\[\int_{-\infty}^{z}\operatorname{Bi}\left(t\right)\,\mathrm{d}t=\int_{0}^{z}%
\operatorname{Bi}\left(t\right)\,\mathrm{d}t=\pi\left(\operatorname{Bi}'\left(%
z\right)\operatorname{Gi}\left(z\right)-\operatorname{Bi}\left(z\right)%
\operatorname{Gi}'\left(z\right)\right)\\
=\pi\left(\operatorname{Bi}\left(z\right)\operatorname{Hi}'\left(z\right)-%
\operatorname{Bi}'\left(z\right)\operatorname{Hi}\left(z\right)\right).\]