I need your help with this integral:

K(p)=∫∞0xpAi2x+Bi2xdx,

where Ai, Bi are Airy functions:

Aix=1π∫∞0cos(xz+z33)dz,

Bix=1π∫∞0(sin(xz+z33)+exp(xz−z33))dz.

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.

**Answer**

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

\frac{1}{\operatorname{Ai}^2(x)+\operatorname{Bi}^2(x)}=\frac{\pi}{2i}\left[\frac{g_+'(x)}{g_+(x)}-\frac{g’_-(x)}{g_-(x)}\right]

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).

Therefore

\begin{align}\mathcal{K}(3n)&=\frac{\pi}{2i}\int_0^{\infty}x^{3n}\left[\frac{g_+'(x)}{g_+(x)}-\frac{g’_-(x)}{g_-(x)}\right]dx=\\

&=\frac{\pi}{2i}\lim_{R\rightarrow\infty}\int_{S_R}z^{3n}\frac{\operatorname{Ai}'(z)}{\operatorname{Ai}(z)}dz,

\end{align}

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:

\begin{align}

\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}

\end{align}

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:

\begin{align}

\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}

\end{align}

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.

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