Editing Airy function (section)

==Relation to other special functions==
For positive arguments, the Airy functions are related to the [[Bessel function#Modified Bessel functions|modified Bessel functions]]:
<math display="block">\begin{align}
 \operatorname{Ai}(x) &{}= \frac1\pi \sqrt{\frac{x}{3}} \, K_{1/3}\!\left(\frac{2}{3} x^{3/2}\right), \\
 \operatorname{Bi}(x) &{}= \sqrt{\frac{x}{3}} \left[I_{1/3}\!\left(\frac{2}{3} x^{3/2}\right) + I_{-1/3}\!\left(\frac{2}{3} x^{3/2}\right)\right].
\end{align}</math>
Here, {{math|''I''<sub>±1/3</sub>}} and {{math|''K''<sub>1/3</sub>}} are solutions of <math display="block">x^2y'' + xy' - \left (x^2 + \tfrac{1}{9} \right )y = 0.</math>

The first derivative of the Airy function is
<math display="block"> \operatorname{Ai'}(x) = - \frac{x} {\pi \sqrt{3}} \, K_{2/3}\!\left(\frac{2}{3} x^{3/2}\right) .</math>

Functions {{math|''K''<sub>1/3</sub>}} and {{math|''K''<sub>2/3</sub>}} can be represented in terms of rapidly convergent integrals<ref>M.Kh.Khokonov. Cascade Processes of Energy Loss by Emission of Hard Photons // JETP, V.99, No.4, pp. 690-707 \ (2004).</ref> (see also [[Bessel function#Modified Bessel functions|modified Bessel functions]])

For negative arguments, the Airy function are related to the [[Bessel function]]s:
<math display="block">\begin{align}
 \operatorname{Ai}(-x) &{}= \sqrt{\frac{x}{9}} \left[J_{1/3}\!\left(\frac{2}{3} x^{3/2}\right) + J_{-1/3}\!\left(\frac{2}{3} x^{3/2}\right)\right], \\
 \operatorname{Bi}(-x) &{}= \sqrt{\frac{x}{3}} \left[J_{-1/3}\!\left(\frac{2}{3 }x^{3/2}\right) - J_{1/3}\!\left(\frac23 x^{3/2}\right)\right].
\end{align}</math>
Here, {{math|''J''<sub>±1/3</sub>}} are solutions of
<math display="block">x^2y'' + xy' + \left (x^2 - \frac{1}{9} \right )y = 0.</math>

The [[Scorer's function]]s {{math|Hi(''x'')}} and {{math|-Gi(''x'')}} solve the equation {{math|1=''y''′′ − ''xy'' = 1/π}}. They can also be expressed in terms of the Airy functions:
<math display="block">\begin{align}
 \operatorname{Gi}(x) &{}= \operatorname{Bi}(x) \int_x^\infty \operatorname{Ai}(t) \, dt + \operatorname{Ai}(x) \int_0^x \operatorname{Bi}(t) \, dt, \\
 \operatorname{Hi}(x) &{}= \operatorname{Bi}(x) \int_{-\infty}^x \operatorname{Ai}(t) \, dt - \operatorname{Ai}(x) \int_{-\infty}^x \operatorname{Bi}(t) \, dt.
\end{align}</math>