# Reference to proof of closed form for of $\zeta_{\text{Ai}}(2)$

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 $s$?

• You don't need to be able to evaluate the zeroes of a function in closed form to evaluate the sum of their reciprocal squares. You can get at the latter from residue computations and stuff like that. (I'm also not sure I'd call an expression involving the Gamma function a closed form.) – Qiaochu Yuan Oct 25 '15 at 18:48

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:

Crandall, R. E.: On the Quantum Zeta Function. J. Phys. A: Math. General 29, pp. 6795–6816, 1996.

Section: 4. The quantum bouncer.

The Airy Zeta Function Mathworld page is also relevant.

• that is an amazing paper thx for sharing! – tired Oct 25 '15 at 19:30
• @tired you are welcome. Actually, I think the concept of the Airy zeta function is introduced in this paper. – user153012 Oct 25 '15 at 19:54
• Is it known whether $\zeta_{\text{Ai}}(n)$ is a polynomial in $x$ for each integer $n \geq 2$? It's mentioned as a conjecture in the paper, and it seems like it'd be amenable to the same sort of contour integral calculation that results in the values you cited. – anomaly Oct 25 '15 at 22:16
• @anomaly I've also thought about it. In this book at pp. 61–62. exercise 50. we found that we can express it as a polynomial in $x$. Moreover take a look at the sequence A096632 in OEIS. – user153012 Oct 25 '15 at 22:32
• Awesome, thanks. – anomaly Oct 26 '15 at 1:19