Analytic continuation of Airy function

[jostpuur]

For $x\in\mathbb{R}$ 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 $x$ a complex parameter $z$ with $\textrm{Im}(z)>0$, the integral will converge on $[0,\infty[$, but diverge on $]-\infty,0]$. With $\textrm{Im}(z)<0$ the integral will converge on $]-\infty,0]$, but diverge on $[0,\infty[$. 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 $z$ into the same integral formula that works for real variables $x$. How is the analytic continuation of Airy function studied then?

 

[jostpuur]

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 $]-\infty,\infty[$ so that the convergence is improved.

 

[jackmell]

How is the analytic continuation of Airy function studied then?

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

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?

 

[jostpuur]

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?

That integral converges if $\textrm{Im}(z)=0$, and diverges if $\textrm{Im}(z)\neq 0$, so it isn't giving an analytic continuation to anywhere.

 

[jackmell]

That integral converges if $\textrm{Im}(z)=0$, and diverges if $\textrm{Im}(z)\neq 0$, so it isn't giving an analytic continuation to anywhere.

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:

http://functions.wolfram.com/Bessel-TypeFunctions/AiryAi/07/02/0001/