\[\int_{z}^{\infty}\operatorname{Ai}\left(t\right)\,\mathrm{d}t=\pi\left(%
\operatorname{Ai}\left(z\right)\operatorname{Gi}'\left(z\right)-\operatorname{%
Ai}'\left(z\right)\operatorname{Gi}\left(z\right)\right),\]