On Integrals of the Airy function

Question

Let $Ai$ be the classical Airy function and let $(a_j)_{j\ge 1}$ be the strictly decreasing sequence of its zeroes: we have $a_{j+1}<a_j<\dots <a_2<a_1<0$, $\lim_{j\rightarrow +\infty}a_j=-\infty$. I believe that for all $k\ge 0,$ $$ \int_{-\infty}^{a_{2k+1}} Ai(t) dt<0<\int_{-\infty}^{a_{2k}} Ai(t) dt,\tag{$\ast$} $$ say with the convention $a_0=0$ (we have then $\int_{-\infty}^{a_0} Ai(t) dt=2/3$). However, it does not appear very simple to prove and not so easy to find a reference in the literature. A Mathematica drawing of the curve of the antiderivative of the Airy function vanishing at $-\infty$ leaves no doubt that $(\ast)$ holds true.

Answer 1

There is nothing special about the Airy function here. The situation becomes immediately clear if you recall the following standard lemma (which can be found in some form in most textbooks dealing with second order linear ODE):

If $u(0)=v(0)=0, 0<u'(0)\le v'(0)$ and $u''=-f(x)u, v''=-g(x)v$ where $f>g$ on $(0,+\infty)$, then the graph of $v$ lies above the first hump of the graph of $u$ (i.e., $v>u$ between $0$ and the first positive zero of $u$).

The standard proof is by noticing that $u>v$ slightly to the right of $0$ (Taylor with the usual trick of considering $(1+\varepsilon)v$ instead of $v$ to break the tie at $0$) and considering the Wronskian $W(u,v)=\det\begin{bmatrix}u&v\\u'&v'\end{bmatrix}$ of $u$, $v$. We have $W'=(f-g)uv>0$ and, thereby, $W>W(0)=0$ as long as both $u,v>0$ and if $v$ tries to go below $u$ as long as $u>0$, it has to hit from above, so at the hitting point $v=u>0,v'\le u'$, i.e., $W(u,v)\le 0$, which is impossible.

Now just look at two consecutive humps of the Airy function and think of them as shot from the zero separating them. The lemma shows that a couple of reflections put one hump below the other one. The rest should be clear.