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
$$\int_0^x \operatorname{Bi}(u)^n du$$
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
$$\int_0^x \operatorname{Bi}(u)^n du$$
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:
$$\int_0^z\operatorname{Bi}(t)\mathrm dt=\pi(\operatorname{Bi}(z)\operatorname{Hi}^\prime(t)-\operatorname{Bi}^\prime(z)\operatorname{Hi}(z))$$