Show dominated convergence to exchange the order of differentiation and integration Problem
Let $\text{Ai}: \mathbb{R} \to \mathbb{R}$ with
$$\text{Ai}(x) = \frac{1}{\pi} \Re \int_{0}^{\infty} \omega e^{ -\frac{t^3}{3} + i x \omega t}  dt$$
be the Airy function with $\omega := e^{\frac{i\pi}{6}}$. Prove that $\text{Ai}''(x) = x\text{Ai}(x)$.
My approach
We have $\omega = e^{\frac{i\pi}{6}} = \cos\frac{\pi}{6} + i\sin\frac{\pi}{6} = \frac{\sqrt{3}}{2} + \frac{1}{2} i$ and $-\frac{t^3}{3} + i x \omega t = -\frac{t^3}{3} - \frac{xt}{2} + \frac{xt\sqrt{3}}{2}i$. We also have $\omega^3 = e^{\frac{i\pi}{2}} = i$. Let $f: \mathbb{R}_0^+\times\mathbb{R}\to\mathbb{C}$ with:
\begin{align}
&f(t,x) = \omega e^{ -\frac{t^3}{3} + i x \omega t}\\ &\frac{\partial}{\partial x}f(t,x) = i \omega^2 t e^{ -\frac{t^3}{3} + i x \omega t} &&\left| \frac{\partial}{\partial x}f(t,x) \right| = t e^{-\frac{t^3}{3} - \frac{xt}{2}}\\
&\frac{\partial^2}{\partial^2 x} f(t,x) = - i t^2 e^{ -\frac{t^3}{3} + i x \omega t}
&&\left| \frac{\partial^2}{\partial^2 x}f(t,x) \right| = t^2 e^{-\frac{t^3}{3} - \frac{xt}{2}}
\end{align}
I need to show the following equation in order to progress:
$$ \frac{\partial^2}{\partial^2 x} \int_0^\infty f(t,x) dt = \frac{\partial}{\partial x} \int_0^\infty \frac{\partial}{\partial x} f(t,x) dt = \int_0^\infty \frac{\partial^2}{\partial^2 x} f(t,x) dt $$
For this I need to show dominated convergence for $\frac{\partial}{\partial x}f(t,x)$ and $\frac{\partial^2}{\partial^2 x}f(t,x)$, a.i. I need to find a function $g_1:\mathbb{R}_0^+ \to \mathbb{R}$ with $\left| \frac{\partial}{\partial x}f(t,x) \right| \leq g_1(t)$ for all $t \in \mathbb{R}_0^+$ and $x \in \mathbb{R}$. I also need to find a function $g_2:\mathbb{R}_0^+ \to \mathbb{R}$ with $\left| \frac{\partial^2}{\partial^2 x}f(t,x) \right| \leq g_2(t)$ for all $t \in \mathbb{R}_0^+$ and $x \in \mathbb{R}$. However, $x\in\mathbb{R}$ can be arbitrarely small making $\frac{\partial}{\partial x}f(t,x)$ and $\frac{\partial^2}{\partial^2 x}f(t,x)$ arbitrarely large at some point. I am unable to find functions $g_1(t)$, $g_2(t)$ to show dominated convergence. What am I missing?
Note
This problem was already discussed here, but the one particular point which I am interested in was handwaved.
 A: The goal here is to change the order of integration and differentiation.
Let $a(x) = \int_0^\infty \omega e^{ -\frac{t^3}{3} } g(x,t) dt$, where $g(x,t) = e^{i\omega xt}$.
You want to show that $a'(x) = \int_0^\infty \omega e^{ -\frac{t^3}{3} } { \partial g(x,t) \over \partial x } dt$, so look at the estimate
$| {a(x+h) - a(h) \over h} - \int_0^\infty \omega e^{ -\frac{t^3}{3} } { \partial g(x,t) \over \partial x } dt | \le \int_0^\infty | \omega|  e^{ -\frac{t^3}{3} }| {g(x+h,t)-g(x,t) \over h} - i\omega t g(x,t)|dt $.
If we can show that $\int_0^\infty | \omega|  e^{ -\frac{t^3}{3} }| {g(x+h,t)-g(x,t) \over h} - i\omega t g(x,t)|dt \to 0$ then we see that $a$ is diferentiable and the derivative given as above.
Use the mean value theorem to show $| {g(x+h,t)-g(x,t) \over h} - i\omega t g(x,t)| = | { e^{i\omega ht } -1 \over h } - i \omega t| \le |\omega|(1+t)$ and $| \omega|  e^{ -\frac{t^3}{3} } |\omega|(1+t)$ is an integrable upper bound. Hence the DCT applies.
Rinse & repeat for the second derivative.
