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How do I evaluate this interesting integral with the Airy function:

$$\int_0^x \operatorname{Bi}(u)^2 du$$

More generally, how do I evaluate

$$\int_0^x \operatorname{Bi}(u)^n du$$

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You mean $\mathrm{Bi}(x) = \frac{1}{\pi} \int\limits_0^\infty \left(\exp\left(-\frac{t^3}{3} + xt\right) + \sin\left(\frac{t^3}{3} + xt\right)\right)\, {\rm d}t \;$? So what you tried so far? And why do you think it's interesting? – draks ... Aug 9 '12 at 8:28
up vote 4 down vote accepted

The first one is easy. We know that $\operatorname{Bi}^{\prime\prime}(z)=z\operatorname{Bi}(z)$ from the Airy differential equation, so

$$\begin{align*} \int\operatorname{Bi}(z)^2\mathrm dz&=z\operatorname{Bi}(z)^2-2\int z\operatorname{Bi}(z)\operatorname{Bi}^\prime(z)\mathrm dz\\ &=z\operatorname{Bi}(z)^2-2\int\operatorname{Bi}^\prime(z)\operatorname{Bi}^{\prime\prime}(z)\mathrm dz\\ &=z\operatorname{Bi}(z)^2-\operatorname{Bi}^\prime(z)^2 \end{align*}$$

and then use the initial conditions for the Airy differential equation to yield

$$\int_0^z\operatorname{Bi}(t)^2\mathrm dt=z\operatorname{Bi}(z)^2-\operatorname{Bi}^\prime(z)^2+\frac{\sqrt[3]{3}}{\Gamma\left(\frac13\right)^2}$$

For $n=1$ in your more general integral, the Scorer functions and their derivatives are involved:

$$\int_0^z\operatorname{Bi}(t)\mathrm dt=\pi(\operatorname{Bi}(z)\operatorname{Hi}^\prime(t)-\operatorname{Bi}^\prime(z)\operatorname{Hi}(z))$$

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what do you think about higher order, is it possible? – Minh Nguyen Aug 9 '12 at 9:16
I haven't tried, it seems complicated... – J. M. Aug 9 '12 at 9:28

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