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How can I prove that $$\Omega = \int_{0}^{\infty} \text{Ai}^4(x) \, dx = \frac{\ln(3)}{24 \pi^2}$$ where $\text{Ai}(x)$ is the Airy-function.

Using the Fourier integral representation of the Airy function, we have

$$\Omega = \frac{1}{(2 \pi)^4} \int_{\mathbb{R}^4} \exp \left ( i \sum_{j=1}^{4} \frac{t_j^3}{3} \right ) \left ( \int_{0}^{\infty} \exp \left (ix \sum_{j=1}^{4} t_j \right ) \, dx\right ) \, dt_1 \cdots dt_4$$

From here we may use the fact that that $$\int_{0}^{\infty} e^{isx} \, dx = \pi \delta(s) + i \mathcal{P}(1/s)$$ so that our integral, upon using this identity in the inner $x$-integral, becomes

\begin{align} \Omega &= \frac{1}{(2 \pi)^4} \int_{\mathbb{R}^4} \exp \left( i \sum_{j=1}^{4} \frac{t_j^3}{3}\right) \left [ \pi \delta \left (\sum t_j\right) + i \mathcal{P} \left( 1/ \sum t_j \right) \right ] \, dt_1 \cdots dt_4 \\ &= \frac{1}{16 \pi^3} \int \int \int_{\mathbb{R}^3} \exp \left (i \frac13 (k_1^3 + k_2^3 + k_3^3 + k_4^3) \right ) \, dk_1 dk_2 dk_3 \\ &= \frac{1}{16 \pi^3} \int \int \int_{\mathbb{R}^3} \exp \left (- i (k_1+k_2)(k_2+k_3)(k_3+k_1) \right ) \, dk_1 dk_2 dk_3 \\ \end{align}

The last integral does not converge absolutely, so a contour deformation may be needed (perhaps a rotation $e^{i \theta}$ of the given contour?...) Then, for some $\epsilon > 0$, consider the deformation $k_j \mapsto (1- i\epsilon) k_j$ (analytic continuation of the Airy function). Then maybe, it will help, to define

$$ \Omega_\epsilon = \frac{(1-i\epsilon)^3}{16 \pi^3} \int \int \int_{\mathbb{R}^3} \exp \left (- i (1-i\epsilon)^3(k_1+k_2)(k_2+k_3)(k_3+k_1) \right ) \, dk_1 dk_2 dk_3$$

If I have not made a mistake, I would appreciate a solution that continues this method. After a substitution, the integral seems rather general - I am surprised that I have not been able to find anything on these regularized (gaussian?) exponential integrals.

I strongly believe that $\Omega = \ln(3)/24 \pi^2$, I have verified it numerically. Though I am having trouble numerically verify the last integral; WolframAlpha refuses to try, and mathematica (basic) will abort. The integral is highly oscillatory. $\ln(3)/24 \pi^2$ is such a delightful form, with recognizable constants and the intruiging $24$. Such a simple closed form decieves me into thinking there will be a simple answer (I certainly hope so...).

Thanks.

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    $\begingroup$ If you got $0.0046...$ numerically hoy did you guess it should be $\ln(3)/24 \pi^2$? And what insight turn you to look into this integral? Since in the first link given in the answer below the exponent of 4 is in fact the highest for which the integral of the Airy is listed and apparently known $\endgroup$ Commented Nov 15 at 1:08
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    $\begingroup$ @Dabed, I used mathematica to get $$0.004638029060494605728742986304132271744298186129019121505045993013683287457580177901783781183973026410$$, jammed that into wolfram alpha, and the first possible closed form it gave was $\log(3)/24 \pi^2$... $\endgroup$ Commented Nov 15 at 4:03
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    $\begingroup$ Could you tell me the syntax to use with WA ? By the way, your result is exact for more than 500 decimal places $\endgroup$ Commented Nov 17 at 12:32
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    $\begingroup$ @ClaudeLeibovici At the bottom, wolfram alpha will give "possible closed forms" $\endgroup$ Commented Nov 17 at 13:04

1 Answer 1

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$\newcommand{\d}{\mathrm{d}}$ $\DeclareMathOperator{\Ai}{Ai}$

