# The Apéry's constant and the Airy function

Several weeks ago, while I was playing with a CAS, Wolfram Alpha online calculator I found the closed-form that provided this calculator for $$\int_0^\infty\operatorname{Ai}(x)\log^3(x)dx,\tag{1}$$

involving the Airy function $$\operatorname{Ai}(x)$$, see this special function from the Wikipedia Airy function. But I don't know nor if it is in the literature neither nor any hint to get the corresponding indefinite integral (if it is feasible).

Question. Do you know if the closed-form for $$\int_0^\infty\operatorname{Ai}(x)(\log(x))^3dx,\tag{1}$$ in terms of well-known constants and $$\zeta(3)$$ is in the literature? Provide the reference, and if I can I search it. In other case, can you provide an hint to calculate this kind of definite integral? Many thanks.

• Why not take baby steps? Does the following got a closed form? $$\int_0^\infty\operatorname{Ai}(x) \ln xdx$$ – Zacky Jul 5 '19 at 19:36
• @Zacky Yes – mrtaurho Jul 5 '19 at 19:42
• From W|A the result is supposedly $$\frac1{81}[-52\zeta(3)-(2\gamma+\ln3)((2\gamma+\ln3)^2+4\pi^2)]\approx-2.0101$$ – TheSimpliFire Jul 5 '19 at 19:44
• I feel like there's a clever way to use integration by parts and Airy differential equation here, but I haven't been able to work it out yet – Yuriy S Jul 5 '19 at 22:38
• This is a message of appreciation to all the users who have been interested in this question, who provided us answers or comments. I am going to study this information. – user686930 Jul 6 '19 at 9:54

You may refer to the entry 9.10.17 of DLMF, which describes the Mellin transform of the Airy function $$\operatorname{Ai}$$:

$$\int_{0}^{\infty} t^{\alpha-1}\operatorname{Ai}(t)\,\mathrm{d}t = \frac{\Gamma(\alpha)}{3^{(\alpha+2)/3}\Gamma\big(\frac{\alpha+2}{3}\big)}. \tag{9.10.17}$$

Differentiating both sides with respect to $$\alpha$$ 3-times and plugging $$\alpha = 1$$ gives the answer to your integral in terms of polygamma functions with further simplifications available.

Sketch of proof of $$\text{(9.10.17)}$$. We begin with the integral representation

$$\operatorname{Ai}(x) = \frac{1}{\pi} \int_{0}^{\infty} \cos\left(\frac{t^3}{3} + xt \right) \, \mathrm{d}t.$$

Taking Mellin transform and switching the order of integration,

\begin{align*} \int_{0}^{\infty} x^{\alpha-1}\operatorname{Ai}(x)\,\mathrm{d}x &= \frac{1}{\pi} \operatorname{Re}\bigg[ \int_{0}^{\infty}\int_{0}^{\infty} x^{\alpha-1} e^{\frac{1}{3}it^3 + ixt} \, \mathrm{d}x\mathrm{d}t \bigg] \\ &= \frac{1}{\pi} \operatorname{Re}\bigg[ \int_{0}^{\infty} \frac{\Gamma(\alpha)}{(-it)^{\alpha}} e^{\frac{1}{3}it^3} \, \mathrm{d}t \bigg] \\ &= \frac{1}{\pi} \operatorname{Re}\bigg[ \Gamma(\alpha) i^{\alpha} \frac{\Gamma\big(\frac{1-\alpha}{3}\big)}{3\cdot(-i/3)^{\frac{1-\alpha}{3}}} \bigg] \\ &= \frac{1}{\pi} \operatorname{Re}\bigg[ \Gamma(\alpha) i^{\frac{2\alpha+1}{3}} \frac{\Gamma\big(\frac{1-\alpha}{3}\big)}{3^{\frac{\alpha+2}{3}}} \bigg] \\ &= \frac{\Gamma(\alpha)\Gamma\big(\frac{1-\alpha}{3}\big)}{\pi 3^{\frac{\alpha+2}{3}}}\cos\big( \tfrac{1}{6}\pi + \tfrac{1}{3}\alpha\pi \big). \end{align*}

Now applying the Euler's reflection formula $$\Gamma(s)\Gamma(1-s) = \pi \csc(\pi s)$$ with $$s = \frac{1-\alpha}{3}$$ yields the right-hand side of $$\text{(9.10.17)}$$.

• Do you have an idea on how to prove that? I was also writing an answer with the same method that you said. However I have difficulties on proving that Mellin transform. – Zacky Jul 5 '19 at 20:23
• @Zacky, I added a heuristic computation as to why we expect that answer. – Sangchul Lee Jul 5 '19 at 20:46
• Alternatively, apply the Mellin transform to Airy's equation $y'' - x y = 0$ to get $Y(\alpha+3) = \alpha(\alpha+1)Y(\alpha)$ and then figure out the ratio of Gamma functions that satisfies that functional equation. – eyeballfrog Jul 5 '19 at 20:47
• @eyeballfrog, That is indeed a nice trick. Thank you :) – Sangchul Lee Jul 5 '19 at 20:49
• One could also exploit Ramanujan's Master Theorem utilizing the series representation of the Airy Function. – mrtaurho Jul 5 '19 at 21:02

A little note. Another way to derive the first formula from Sangchul Lee's answer.

We use the fact that:

$$\mathrm{Ai}''(x)=x \mathrm{Ai}(x)$$

Let's consider the following integral:

$$I_n(y)=\int_0^\infty \mathrm{Ai}(y x) x^n dx$$

$$I'_n(y)=\int_0^\infty \mathrm{Ai}'(y x) x^{n+1} dx$$

$$I''_n(y)=\int_0^\infty \mathrm{Ai}(y x) x^{n+3} dx=I_{n+3}(y)$$

Now we can rewrite it:

$$I_n(y)= \frac{1}{y^{n+1}}\int_0^\infty \mathrm{Ai}(x) x^n dx=\frac{I_n(1)}{y^{n+1}}$$

$$I''_n(y)= \frac{(n+1)(n+2) I_n(1)}{y^{n+3}}=I_{n+3}(y)$$

So we have obtained an explicit recurrence relation for $$y=1$$:

$$I_{n+3}=(n+1)(n+2) I_n \tag{1}$$

We know that:

$$I_0=\frac{1}{3}$$ (a well known integral)

$$I_1=-\mathrm{Ai}'(0)=\frac{1}{3^{1/3} \Gamma(1/3)}$$

$$I_2=\int_0^\infty \mathrm{Ai}(x) x^2 dx= \int_0^\infty \mathrm{Ai}''(x) x dx=\mathrm{Ai}'(x) x \big|_0^\infty-\int_0^\infty \mathrm{Ai}'(x) dx=\mathrm{Ai}(0)=\frac{1}{3^{2/3} \Gamma(2/3)}$$

So, we now know the general recurrence relation for $$I_n$$ and three initial conditions needed to find any term.

After we derive the general expression for $$I_n$$, it's easy enough to use continuity of all the functions involved to generalize for non-integer parameters and obtain the desired Mellin transform.

• That's neat! (+1) – mrtaurho Jul 6 '19 at 12:03

Using the various provided answers, I present to you the craziest integral I've ever seen. Behold:

$$\int_0^\infty \mathrm{Ai}(x)\ln^3x\ dx=-\frac1{81}\left[(4\pi^2+\beta^2)\beta+52\zeta(3)\right]$$ where $$\beta=2\gamma+\ln3$$.

• "the craziest integral you have ever seen" Come to Pau and visit me : I have a zoo of monsters ! Cheers :-) – Claude Leibovici Jul 6 '19 at 5:05