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While I've been thinking about this question, I've found that for all $n \geq 1$ integer values, we have $$ \mathcal{I}_n = \int_{-\infty}^\infty \frac{\operatorname{Ai}^2(x+a_n)}{x^2}dx \stackrel{?}{=} 1, $$ where $\operatorname{Ai}$ is an Airy function and $a_n$ are the zeros of the $\operatorname{Ai}$ function.

How to prove this identity?

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user153012
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Integrating by parts we get $$ \mathcal{I}_n = \int_{-\infty}^\infty \frac{\operatorname{Ai}^2(x+a_n)}{x^2}dx =2\cdot \int_{-\infty}^\infty \frac{\operatorname{Ai}(x+a_n)\operatorname{Ai}^\prime(x+a_n)}{x}dx\qquad (1) $$ [due to the asymptotics $$\operatorname{Ai}(x)\sim \frac{e^{-\frac{2}{3}x^{3/2}}}{2\sqrt{\pi}x^{1/4}},\qquad \operatorname{Ai}(-x)\sim \frac{\sin\left(\frac{2}{3}x^{3/2}+\frac{\pi}{4}\right)}{\sqrt{\pi}x^{1/4}},\qquad x\to+\infty $$ the boundary terms in $(1)$ vanish and the integral in the rhs is convergent.]

Now consider the more general integral viewed as Cauchy principal value: $$ I(a)=\int_{-\infty}^\infty \frac{\operatorname{Ai}(x+a)\operatorname{Ai}^\prime(x+a)}{x}dx. $$ We will need two integral representations(see the book Vallee, Soares "Airy functions and applications to physics"): $$ \operatorname{Ai}^2(x)=\frac{1}{2\pi^{3/2}}\int_0^\infty \cos\left(\frac{t^3}{12}+tx+\frac{\pi}{4}\right)\frac{dt}{\sqrt{t}},\ \ \qquad (2) $$ $$ \operatorname{Ai}(x)\operatorname{Bi}(x)=\frac{1}{2\pi^{3/2}}\int_0^\infty \sin\left(\frac{t^3}{12}+tx+\frac{\pi}{4}\right)\frac{dt}{\sqrt{t}}.\qquad(3) $$ Representing the derivative $\operatorname{Ai}(x+a)\operatorname{Ai}^\prime(x+a)$ using $(2)$ gives $$ I(a)=\frac{-1}{4\pi^{3/2}}\int_{-\infty}^\infty\frac{dx}{x}\int_0^\infty \sin\left(\frac{t^3}{12}+tx+ta+\frac{\pi}{4}\right)\sqrt{t}\ {dt}=\\ \frac{-1}{4\pi^{3/2}}\int_0^\infty \sqrt{t}\ {dt}\int_{-\infty}^\infty\frac{dx}{x}\left[\sin\left(\frac{t^3}{12}+ta+\frac{\pi}{4}\right)\cos tx+\cos\left(\frac{t^3}{12}+ta+\frac{\pi}{4}\right)\sin tx\right]=\\ \frac{-1}{4\pi^{3/2}}\int_0^\infty \sqrt{t}\left[\sin\left(\frac{t^3}{12}+ta+\frac{\pi}{4}\right)\cdot 0+\cos\left(\frac{t^3}{12}+ta+\frac{\pi}{4}\right)\cdot \pi\right]dt=\\ \frac{-1}{4\pi^{1/2}}\frac{d}{da}\int_0^\infty \sin\left(\frac{t^3}{12}+ta+\frac{\pi}{4}\right)\frac{dt}{\sqrt{t}}=\\ -\frac{\pi}{2}\frac{d}{da}\operatorname{Ai}(a)\operatorname{Bi}(a)=-\frac{\pi}{2}\left[\operatorname{Ai}'(a)\operatorname{Bi}(a)+\operatorname{Ai}(a)\operatorname{Bi}'(a)\right]. $$ [in the third line the following Cauchy principal values have been used $\int_{-\infty}^\infty \frac{\cos xt}{x} dx=0$ and $\int_{-\infty}^\infty\frac{\sin xt}{x} dx=\pi\cdot \text{sgn}(t)$].

Finally using the value of the Wronskian $ W \left\{ \operatorname{Ai}(a),\operatorname{Bi}(a) \right\}=\frac{1}{\pi} $ and substituting $a=a_n$ $$ I(a_n)=-\frac{\pi}{2}\left(2\cdot \operatorname{Ai}(a)\operatorname{Bi}'(a)-\frac{1}{\pi}\right)_{a=a_n}=\frac{1}{2}, $$ hence $\mathcal{I}_n=2I(a_n)=1$.

Cave Johnson
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  • Thank you for your answer. +1. It seems to me that if $a_n$ is a root of the $\operatorname{Bi}'$ function, then $\mathcal{I}_n$ also equals to $1/2$. However, I've tested it with Mathematica, and numerically it isn't $1/2$. I've implemented the function $\operatorname{Bi}'$ as offered here. What do you think? Maybe I've misunderstood something. – user153012 Jan 10 '16 at 14:14
  • @user153012 note that $\mathcal{I}_n$ exists only when $a_n$ is zero of $\text{Ai}$, but $I(a)$ can be defined for all real $a$. – Cave Johnson Jan 10 '16 at 14:53
  • So it could be, that $I(a)$ equals to $1/2$, and $\mathcal{I}_n$ does not exist. Is it true, that $\mathcal{I}_n$ exists if and only if $a_n$ is a zero of $\operatorname{Ai}$, and then $\mathcal{I}_n=1/2$? I mean: Do $\mathcal{I}_n$ characterize the zeros of $\operatorname{Ai}$ function? – user153012 Jan 10 '16 at 15:06
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    @user153012 It is true that $\mathcal{I}_n$ exists if and only if $a_n$ iz zero of $\text{Ai}$ (othervise the integral diverges), and then $\mathcal{I}_n=2I(a_n)=1$. It could be that $I(b_n)=\frac{1}{2}$ for some $b_n$ which is not zero of $\text{Ai}$ (for example $b_n$ is zero of $\text{Bi}'$ as you noted), but $\mathcal{I}_n$ is not defined for such $b_n$. – Cave Johnson Jan 10 '16 at 15:46