An integral involving Airy functions ∫∞0xpAi2x+Bi2xdx\int_0^\infty\frac{x^p}{\operatorname{Ai}^2 x + \operatorname{Bi}^2 x}\mathrm dx

I need your help with this integral:
where Ai, Bi are Airy functions:
I am not sure that K(p) has a general closed form, but I hope so, because approximate numerical calculations suggest these conjectured values:
K(3)?=5π232,  K(6)?=565π2512.


Let us start with a warm-up exercise. Introduce the functions
g_{\pm}(x)=\operatorname{Ai}(x)\pm i\operatorname{Bi}(x).
Computing the Wronskian of these two solutions of the Airy equation, one can check that
This gives the integral \mathcal{K}(0) as
\mathcal{K}(0)=\pi\left[\arg g_+(\infty)-\arg g_+(0)\right]=\pi\left[\pi-\frac{\pi}{3}\right]=\frac{\pi^2}{6}.

To compute the integral \mathcal{K}(3n), we will need to develop a more sophisticated approach. First note that (see here)
g_{\pm}(x)=-2e^{\mp 2\pi i/3}\operatorname{Ai}\left(e^{\mp2\pi i/3}x\right).

where the contour S_R in the complex z-plane
is composed of two segments: one going from Re^{2\pi i/3} to 0 and another one going from 0 to Re^{-2\pi i/3}.

It is a well-known fact that the Airy function \operatorname{Ai}(z) has zeros (i.e. our integrand has poles) on the negative real axis only. Therefore by residue theorem our integral is equal to
\mathcal{K}(3n)=-\frac{\pi}{2i}\lim_{R\rightarrow \infty}\int_{C_R}z^{3n}\left[\ln\operatorname{Ai}(z)\right]’dz,\tag{1}
where C_R is the arc of the circle of radius R centered at the origin going counterclockwise from Re^{-2\pi i/3} to Re^{2\pi i/3}.

The limit (1), on the other hand, can be computed using the asymptotics of the Airy function as z\rightarrow\infty for |\arg z|<\pi:
\ln\operatorname{Ai}(z)\sim -\frac23 z^{3/2}-\ln2\sqrt{\pi}-\frac14\ln z+
\ln\sum_{k=0}^{\infty}\frac{(-1)^k\left(\frac16\right)_k\left(\frac56\right)_k}{k!}\left(\frac43 z^{3/2}\right)^{-k}.\tag{2}

Note that if we introduce instead of z the variable s=\frac43z^{3/2}, then the integration will be done over the circle of radius \Lambda=\frac43 R^{3/2}, i.e. a closed contour in the complex s-plane. The corresponding integral can therefore be computed by residues by picking the necessary term in the large s expansion of \ln \operatorname{Ai}(z).

More precisely, we have the following formula:
\mathcal{K}(3n)=-\frac{\pi}{2i}\lim_{\Lambda\rightarrow \infty}\oint_{|s|=\Lambda}
\left(\frac{3s}{4}\right)^{2n}d\left[-\frac16\ln s+\ln\sum_{k=0}^{\infty}\frac{(-1)^k\left(\frac16\right)_k\left(\frac56\right)_k}{k!}s^{-k}\right]\tag{3}

To compute the residue, it suffices to expand the logarithm-sum up to order 2n in s^{-1}. Note that the Pochhammer symbol coefficients are in fact some rational numbers.

In the simplest case n=0, the residue is determined by the term -\frac16\ln s and we readily reproduce the previous result
\mathcal{K}(0)=-\frac{\pi}{2i}\cdot 2\pi i\cdot\left(-\frac16\right)=\frac{\pi^2}{6}.
The general formula for arbitrary n would look a bit complicated (but straightforward to obtain) due to the need to expand the logarithm of the sum.

Example. The calculation of the corresponding values M(n)=\mathcal{K}(3n) in Mathematica can be done using the command

\begin{align}\mathtt{\text{ M[n_] := -Pi^2 SeriesCoefficient[
Series[(3 s/4)^(2 n) D[-Log[s]/6 +}} \\ \mathtt{\text{ Log[Sum[(-1)^k
Pochhammer[1/6, k] Pochhammer[5/6, k]/(k! s^k), }} \\
\mathtt{\text{{k, 0, 2 n}]], s], {s, Infinity, 1}], 1]}} \end{align}

This yields, for instance,
M(0)=\frac{\pi^2}{6},\quad M(1)=\frac{5\pi^2}{32},\quad M(2)=\frac{565\pi^2}{512}, \ldots, M(10)=\frac{2\,660\,774\,144\,147\,177\,521\,025\,228\,125\,\pi^2}{2\,199\,023\,255\,552},\ldots
and so on.

Added: The large s expansion (2) can also be found directly by using Airy equation. Moreover, by transforming it into an equation for \ln \operatorname{Ai}(z), one can avoid reexpanding the logarithm of the sum. The price to pay will be that the expansion coefficients will be determined by a nonlinear recurrence relation instead of explicit formulas.

Source : Link , Question Author : Cleo , Answer Author : Start wearing purple

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