The equality

$$\int_0^{\infty}\Ai(z)^4\d z = \frac{\ln 3}{24\pi^2}$$

is established by Laurenzi (1993, Appendix A; cf DLMF 9.11.18). To summarize, his strategy is to

  1. prove, using integration by parts and the Airy differential equation, that the integrals $$I_k\stackrel{\triangle}{=}\int_0^{\infty}z^k\Ai(z)^2\d z$$ satisfy the recurrence relation (Eq. 32) $$(2k+1)I_k= \tfrac{1}{2}k(k-1)(k-2)I_{k-3}$$ for all $k\geq 3$, whence $$I_k=\frac{(2/3)^{2/3}\Gamma(k+1)}{4\sqrt{3\pi}(12)^{k/3}\Gamma(k/3 + 7/6)};$$
  2. expand one of the factors of $\Ai(z)^2$ in the original integrand as a sum of hypergeometric $\,_2F_3$ series $$\Ai(z)^2=\frac{1}{12^{5/6}\pi^{3/2}}\sum_{k=0}^{\infty} (12z^3)^k\left(\frac{\Gamma(k+1/6)}{(3k)!}- 2(12)^{1/3}\frac{\Gamma(k+1/2)}{(3k+1)!}z+ 12^{2/3}\frac{\Gamma(k+5/6)}{(3k+2)!}z^2\right)$$ and integrate termwise to get $$\int_0^{\infty}\Ai(z)^4\d z = \frac{1}{72\pi^2}\sum_{k=0}^{\infty}\left(\frac{1}{k+1/6} - \frac{2}{k+1/2} + \frac{1}{k+5/6}\right)\text{;}$$ finally,
  3. express the infinite sum using the digamma function $\psi(z)$ and use that function's multiplication formula to reduce to the result.

More generally, Abramochkin and Razueva (2016) were able to express the Mellin transform in terms of the hypergeometric function

$$\int_0^{\infty}z^{s-1}\Ai(z)^4\d z = \frac{\Gamma(s)}{48^{(s+2)/3}\pi^{3/4}\Gamma(\tfrac{2s+7}{6})}F(\tfrac{s+2}{3},\tfrac{1}{2};\tfrac{2s+7}{6}\mid\tfrac{1}{4})$$

by starting with the integral representation $$\Ai(z)^2 = \pi^{-3/2}\int_0^{\infty}\cos\left(\tfrac{u^6}{12}+u^2 z + \tfrac{\pi}{4}\right)\d u\text{,}$$

using relations like

$$\int_0^{\infty}\left\{\begin{matrix}\cos tz\\ \sin tz\end{matrix}\right\}z^{s-1}\d z = \frac{\Gamma(s)}{t^s}\begin{cases}\cos \tfrac{\pi s}{2} \\ \sin \tfrac{\pi s}{2}\end{cases}$$ to evaluate the integral ($0 < s < 1$, $t>0$), and making further manipulations.

Abramochkin, E.G., and E.V. Razueva. 2016. β€œMellin Transform of Quartic Products of Shifted Airy Functions.” Integral Transforms and Special Functions 27 (6): 454–67. (arXiv πŸ”“)

Laurenzi, Bernard J. 1993. β€œMoment Integrals of Powers of Airy Functions.” Zeitschrift fΓΌr Angewandte Mathematik und Physik 44 (5): 891–908. (ResearchGate πŸ”“)

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  • $\begingroup$ thanks! this answers my question :) I will check out the paper. do you think it would be possible to evaluate the integral using the method in my post (with the oscillatory exp integral)? $\endgroup$ Commented Nov 11 at 19:54
  • $\begingroup$ @MaximeJaccon I have edited my answer. I think that, while something like what you're doing may be feasible, you probably have to regulate the integral to make changing the order of integration valid. $\endgroup$ Commented Nov 11 at 20:51
  • $\begingroup$ thanks for the 2016 paper! tmrw, i will bounty this question to ask about how may one can continue from the oscillatory integral. if there are no other answers, I will accept your answer. $\endgroup$ Commented Nov 12 at 13:04

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