Showing Integral form satisfies the Airy function

[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:
<br /> Ai(x) = \frac{1}{\pi} \int_0^\infty \cos\left(\tfrac13t^3 + xt\right)\, dt,<br />

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

 

[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&#039;&#039;-xy = -\frac{1}{\pi} \int_0^\infty (t^2+x)\cos\left(\tfrac13t^3 + xt\right)\, dt

Why is RHS identically zero ?

 

[elibj123]

make a change of variables:

u=\frac{1}{3}t^{3}+xt =&gt; 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

 

[thrillhouse86]

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