Show that $\displaystyle y(z)=\frac{1}{2\pi i}\int_{C}{exp(zt-\frac{t^{3}}{3})dt} $ is a solution of $\displaystyle y''(z)=zy(z)$
I think you need an i in there huh? Like:
$\displaystyle A(s)=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{i(z^3/3+sz)}dz$
then $\displaystyle A''-sA=0$
Differentiate under the integral sign and use Cauchy's Theorem to express the Airy integral in terms of a contour integral over a square region bordering the real axis of height say i. Then since the integrand is analytic, we can express the Airy integral in terms of the horizontal leg at $\displaystyle z=x+i$ since the integral over the two vertical legs tend to zero. Then:
$\displaystyle Ai(s)=\frac{1}{2\pi}\int_{\infty}^{-\infty} e^{i(z^3/3+sz)}dz,\quad z=x+i$
Now:
$\displaystyle A''-sA=-\frac{1}{2\pi}\int_C (z^2+s)e^{i(z^3/3+sz)}dz;\quad z=x+i$
but that integrand is the differential of $\displaystyle f(z)=ie^{i(z^3/3+sz)}$ and since the integrand is analytic, we can just evaluate it at the endpoints:
$\displaystyle \frac{1}{2\pi}\int_{\infty}^{-\infty}df$
and note if you substitute $\displaystyle z=x+i$ into $\displaystyle f$, it's order is $\displaystyle O(e^{-x^2})$ so we have:
$\displaystyle O(e^{-x^2})\biggr|_{\infty}^{-\infty}$
which is zero.