5
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ the solution of the given equation is the Airy-function $\endgroup$ – Dr. Sonnhard Graubner Dec 28 '17 at 12:57
  • 2
    $\begingroup$ 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) $\endgroup$ – tired Dec 29 '17 at 0:20
2
$\begingroup$

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!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.