Showing Integral form satisfies the Airy function

  • #1

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

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
 

Answers and Replies

  • #2
Assume that
[tex]y = \frac{1}{\pi} \int_0^\infty \cos\left(\tfrac13t^3 + xt\right)\, dt,[/tex]

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

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

Why is RHS identically zero ? :cry:
 
Last edited:
  • #3
240
2
make a change of variables:

[tex]u=\frac{1}{3}t^{3}+xt => du=(t^{2}+x)dt [/tex]

you will simply have:

[tex]-\frac{1}{\pi}\int^{\infty}_{0}cos(u)du[/tex]

that technically doesn't converge to zero. So that's the most far you can get
 
  • #4
surely the positive and negative components of the cos function will add up to zero when you integrate ?
 

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