# 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 $\operatorname{Ai}$, $\operatorname{Bi}$ are Airy functions:

I am not sure that $\mathcal{K}(p)$ has a general closed form, but I hope so, because approximate numerical calculations suggest these conjectured values:

Computing the Wronskian of these two solutions of the Airy equation, one can check that

This gives the integral $\mathcal{K}(0)$ as

To compute the integral $\mathcal{K}(3n)$, we will need to develop a more sophisticated approach. First note that (see here)

Therefore

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

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 $|\arg z|<\pi$:

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:

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

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

This yields, for instance,

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.