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Analytic continuation of Airy function

  1. May 19, 2015 #1
    For [itex]x\in\mathbb{R}[/itex] we can set

    \textrm{Ai}(x) = \frac{1}{2\pi} \int\limits_{-\infty}^{\infty} e^{i\big(\frac{t^3}{3} + tx\big)}dt

    If we substitute in place of [itex]x[/itex] a complex parameter [itex]z[/itex] with [itex]\textrm{Im}(z)>0[/itex], the integral will converge on [itex][0,\infty[[/itex], but diverge on [itex]]-\infty,0][/itex]. With [itex]\textrm{Im}(z)<0[/itex] the integral will converge on [itex]]-\infty,0][/itex], but diverge on [itex][0,\infty[[/itex]. The Wikipedia page tells me that a complex analytic version of the Airy function exists, but apparently it cannot be defined simply by substituting a complex variable [itex]z[/itex] into the same integral formula that works for real variables [itex]x[/itex]. How is the analytic continuation of Airy function studied then?
  2. jcsd
  3. May 24, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
  4. May 24, 2015 #3
    I realized that the complex Airy function can be defined by setting

    \textrm{Ai}(z) = \frac{\sqrt{3}-i}{4\pi}\int\limits_{-\infty}^0 e^{\frac{t^3}{3} + \big(\frac{1}{2}+\frac{i\sqrt{3}}{2}\big)zt}dt + \frac{\sqrt{3}+i}{4\pi}\int\limits_0^{\infty} e^{-\frac{t^3}{3}+ \big(-\frac{1}{2}+\frac{i\sqrt{3}}{2}\big)zt}dt

    The idea has been that we try to define the function by a formula

    \textrm{Ai}(z) = \frac{1}{2\pi}\int e^{i\big(\frac{\zeta^3}{3}+z\zeta\big)}d\zeta

    but modify the original integration path [itex]]-\infty,\infty[[/itex] so that the convergence is improved.
  5. May 25, 2015 #4

    ##g(z)=\frac{1}{2\pi i}\int_{-i\infty}^{i\infty} e^{t^3/3-zt}dt##

    where the path is along the imaginary axis.

    Can you show this integral function reduces to the airy function when z is real (let t=iy)? Now, if ##g(z)## converges for some range of complex ##z## and it reduces to the Airy function for real z, then isn't ##g(z)## the analytic continuation of the airy function?
    Last edited: May 25, 2015
  6. May 26, 2015 #5
    That integral converges if [itex]\textrm{Im}(z)=0[/itex], and diverges if [itex]\textrm{Im}(z)\neq 0[/itex], so it isn't giving an analytic continuation to anywhere.
  7. May 26, 2015 #6
    Ok sorry. I see that now. I don't wish to make it worst and I probably should just stay out of it now but I like Complex Analysis so let me suggest another option and then I'll leave it to others to help you: Wolfram offers another contour integral:

    ##Ai(z)=\frac{1}{2\pi i}\mathop{\int}\limits_{\infty e^{-\pi i/3}}^{\infty e^{\pi i/3}} e^{t^3/3-zt} dt##

    I assume they mean the pie-slice, from -pi/3 to pi/3, a closed-contour, then let ##t=re^{\pm \pi i/3}## along the straight-line contours and probably we can show the integral along the circular part goes to zero as r goes to infinity but not sure. Then use the Residue Theorem. And it looks a bit tough to show that converges for a range of z and then will reduce to the airy function for real z but it looks encouraging though.

    Here's the link to the Wolfram functions:

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