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?

• the solution of the given equation is the Airy-function – Dr. Sonnhard Graubner Dec 28 '17 at 12:57
• you can make $x$ slightly complex ($x\rightarrow x-i \delta$ with $\delta >0$). then your argument should go through pretty smooth (taking limit $\delta \rightarrow 0_+$ in the end) – tired Dec 29 '17 at 0:20

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 $$I(\nu = 1) = 0$$