# Simplifying an integral expression involving Airy functions

## Main Question or Discussion Point

Hello,
Is there some way to express the following integral in terms of some simpler functions?

$f(x,s) = \int^{\infty}_{-\infty} dk\, e^{-ks} \text{Ai}(-k) \text{Ai}(x-k)$

where the parameter $s \in (0,1)$ and the coordinate $x \in (-\infty,+\infty)$

The best I can come up with is to integrate numerically, but it takes time to get a good resolution :(

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arildno
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I made a google search, and came up with the following article by Vallee, Soares, de Izarra: "An integral representation for the product of Airy functions", published in ZAMP (1997), pages 156-160
On the following preview image, eq. (12), at the bottom of preview, seems sufficiently close, for some parameter choices, to be worthwhile for you to take a closer look at (the actual results are NOT, unfortunately, part of the preview).

In any case, even if that article does not contain what you seek, perhaps the authors might give you valuable advice?

Best of luck!

1 person
Thanks for your effort! I downloaded the full article and it does not contain the formula I am looking for :( There is a similar one when the 's' in the exponent is imaginary, and they give some general strategy to calculate those integrals, but I need to work it all out by myself and I suspect that it's gonna take me a whole lot of time :(

arildno
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Thanks for your effort! I downloaded the full article and it does not contain the formula I am looking for :( There is a similar one when the 's' in the exponent is imaginary, and they give some general strategy to calculate those integrals, but I need to work it all out by myself and I suspect that it's gonna take me a whole lot of time :(
Not a bull's eye, then?
Okay, I hope you have zoomed in a bit closer on your problem..

well I'm integrating the thing numerically, but when x->0, the analytical result is a dirac delta function, but in terms of numerical integration i have to include very high k values to smooth out unphysical oscillations, which is heavy duty on my pc :(

arildno
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Aah, so you are to get that Dirac delta perversity, then?
No wonder why that conventionalist minded computer grumbles and writhes in pain.

Couldn't you, in order to avoid that particular numerical problem, try to devise some clever asymptotic matching scheme? (I'm not at all sure it will work, it's more than a decade since I did these sorts of things, so my suggestion might prove silly or impractical)

Aah, so you are to get that Dirac delta perversity, then?
No wonder why that conventionalist minded computer grumbles and writhes in pain.

Couldn't you, in order to avoid that particular numerical problem, try to devise some clever asymptotic matching scheme? (I'm not at all sure it will work, it's more than a decade since I did these sorts of things, so my suggestion might prove silly or impractical)
yeah I'm trying that right now. Going to adjust my range of k-integration according to the value of s I'm considering, should do the trick. But it would still be cool to have an analytical expression...

arildno
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In my experience, functions that behave like beasts in some region of the domain are rarely tamed by analytical means.
Anyways, I would suggest that you pop off an e-mail to one of the authors of the 1997 article, asking if they happen to know of any work done on your specific problem. There's a jungle out there, also in the realm of published materials.

Avodyne
Using the results in that paper, I believe I have derived

$f(y,s) = {\exp\left[{s^4-6s^2y-3y^2\over12s}\right]\over2\sqrt{\pi}\sqrt{s}}$

Start with eq.(20), but replace the factor of Ai(x) with exp(-sx), and take z=0. Then I used Mathematica to do the x integral in eq.(21). Mathematica would only do it if I shifted and rescaled x so that the argument of the Airy function was just -x. Then it gave a horrible mess, but using FullSimplify followed by PowerExpand gave a simple result, and then the xi integral was simple, giving the result above.

No promises that I did it right!

EDIT: actually, if we replace s by i*s, then that x integral is just the Fourier transform of Ai(-x). But then the result is analytic in s in some domain which includes imaginary s. So I didn't need Mathematica to do that integral.

