I was stunned to find out we can evaluate the Airy Zeta Function at $s = 2$ exactly: $$ \zeta_{\text{Ai}}(2) \equiv \sum_{k=1}^\infty a_k^{-2}=\frac{3^{5/3}}{4\pi^2}\Gamma^4\left(\frac23\right) $$
How on earth was this evaluated? When I learned about the solution to the Basel problem, I was amazed, but I could see how the properties of the sum could lead it to having a solution. This, on the other hand, is a different story. The zeros of the Airy function can't even be expressed in closed form, how can we possibly evaluate the sum of their reciprocal squares?
Point is, this is an amazing result, and I would love to read a proof of it, but I can't find one. Does anyone have a link to a proof of this, and possibly of other closed forms for $\zeta_{\text{Ai}}(s)$ for integer values of $s$?
If $$ x = \frac{1}{2\pi\operatorname{Ai}(0)\operatorname{Bi}(0)} = \frac{3^{5/6}}{2\pi}\left(\Gamma\left(\tfrac23\right)\right)^2, $$ then $$ \begin{align} \zeta_{\operatorname{Ai}}(2) &= x^2,\\ \zeta_{\operatorname{Ai}}(3) &= \tfrac12\left(2x^3-1\right),\\ \zeta_{\operatorname{Ai}}(4) &= \tfrac13\left(3x^4-x\right),\\ \zeta_{\operatorname{Ai}}(5) &= \tfrac{1}{12}\left(12x^5-5x^2\right),\\ \zeta_{\operatorname{Ai}}(6) &= \tfrac{1}{20}\left(20x^6-10x^3+1\right),\\ \zeta_{\operatorname{Ai}}(7) &= \tfrac{1}{180}\left(180x^7-105x^4+13x\right). \end{align} $$ You could find the evaluations in the paper:
Section: 4. The quantum bouncer.
The Airy Zeta Function Mathworld page is also relevant.