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I met the following integral when I was reading a paper: $$\int_0^\infty Ai(y)dy=\frac{1}{3},$$ where $$Ai(y)=\frac{1}{\pi}\int_0^\infty \cos(\alpha y+\frac{\alpha^3}{3})d\alpha.$$

The paper adopted one asymptotic result of the Airy function, $$\int_0^x Ai(y)dy \sim \frac{1}{3}-\frac{1}{2}\pi^{-\frac{1}{2}}x^{-\frac{3}{4}}\exp{\left(-\frac{2}{3}x^\frac{3}{2}\right)}, \quad \text{for large x}.$$ You may find this result on Eq.(10.4.82), Handbook of Mathematical Functions, Abramowitz and Stegun. A similar result is given in Eq.(10.4.83), $$\int_0^x Ai(-y)dy \sim \frac{2}{3}-\frac{1}{2}\pi^{-\frac{1}{2}}x^{-\frac{3}{4}}\cos{\left(\frac{2}{3}x^\frac{3}{2}+\frac{\pi}{4}\right)}, \quad \text{for large x}.$$

However, I was wondering whether we can see the first result by calculating the integrals directly, and $$\int_0^\infty Ai(-y)dy=\frac{2}{3}$$ as well.

Any advice or references are appreciated.

gouwangzhangdong
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1 Answers1

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You can obtain it using the integral representation $$ \operatorname{Ai}(y) = \frac{{\sqrt 3 }}{{2\pi }}\int_0^{ + \infty } {\exp \left( { - \frac{{t^3 }}{3} - \frac{{y^3 }}{{3t^3 }}} \right)\,\mathrm{d}t} ,\quad |\arg y|<\tfrac{\pi}{6} $$ (cf. $(9.5.6)$). Indeed, by the Fubini theorem and the Gauss multiplication theorem for the gamma function, it is found that your integral is \begin{align*} & \frac{{\sqrt 3 }}{{2\pi }}\int_0^{ + \infty } {\exp \left( { - \frac{{t^3 }}{3}} \right)\int_0^{ + \infty } {\exp \left( { - \frac{{y^3 }}{{3t^3 }}} \right)\,\mathrm{d}y} \,\mathrm{d}t} \\ &\mathop = \limits^{x = y^3 /(3t^3 )} \frac{1}{{2\pi }}\frac{1}{{3^{1/6} }} \int_0^{ + \infty } {\exp \left( { - \frac{{t^3 }}{3}} \right)t \int_0^{ + \infty } \mathrm{e}^{ - x} x^{1/3 - 1} \,\mathrm{d}x \,\mathrm{d}t} \\ & = \frac{1}{{2\pi }}\frac{1}{{3^{1/6} }}\Gamma\! \left( {\frac{1}{3}} \right)\int_0^{ + \infty } {\exp \left( { - \frac{{t^3 }}{3}} \right)t\,\mathrm{d}t} \\ &\mathop = \limits^{s = t^3 /3} \frac{1}{{2\pi }}\frac{1}{{\sqrt 3 }}\Gamma\! \left( {\frac{1}{3}} \right)\int_0^{ + \infty } {\mathrm{e}^{ - s} s^{2/3 - 1} \,\mathrm{d}s} \\ & = \frac{1}{{2\pi }}\frac{1}{{\sqrt 3 }}\Gamma\! \left( {\frac{1}{3}} \right)\Gamma\! \left( {\frac{2}{3}} \right) = \frac{1}{3}. \end{align*} For a more general result, see $(9.10.17)$. It can be derived the same way.

Gary
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    How can I see the connection between your formula of $Ai(y)$ with mine? Meanwhile, how can I get $\int_0^\infty Ai(-y)dy=\frac{2}{3}$. – gouwangzhangdong Feb 03 '21 at 06:00
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    @gouwangzhangdong View $Ai(y)$ as the Fourier transform of $e^{i \alpha^3/3}$ to find $\int_{-\infty}^\infty Ai(y)e^{iyx}dx=e^{-ix^3/3}$. Then just plug in $x=0$ and use Gary's result. – Po1ynomial Apr 01 '23 at 15:08
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    I was trying to figure out how to derive that integral representation when I realized that it comes from the integral representation of the modified Bessel function of the second kind that I used here to evaluate the Mellin transform of $\operatorname{Ai}(x)$. We just need to replace $K_{1/3}$ with $K_{-1/3}$. – Random Variable Apr 03 '23 at 19:45