Last edited:
1 person
Using the results in that paper, I believe I have derived

$f(y,s) = {\exp\left[{s^4-6s^2y-3y^2\over12s}\right]\over2\sqrt{\pi}\sqrt{s}}$

Start with eq.(20), but replace the factor of Ai(x) with exp(-sx), and take z=0. Then I used Mathematica to do the x integral in eq.(21). Mathematica would only do it if I shifted and rescaled x so that the argument of the Airy function was just -x. Then it gave a horrible mess, but using FullSimplify followed by PowerExpand gave a simple result, and then the xi integral was simple, giving the result above.

No promises that I did it right!

EDIT: actually, if we replace s by i*s, then that x integral is just the Fourier transform of Ai(-x). But then the result is analytic in s in some domain which includes imaginary s. So I didn't need Mathematica to do that integral.
Wow, this is correct!! Could you do the same for this integral as well?

$g(x,s) = \int^{\infty}_{-\infty} dk\, e^{-ks} \text{Ai}(x-k)$

arildno
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Nice that the paper was useful after all!
(Slapping my own back, )

woops, never mind, found it myself:

$g(x,s) \propto \exp(-s^3/3-sx)$

Anyway, thanks for the inspiration

Nice that the paper was useful after all!
(Slapping my own back, )
well I guess that paper is way beyond my own level, that's why I have a hard time applying it...

arildno
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well I guess that paper is way beyond my own level, that's why I have a hard time applying it...
I am too rusty to follow that paper myself, so kudos to Avodyne who could.

Avodyne
I've realized that you can do this integral without using the results in the paper. The key is to let $s$ be imaginary, and then analytically continue to real $s$ later.

$f(x,-iz) = \int^{\infty}_{-\infty} dk\, e^{ikz} \text{Ai}(-k) \text{Ai}(x-k)$

and substitute in

$\text{Ai}(-k) = {1\over2\pi}\int^{\infty}_{-\infty} dt\, \exp\bigl[it^3\!/3 + i(-k)t\bigr]$

$\text{Ai}(x-k) = {1\over2\pi}\int^{\infty}_{-\infty} du\, \exp\bigl[iu^3\!/3 + i(x-k)u\bigr]$

Then do the integral over $k$, which yields $2\pi\delta(z-t-u)$. Then integrate over $u$, which (because of the delta function) just replaces $u$ with $z-t$ everywhere. Now we have

$f(x,-iz) = {1\over2\pi}\int^{\infty}_{-\infty} dt\,\exp\bigl[i\bigl(zt^2-(z^2+x)t + z^3\!/3+zx\bigr)\bigr]$

At this point, let $z=is$. Now we have

$f(x,s) = {1\over2\pi}\int^{\infty}_{-\infty} dt\,\exp\bigl[-st^2+i(s^2-x)t + s^3\!/3-sx\bigr]$

This is a simple gaussian integral which yields the result I gave above.

I've realized that you can do this integral without using the results in the paper. The key is to let $s$ be imaginary, and then analytically continue to real $s$ later.

$f(x,-iz) = \int^{\infty}_{-\infty} dk\, e^{ikz} \text{Ai}(-k) \text{Ai}(x-k)$

and substitute in

$\text{Ai}(-k) = {1\over2\pi}\int^{\infty}_{-\infty} dt\, \exp\bigl[it^3\!/3 + i(-k)t\bigr]$

$\text{Ai}(x-k) = {1\over2\pi}\int^{\infty}_{-\infty} du\, \exp\bigl[iu^3\!/3 + i(x-k)u\bigr]$

Then do the integral over $k$, which yields $2\pi\delta(z-t-u)$. Then integrate over $u$, which (because of the delta function) just replaces $u$ with $z-t$ everywhere. Now we have

$f(x,-iz) = {1\over2\pi}\int^{\infty}_{-\infty} dt\,\exp\bigl[i\bigl(zt^2-(z^2+x)t + z^3\!/3+zx\bigr)\bigr]$

At this point, let $z=is$. Now we have

$f(x,s) = {1\over2\pi}\int^{\infty}_{-\infty} dt\,\exp\bigl[-st^2+i(s^2-x)t + s^3\!/3-sx\bigr]$

This is a simple gaussian integral which yields the result I gave above.
Yep, now that you've shown it, it seems easy :D