Formula 10.4.44

$\textrm{Hi}(z)=\pi^{-1}\int_0^\infty\exp\left(-\frac{1}{3}t^3+zt\right)dt \\ \qquad~ =\frac{2}{3} \textrm{Bi}~(z)+\int_0^z[\textrm{Ai}~(t)~\textrm{Bi}~(z)-\textrm{Ai}(z) \textrm{Bi}(t)]dt $

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