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

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

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. “Mellin Transform of Quartic Products of Shifted Airy Functions.” Integral Transforms and Special Functions 27 (6): 454–67. (arXiv 🔓)

Laurenzi, Bernard J. 1993. “Moment Integrals of Powers of Airy Functions.” Zeitschrift für Angewandte Mathematik und Physik 44 (5): 891–908. (ResearchGate 🔓)

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

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. “Mellin Transform of Quartic Products of Shifted Airy Functions.” Integral Transforms and Special Functions 27 (6): 454–67. (arXiv 🔓)

Laurenzi, Bernard J. 1993. “Moment Integrals of Powers of Airy Functions.” Zeitschrift für Angewandte Mathematik und Physik 44 (5): 891–908. (ResearchGate 🔓)

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

The equality

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

is established by Laurenzi (1993, Appendix A). 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.

Laurenzi, Bernard J. 1993. “Moment Integrals of Powers of Airy Functions.” Zeitschrift für Angewandte Mathematik und Physik 44 (5): 891–908. (ResearchGate 🔓)

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

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

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. “Mellin Transform of Quartic Products of Shifted Airy Functions.” Integral Transforms and Special Functions 27 (6): 454–67. (arXiv 🔓)

Laurenzi, Bernard J. 1993. “Moment Integrals of Powers of Airy Functions.” Zeitschrift für Angewandte Mathematik und Physik 44 (5): 891–908. (ResearchGate 🔓)