Prove that $\int_{-\infty}^\infty \frac{\operatorname{Ai}^2(x+a_n)}{x^2}dx = 1$ 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?

 A: 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 $\mathcal{I}_n=2I(a_n)=1$.
