Solution to the Airy equation with real part I'm given a solution to the Airy equation:
$Ai:\mathbb{R}\rightarrow\mathbb{R}, Ai(x)=\frac{1}{\pi}Re\int_{0}^{\infty}\omega\exp(-\frac{t^3}{3}+ix\omega t)dt$
, where $\omega = e^{\frac{i\pi}{6}}$
Now I neeed to show the Airy equation:
$\frac{d^2Ai}{dx^2} -xAi = 0$ for $x \in \mathbb{R}$
I know $Re(\omega) = \frac{\sqrt{3}}{2}$ and $Re(ix\omega t) = \cos(\frac{\sqrt{3}xt}{2})$.
I know how to solve it for the cos variant:
$Ai:\mathbb{R}\rightarrow\mathbb{R}, Ai(x)=\frac{1}{\pi}\int_{0}^{\infty}\cos(\frac{t^3}{3}+xt)dt$
but I can't seem to get it with the Real part version. Is there a way to convert this solution to the cos version? I tried applying eulers identity first but I'm not sure how to deal with the Real part in front of the Integral.
 A: The function you are given comes from distorting a contour of integration using Calculus of residues technics. In the reference I gave you in the comments,  they use a contour that starts at the origin with a straight segment up to point $R$, then a moving along the circle of radius $R$ up to and angle $\theta$ and then moving back to the origin along the line with parallel to $e^{i\theta}$. After that, they let $R\rightarrow\infty$.
There are many other ways to distort the contour for the Airy function. In any case, for the problem in the OP you don't need to identify which distortion is used. It suffices to differentiate and see that the function given to you satisfies Airy's equation.
Notice that the function, as is given to you, is actually analytic in $\omega$ since the term $e^{-t^3/3}$ decays to $0$ faster than any polynomial. That means that one can exchange the order of differentiation and integration without any qualms (the justification being dominated convergence).
Let
$ F(x)=\int_{0}^{\infty}\omega\exp(-\frac{t^3}{3}+ix\omega t)dt$
, where $\omega = e^{\frac{i\pi}{6}}$
$$F'(x)=\int^\infty_0\exp(-\tfrac{t^3}{3})i\omega^2 t\exp(ix\omega t)\,dt$$
$$\begin{align}
F''(x)&= -\int^\infty_0\exp(-\tfrac{t^3}{3})\omega^3 t^2\exp(ix\omega t)\,dt\\
&=-i\int^\infty_0\exp(-\tfrac{t^3}{3}) t^2\exp(ix\omega t)\,dt
\end{align} $$
Integration by parts yields
$$\begin{align}
x\,F(x)&=-i\int^\infty_0 \exp(-\tfrac{t^3}{3})i\omega x\exp(ix\omega t)\,dt \\
&=-i\Big(\exp(-\tfrac{t^3}{3})\exp(ix\omega t)|^\infty_0+\int^\infty_0\exp(-\tfrac{t^3}{3})t^2\exp(ix\omega t)\,dt\Big)\\
&=i + -i\int^\infty_0\exp(-\tfrac{t^3}{3})t^2\exp(ix\omega t)\,dt
\end{align}
$$
Taking real parts and putting things together you obtain the result.
