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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?2012-08-09

1 Answers 1

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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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    I haven't tried, it seems complicated...2012-08-09