Proving that the function $f(x) = \int_0^\infty \cos (w^3/3 - x w ) d w$ satisfies the equation $f'' + x f = 0$ It is assumed that $x$ is real. 
Formally, we have 
$$ f'' = \int_0^\infty  -\cos (w^3/3 - x w ) w^2  d w , $$
and hence 
$$f'' + x f  = \int_0^\infty  \cos (w^3/3 - x w ) (-w^2 + x )  d w \\ = -\int_0^\infty  \cos (w^3/3 - x w ) d(w^3/3 - x w ) \\ = - \sin(w^3/3- x w ) |_0^\infty .  $$
The problem is that $\sin(w^3/3- x w ) $ does not converges as $w\rightarrow \infty$. 
Apparently, the problem is rooted in the fact that the expression for $f''$ is not well defined---it does not converge. 
So, could anyone give a simple, elementary proof? 

Your derivation is fine and leads to an ambiguous form, as you state. To disambiguate this, consider the following.
Your integral is 
$$
-\int_0^\infty  \cos (w^3/3 - x w ) d(w^3/3 - x w ) 
$$
If you make the transformation $t = w^3/3 - x w$ you have, for any given (finite) $x$, 
$$
-\int_0^\infty  \cos (t) d t 
$$
which is a special case of 
$$
I(\nu) = -\int_0^\infty  \cos (\nu \, t) d t 
$$
for $\nu = 1$. 
By symmetry, this is
$$
I(\nu) = -\frac12 \int_{-\infty}^\infty  \cos (\nu \, t) d t = -\frac12 \Re {\Large(} \int_{-\infty}^\infty  e^{j \nu \, t} d t \Large)
$$
, i.e. the real part of a complex integral.
To disambiguate your expression you can now interpret the last integral in a distribution sense. Indeed, we have the Fourier transformation of the delta distribution which can be (symbolically) stated as 
$$
\delta (\nu ) = \int_{-\infty}^\infty  e^{j \nu \, t} d t
$$
Hence 
$$
I(\nu) =  -\frac12 \Re {\Large(} \delta (\nu ) \Large)
$$
and in particular, the desired 
$$
I(\nu = 1) =  0
$$
Indeed, many treatments of the Airy function (we have the integral representatin of the Airy function here) go via the Fourier Transform.
Happy New Year 2018! 
