Suppose that $z>0$ and let us write $z=v^2$ where $v>0$, the question is how to show that from $Ai(z)=\frac{1}{\pi}\int_0^\infty \cos(\frac{1}{3}t^3+zt)dt$ that:
$$Ai(v^2)=\frac{1}{2\pi i}\int_I \exp(v^2 s-1/3 s^3)ds$$
where $I$ is the imaginary axis from $-\infty i$ to $+\infty i$.
I first tried the identity: $\cos(\ldots)= \frac{1}{2} (\exp(i\ldots)+\exp(-i\ldots))$, but I don't see how to continue from there?
Thanks!
Best Answer
\begin{align*} Ai(v^2)&=\frac1\pi\int_0^\infty \cos\left(\frac13t^3+v^2t\right)\,dt\\&=\frac1{2\pi}\int_\mathbb R\cos\left(\frac13t^3+v^2t\right)\,dt\\&=\frac1{2\pi}\int_\mathbb R\cos\left(\frac13t^3+v^2t\right)+i \sin\left(\frac13t^3+v^2t\right)\,dt\\&=\frac1{2\pi i}\int_\mathbb R e^{i (\frac13t^3+v^2t)}\,d(it)\\&=\frac1{2\pi i}\int_Ie^{v^2s-\frac13s^3}\,ds. \end{align*}