int_{-\infty}^{\infty}\mrm{H}\pars{x}\mrm{Ai}\pars{x}\,\dd x\qquad\qquad
\pars{~\substack{\ds{\mrm{H}:\mathbb{R}\setminus\braces{0} \to \mathbb{R}}}
\\[3mm] {Heaviside\ Step\ Function}~}
\\[5mm] = &\
\int_{-\infty}^{\infty}\overbrace{\pars{\int_{-\infty}^{\infty}
{\expo{\ic \tau x} \over \tau - \ic 0^{+}}
\,{\dd \tau \over 2\pi\ic}}}^{\ds{\mrm{H}\pars{x}}}\
\mrm{Ai}\pars{x}\,\dd x\qquad\
\pars{~\substack{\mbox{Note that}
\\[2mm]
\ds{\left.\vphantom{\Large A}\mrm{H}\pars{x}\right\vert_{\ x\ \not=\ 0} =
\lim_{\epsilon \to 0^{+}}\int_{-\infty}^{\infty}
{\expo{\ic \tau x} \over \tau - \ic\epsilon}
\,{\dd \tau \over 2\pi\ic}}}~}
\\[5mm] = &\
\int_{-\infty}^{\infty}{1 \over \tau - \ic 0^{+}}
\bracks{\int_{-\infty}^{\infty}\mrm{Ai}\pars{x}\expo{\ic\tau x}\,\dd x}
\,{\dd\tau \over 2\pi\ic} =
\int_{-\infty}^{\infty}{\exp\pars{-\ic\tau^{3}/3} \over \tau - \ic 0^{+}}
\,{\dd\tau \over 2\pi\ic}
\\[5mm] = &\
\mrm{P.V.}\int_{-\infty}^{\infty}{\exp\pars{-\ic\tau^{3}/3} \over \tau}
\,{\dd\tau \over 2\pi\ic} + {1 \over 2} =
-\,{1 \over \pi}\int_{0}^{\infty}{\sin\pars{\tau^{3}/3} \over \tau}\,\dd\tau + 
{1 \over 2}
\\[5mm] \stackrel{\large\tau^{3}/3\ \mapsto\ \tau}{=}\,\,\,&
-\,{1 \over 3\pi}\int_{0}^{\infty}{\sin\pars{\tau} \over \tau}\,\dd\tau +
{1 \over 2} =
-\,{1 \over 3\pi}\,{\pi \over 2} + {1 \over 2} = \bbx{1 \over 3}
\end{align}
A: Given the integral representation it follows that $\text{Ai}(x)$ fulfills the differential equation $y''=x y$.
In particular $\text{Ai}(x)$ is an entire function and 
$$\text{Ai}(x)=\frac{1}{\pi}\int_{0}^{+\infty}\cos\left(\frac{t^3}{3}+xt\right)\,dt \tag{1}$$
$$\begin{eqnarray*}(\mathcal{L}\text{Ai})(s)&=&\frac{1}{\pi}\int_{0}^{+\infty}\int_{0}^{+\infty}\cos\left(\frac{t^3}{3}+xt\right)e^{-sx}\,dt\,dx\\&=&\frac{1}{\pi}\int_{0}^{+\infty}\frac{s\cos\left(\frac{t^3}{3}\right)-t\sin\left(\frac{t^3}{3}\right)}{s^2+t^2}\,dt \tag{2}\end{eqnarray*}$$
It is not difficult to show that
$$ \lim_{s\to 0^+}\int_{0}^{+\infty}\cos\left(\frac{t^3}{3}\right)\frac{s\,dt}{s^2+t^2} = \frac{\pi}{2}\tag{3}$$
$$ \lim_{s\to 0^+}\int_{0}^{+\infty}\sin\left(\frac{t^3}{3}\right)\frac{t\,dt}{s^2+t^2} = \frac{\pi}{6}\tag{4}$$
and it follows that
$$ \int_{0}^{+\infty}\text{Ai}(x)\,dx = \lim_{s\to 0^+}\left(\mathcal{L}\text{Ai}\right)(s) = \frac{1}{\pi}\left(\frac{\pi}{2}-\frac{\pi}{6}\right)=\color{red}{\frac{1}{3}}\tag{5} $$
as wanted.
