# Analytic continuation of Airy function

• jostpuur
In summary, the conversation discusses how to define the complex Airy function and its analytic continuation. It is stated that substituting a complex variable into the integral formula for real variables does not work. Instead, another integral function along the imaginary axis is proposed as a possible analytic continuation. However, it is noted that this function only converges for purely imaginary z and does not provide an analytic continuation to any other points. Another option is suggested, using a closed-contour integral along a pie-slice, which may converge for a range of complex z and reduce to the Airy function for real z.
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?

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.

jostpuur said:
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?

Last edited:
jackmell said:
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.

jostpuur said:
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/

## 1. What is the Airy function and why is it important in analytic continuation?

The Airy function is a special function that arises in many areas of mathematics and physics, including the study of differential equations and complex analysis. It is important in analytic continuation because it provides a way to extend the function to values outside of its original domain, allowing for a deeper understanding and analysis of its behavior.

## 2. What is analytic continuation and how does it relate to the Airy function?

Analytic continuation is the process of extending a function to values outside of its original domain using complex analysis techniques. In the case of the Airy function, analytic continuation allows us to evaluate the function at complex numbers and explore its behavior in the complex plane.

## 3. Can the Airy function be represented by a single analytic expression?

Yes, the Airy function can be represented by a single analytic expression called the Airy integral, which involves a contour integral over a certain path in the complex plane. This expression is used to calculate the value of the function at complex numbers and is the key to its analytic continuation.

## 4. How is the analytic continuation of the Airy function used in practical applications?

The analytic continuation of the Airy function has many practical applications, including in the study of fluid mechanics, quantum mechanics, and statistical mechanics. It is also used in the development of numerical methods for solving differential equations and in the analysis of complex physical systems.

## 5. Are there any unresolved questions or challenges in the field of analytic continuation of the Airy function?

Yes, there are still many open questions and challenges in this field. One major challenge is the development of efficient and accurate numerical methods for computing the values of the Airy function at complex numbers. Additionally, there is ongoing research on the behavior and properties of the Airy function under different conditions, which can have important implications for its analytic continuation.

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