I believe that the function AiryAi[x] satisfies the ODE y''[x]-x*y[x]==0 (see Encyclopedia of Mathematics and Wiki and Weisstein, Eric W. "Airy Functions." From MathWorld--A Wolfram Web Resource and the documentation), but, being a doubting Thomas, I try to check it for x \[Element] Reals by

FullSimplify[y''[x] - x*y[x] /. {y[x] -> 1/Pi*Integrate[Cos[t^3/3 + x*t], {t, 0, Infinity},
Assumptions->x\[Element] Reals], y''[x] ->FullSimplify[D[1/Pi*Integrate[Cos[t^3/3 + x*t],{t, 0, Infinity}],  
Assumptions -> x \[Element] Reals], {x, 2}],Assumptions -> x >\[Element] Reals]}

Piecewise[{{-1/2*((AiryAi[-x] - AiryAi[x] + 2*x*AiryAiPrime[x])*Derivative[2][Abs][x])/x, x < 0}}, ((AiryAi[-x] - AiryAi[x] + 2*x*AiryAiPrime[x])* Derivative[2][Abs][x])/(2*x)]


Piecewise[{{-1/ 2*((AiryAi[-x] - AiryAi[x] + 2*x*AiryAiPrime[x])*
Derivative[2][Abs][x])/x, x < 0}}, ((AiryAi[-x] - AiryAi[x] + 2*x*AiryAiPrime[x])*
Derivative[2][Abs][x])/(2*x)] /. x -> 1.2 

0.0422432 (Abs^\[Prime]\[Prime])[1.2]

It seems that

FullSimplify[D[1/Pi*Integrate[Cos[t^3/3 + x*t], {t, 0, Infinity},  Assumptions -> x \[Element] Reals], x], 
Assumptions -> x \[Element] Reals]

which results in


does not differentiate the improper integral with respect to x, but uses table derivatives in view of

1/Pi*Integrate[-t*Sin[t^3/3 + x*t], {t, 0, Infinity}, Assumptions -> x \[Element] Reals]

(1/(6 \[Pi]))(\[Pi] (x - Abs[x]) BesselJ[-(2/3), 2/3 Abs[x]^(3/2)] - \[Pi] (x - Abs[x]) BesselJ[2/3, 2/3 Abs[x]^(3/2)] - Sqrt[3] (x + Abs[x]) BesselK[-(2/3), 2/3 Abs[x]^(3/2)])

and the latest formula leads to incorrect results. Also

1/Pi*Integrate[Cos[t^3/3 + x*t], {t, 0, Infinity}]

ConditionalExpression[(\[Pi] (-x+Abs[x]) BesselJ[-(1/3),2/3 Abs[x]^(3/2)]+\[Pi] (-x+Abs[x]) BesselJ[1/3,2/3 Abs[x]^(3/2)]+Sqrt[3] (x+Abs[x]) BesselK[-(1/3),2/3 Abs[x]^(3/2)])/(6 \[Pi] Sqrt[Abs[x]]),x\[Element]Reals]

leads to incorrect results in view of

D[ConditionalExpression[(\[Pi] (-x + Abs[x]) BesselJ[-(1/3), 
   2/3 Abs[x]^(3/2)] + \[Pi] (-x + Abs[x]) BesselJ[1/3, 
    2/3 Abs[x]^(3/2)] +  Sqrt[3] (x + Abs[x]) BesselK[-(1/3), 
    2/3 Abs[x]^(3/2)])/(6 \[Pi] Sqrt[Abs[x]]), x \[Element] Reals], {x, 2}] // FullSimplify

ConditionalExpression[ Piecewise[{{x*AiryAi[x] - ((AiryAi[-x] - AiryAi[x] + 2*x*AiryAiPrime[x])* Derivative[2][Abs][x])/(2*x), x < 0}, {x*AiryAi[x] + ((AiryAi[-x] - AiryAi[x] + 2*x*AiryAiPrime[x])*Derivative[2][Abs][x])/(2*x), x > 0}}, Indeterminate], Element[x, Reals]].

BTW, I don't find that integral in Prudnikov, Gradshtein&Ruezhik, and Abramowitz&Stegun. The only known to me proof by hand (see here) is tricky and complicated and the one cannot be mimicked in Mathematica. It is known that AiryAi[z] is an entire function, but its power series expansion is too complex to verify the ODE under investigation. The command FourierTransform[y''[x] - x*y[x] == 0, x, s] returns the input as well as LaplaceTransform[y''[x] - x*y[x] == 0, x, s].

