# Showing Integral form satisfies the Airy function

1. Dec 10, 2009

### thrillhouse86

Hey All,

Can someone please give me the gist of how to show that the integral form of the Airy function for real inputs:
$$Ai(x) = \frac{1}{\pi} \int_0^\infty \cos\left(\tfrac13t^3 + xt\right)\, dt,$$

satisfies the Airy Differential Equation: y'' - xy = 0

I tried differentiating twice wrt to the x variable (assuming I could just bring it inside the integration) and then subbing back into the ODE but that failed.

Regards,
Thrillhouse

2. Dec 18, 2009

### matematikawan

Assume that
$$y = \frac{1}{\pi} \int_0^\infty \cos\left(\tfrac13t^3 + xt\right)\, dt,$$

and the operations integrate and differentiate can be interchange (??), I obtain

$$y''-xy = -\frac{1}{\pi} \int_0^\infty (t^2+x)\cos\left(\tfrac13t^3 + xt\right)\, dt$$

Why is RHS identically zero ?

Last edited: Dec 18, 2009
3. Dec 18, 2009

### elibj123

make a change of variables:

$$u=\frac{1}{3}t^{3}+xt => du=(t^{2}+x)dt$$

you will simply have:

$$-\frac{1}{\pi}\int^{\infty}_{0}cos(u)du$$

that technically doesn't converge to zero. So that's the most far you can get

4. Dec 19, 2009

### thrillhouse86

surely the positive and negative components of the cos function will add up to zero when you integrate ?