The Airy function is defined as follows:(adsbygoogle = window.adsbygoogle || []).push({});

[tex]

\textrm{Ai}(x) = \frac{1}{\pi}\int\limits_0^{\infty} \cos\big(\frac{1}{3}t^3 + xt\big)dt

[/tex]

It seems pretty probable, that its derivative is this:

[tex]

\textrm{Ai}'(x) = -\frac{1}{\pi}\int\limits_0^{\infty} t\sin\big(\frac{1}{3}t^3 + xt\big)dt

[/tex]

I'm interested to know how this result is proven. It's not as simple as it might seem on quick glance. Only thing that's trivial is this:

[tex]

\textrm{Ai}'(x) = \lim_{\Delta x\to 0}\frac{1}{\pi}\int\limits_0^{\infty}

\frac{\cos\big(\frac{1}{3}t^3 + (x+\Delta x)t\big) - \cos\big(\frac{1}{3}t^3 + xt\big)}{\Delta x}dt

= -\lim_{\Delta x\to 0}\frac{1}{\pi}\int\limits_0^{\infty} t\sin\big(\frac{1}{3}t^3 + (x+\xi_{t,\Delta x})t\big) dt

[/tex]

Where [itex]|\xi_{t,\Delta x}|\leq |\Delta x|[/itex]. So the non-trivial task is to prove that the order of limit and integration can be changed. The integral does not converge absolutely, but instead relies on cancellation of accelerating oscillations. If we have no better knowledge about [itex]\xi_{t,\Delta x}[/itex], it could behave in such manner that it starts picking more positive parts of the integrand than negative parts, and have surprising effect on the value of the integral.

EDIT 1:

At first I thought it would be simple thing to prove that

[tex]

\textrm{Ai}''(x) = x\textrm{Ai}(x)

[/tex]

after it has been shown that the derivative and integral can be commuted, but actually that's not the only problem. If I assume that I can commute [itex]D_x^2[/itex] with the integral, I get the following:

[tex]

\textrm{Ai}''(x) = -\frac{1}{\pi}\int\limits_0^{\infty} t^2\cos\big(\frac{1}{3}t^3 + xt\big)dt

= -\frac{1}{\pi}\int\limits_0^{\infty}\Big( D_t\sin\big(\frac{1}{3}t^3 + xt\big) \;-\; x\cos\big(\frac{1}{3}t^3 + xt\big)\Big)dt

[/tex]

But [itex]\lim_{t\to\infty}\sin(\frac{1}{3}t^3 + xt)[/itex] does not exist and is not zero, so actually we can't conclude that the integral of the derivative would be zero. So how do you compute the second derivative then?

EDIT 2:

hmhmh... I just noticed that

[tex]

\int\limits_0^{\infty} t^{\alpha} \cos\big(\frac{1}{3}t^3 + xt\big)dt = \frac{1}{3}\int\limits_0^{\infty} u^{(\alpha - 2)/3} \cos\big(\frac{1}{3}u + xu^{1/3}\big) du

[/tex]

seems to converge only when [itex]-1 < \alpha < 2[/itex], and diverges for [itex]2 \leq \alpha[/itex].

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# How to prove derivative of Airy function

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