Is it possible to check the ODE for AiryAi[x] in Mathematica without vicious circle?

Edit. Bad copy&paste after "...in view of"

  • $\begingroup$ eq = y''[x] - x*y[x] == 0; sol = DSolve[y''[x] - x*y[x] == 0, y, x]; eq /. sol // FullSimplify. $\endgroup$ Commented Mar 31, 2023 at 9:11
  • $\begingroup$ sol = DSolve[y''[x] - x*y[x] == 0, y, x] results in {{y -> Function[{x}, AiryAi[x] C[1] + AiryBi[x] C[2]]}}. I think this formula is implemented in Mathematica as a table value. Vicious circle. $\endgroup$
    – user64494
    Commented Mar 31, 2023 at 10:29

3 Answers 3


What about

airyai[x_  ] :=1/Pi* Integrate[Cos[t^3/3 + x*t], {t, 0, Infinity}, 
Assumptions -> x >=  0 ] 
y''[x] - x*y[x] == 0 /. y -> Function[x, airyai[x]] 
(* True *)

airyaiN[x_  ] :=1/Pi* Integrate[Cos[t^3/3 + x*t], {t, 0, Infinity}, 
Assumptions -> x <  0 ] 
y''[x] - x*y[x] == 0 /. y -> Function[x, airyaiN[x]]
(* True*) 

Dont't know why Mathematica cann't handle Assumptions->Element[x,Reals] in this way, but

erg = Assuming[Element[x, Reals], 1/Pi* Integrate[Cos[t^3/3 + x*t], {t, 0, Infinity}  ] ]

confirms AiryAi[x]==1/Pi* Integrate[Cos[t^3/3 + x*t], {t, 0, Infinity}] for real x

  • $\begingroup$ Thank you for your interest to the question. It seems that airyai[x_ ] uses a table integral, not calculating it, since the result is not in accordance with the result of 1/Pi*Integrate[Cos[t^3/3 + x*t], {t, 0, Infinity}, Assumptions -> x \[Element] Reals], i.e. (1/(6 \[Pi] Sqrt[ Abs[x]]))(\[Pi] (-x + Abs[x]) BesselJ[-(1/3), 2/3 Abs[x]^(3/2)] + \[Pi] (-x + Abs[x]) BesselJ[1/3, 2/3 Abs[x]^(3/2)] + Sqrt[3] (x + Abs[x]) BesselK[-(1/3), 2/3 Abs[x]^(3/2)]). That moment is noticed in my question. PS. I'd like to add that airyai[1] produces AiryAi[1]. $\endgroup$
    – user64494
    Commented Mar 31, 2023 at 8:58
  • $\begingroup$ Added by you simplification is doubtful: compare Plot[1/(6 \[Pi] Sqrt[ Abs[x]]) (\[Pi] (-x + Abs[x]) BesselJ[-(1/3), 2/3 Abs[x]^(3/2)] + \[Pi] (-x + Abs[x]) BesselJ[1/3, 2/3 Abs[x]^(3/2)] + Sqrt[3] (x + Abs[x]) BesselK[-(1/3), 2/3 Abs[x]^(3/2)]) /. x -> I*x, {x, -2., 2.}] with Plot[Re[AiryAi[I*x]], {x, -2, 2}]. Also let us put airyai1[x_] := 1/Pi*Integrate[Cos[t^3/3 + x*t], {t, 0, Infinity}, Assumptions -> Im[x] > 0] and then airyai1[1 + I] says the integral diverges. $\endgroup$
    – user64494
    Commented Mar 31, 2023 at 10:16
  • $\begingroup$ No doubts discernible: Plot[Evaluate[{airyai[x], AiryAi[x]}], {x, -10, 10}, PlotStyle -> {Automatic, Dashed}] $\endgroup$ Commented Mar 31, 2023 at 10:53
  • $\begingroup$ My conclusion: vicious circle is formed here by Mathematica developers (compare with the linked verification by hand). That's all. $\endgroup$
    – user64494
    Commented Mar 31, 2023 at 11:25

Here is my answer to the question. The AiryAi[z] is an entire function presented as the sum of the power series centered at zero (for example,see MathWorld )

1/3^(2/3)/Pi*Sum[Gamma[(n + 1)/3]/n!*(3^(1/3)*z)^n*Sin[2*(n + 1)*Pi/3], {n, 0, Infinity}]

Making use of it, we may interchange the summation and differentiation. The the term with z^(n+1) in D[AiryAi[z],{z,2}] is

D[Gamma[(n + 1)/3]/n!*(3^(1/3)*z)^n*Sin[2*(n + 1)*Pi/3], {z, 2}] /. n -> n + 3

(3^((3 + n)/3) (2 + n) (3 + n) z^(1 + n) Gamma[(4 + n)/3] Sin[2/3 (4 + n) \[Pi]])/(3 + n)!

