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 $\int_{0}^{\infty} \mathrm{d}x \mathrm{Ai}(x)=\frac{1}{3}$?

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.

• Thank you for your help! Is this method a common technique to evaluate integral from 0 to $\infty$? – ZHW Jun 22 '17 at 18:47
• @ZHW: I'd say so but I am partial to that: every time I see an integral of the form $\int_{0}^{+\infty}(\ldots)\,dx$ I think about simplifying it through the Laplace (inverse) transform. – Jack D'Aurizio Jun 22 '17 at 18:49
• @JackD'Aurizio Hi Jack. I hope that you're doing well. In the answer I posted, I took a few liberties with formal manipulations without justifying them. It seems you did likewise. So, here is a question. (1) How does one justify the interchange of integration you used in your development? Fubini-Tonelli does not apply here. So, what other ways could one proceed? – Mark Viola Jul 31 '18 at 15:54

\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!

• Thank you for the help! For those two integrals in your first answer, I suppose the $\lim_{L\rightarrow \infty}$ operation is taken before the integration, in order to get rid of terms with $t^3/3$. Am I correct about this? Besides, may I know more about distributions in your second answer? I am not so familiar with it, but would like to learn more. – ZHW Jun 22 '17 at 19:31
• You're welcome. The limit is taken after the integration. We cannot bring the limit inside the integral since the limit of the integrand fails to exist. HERE is an article on distributions. – Mark Viola Jun 22 '17 at 19:36

$\newcommand{\bbx}{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}{\displaystyle{#1}} \newcommand{\expo}{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}{\mathcal{#1}} \newcommand{\mrm}{\mathrm{#1}} \newcommand{\pars}{\left(\,{#1}\,\right)} \newcommand{\partiald}[]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}{\left\vert\,{#1}\,\right\vert}$

$\ds{\mrm{Ai}\pars{x} = \int_{-\infty}^{\infty}{\dd\tau \over 2\pi}\,\exp\pars{-\ic\tau x} \exp\pars{-\ic\,{\tau^{3} \over 3}}\,, \qquad\qquad\int_{0}^{\infty}\mrm{Ai}\pars{x}\,\dd x:\ {\large ?}}$.

\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}

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 $\frac{1}{3}$.

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}