I am given a set of independent solutions of the Airy equation
$$ y''(z) - z~ y (z) = 0,$$
as contour integrals
$$ y_{1,2}(z) = \int_{C_{1,2}} e^{i\left( t^3/3 + z t\right)} dt, $$
where $C_{1}$ is the path along the real axes starting from negative infinity to the positive, and $C_2$ starts from infinity from direction $5\pi/6$ passes through the point (0,0) and ends in the infinity in direction $3\pi/2$. Note that these contours are not in typical regions for Airy function.
How do I show that these are indeed the solutions of the Airy equation and that they are independent?
If I just plug them into the equation $$ y_{1,2}''(z) - z~ y_{1,2} (z) = - \int_{C_{1,2}} \left( t^2 + z \right)e^{i\left( t^3/3 + z t\right)} dt = i \int_{C_{1,2}} \frac{d}{dt} e^{i\left( t^3/3 + z t\right)} dt, $$ and try using the fundamental theorem of calculus, it does not seem to help me much. How do I proceed?
$C_2$ is the easier of the two, and is where the Fundamental Theorem of Calculus comes in. Suppose $t = Re^{5i\pi/6}$. What's $\lim_{R\rightarrow\infty}e^{i(t^3/3+zt)}$? What about when $t = Re^{3i\pi/2}$?
For $C_1$, you should look at the wedge-shaped contours $\{se^{5i\pi/6}|0\le s\le R\} \cup \{Re^{i\theta}|5\pi/6\le\theta\le\pi\}\cup [-R, 0]$ and $[0, R]\cup \{Re^{i\theta}|0\le\theta\le\pi/6\} \cup \{se^{i\pi/6}|R\ge s\ge 0\}$. The radial contours along $5\pi/6$ and $\pi/6$ can still be handled by FTC. For the circle contours, show that the integrand decays quickly with $R$, and thus the integral over the whole arc vanishes as $R\rightarrow \infty$.