The term with z^(n+1) in z*AiryAi[z] equals

z*Gamma[(n + 1)/3]/n!*(3^(1/3)*z)^n*Sin[2*(n + 1)*Pi/3]


FullSimplify[(3^((3 + n)/3) (2 + n) (3 + n) z^(1 + n)
Gamma[(4 + n)/3] Sin[2/3 (4 + n) \[Pi]])/(3 + n)! -
(3^(n/3) z^(1 + n) Gamma[(1 + n)/3] Sin[2/3 (1 + n) \[Pi]])/n!, 
 Assumptions -> n \[Element] Integers && n >= 0]


finishes the job.


You are in need to do the general methodology for solving an ODE.

  1. $y[x]''-x y[x] ==0$ is a homogenenous ODE, right-hand-side is zero.
  2. $y[x]''-x y[x] ==0$ is a linear ODE in $y[x]$.
  3. $y[x]''-x y[x] ==0$ is linear in $x$.
  4. $y[x]''-x y[x] ==0$ is a second order ODE, so only the second derivative and the function appear with a linear polynomial with only the quadratic term with nonzero coefficient as linear coefficient function.
  5. This can be solved by separation. $\frac{y[x]''}{y[x]} == x$ where the function itself is nonzero. This implies that the function may a have zeros that have to be treated separately. Separation is a good property but makes it more difficult to proof if the found solution is unique.
  6. $y[x]''-x y[x] ==0$ is linear in the function. So we can write a operator version of our ODE: $(\frac{d^2}{dx^2}-x)y[x] ==0$. From this we can derive a characteristic polynomial for eigenvalues $(\lambda^2-x)y[x] ==0$ but this is strange to that formalism because of $\lambda$ depends on $x$.

There is a theory for ODEs with the square roots depends on the coefficient functions. The is mainly the solution theory for the Airy ODE. This leads to the solution with the cosine and the integral formulation. That is not so nice and the straight forward methodologies are applied only. This is contour integral that is path independent but You are in need of a path so the will lead to variations in how the solution is written. Your wikipedia reference displays this well and the references therein. But this is just a general solution. That is not explained nor definition in Your reference. https://encyclopediaofmath.org/wiki/Euler_equation give some insight into the missing or left out steps by the wikipedia authors team.

  1. The consequence is only that a solution has to exist and we will have two of them that are linear independent. That follows from the second order derivative given.
  2. For the uniqueness we are in need of boundary conditions.
  3. You gave in Your question the restriction that $x \in \mathbb R$. From that we can have a behaviour of the solutions at $\infty$ and $-\infty$.
  4. A meaningful boundary condition can be derived for $x==0$ with the Wronskian of both the divergent and convergent solution.


The ODE is given on https://encyclopediaofmath.org/wiki/Airy_equation. They relate the general solution in terms of the Bessel functions. The general solutions have still two free parameters to be adapted to the boundary conditions. These are $\{c_{1},c_{2}\}$. So the general solutions is really mighty. A more in depth discussion will show that the general solutions consists of an incoming and an outgoing part, divergent and convergent for infinity.

The given reference mentions the problem for complex $x$ too.

Your starting reference is https://encyclopediaofmath.org/index.php?title=Airy_functions. That has the keyword particular solutions.

So a general solution of an ODE consist of two parts the general solutions and of particular solutions. I mentioned the zeros of the functions itself and the zero of the polynomial that is the coefficient function.

So what they expect You to know ahead of reading the articles with all of Your reference is that You are firm and professional with https://en.wikipedia.org/wiki/Ordinary_differential_equation. That is for sure plenty of work time. A mayor point is to know what it is with https://en.wikipedia.org/wiki/Picard%E2%80%93Lindel%C3%B6f_theorem for local existence and uniqueness. Here is another reference https://encyclopediaofmath.org/wiki/Cauchy-Lipschitz_theorem.

The existence is already proven with my list of steps. The uniqueness needs boundary conditions. That are two parts. You gave an ODE that suffices for local existence. With additional boundary conditions the question of uniquess appears and has to be solved!

  • $\begingroup$ Thank you. I will be waiting for a real verification of AiryAi[x] as a solution of the ODE under consideration with Mathematica. $\endgroup$
    – user64494
    Commented Mar 31, 2023 at 10:02

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