# Showing Integral form satisfies the Airy function

## Main Question or Discussion Point

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

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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:
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

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