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