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I need your help with this integral: $$\mathcal{K}(p)=\int_0^\infty\frac{x^p}{\operatorname{Ai}^2 x + \operatorname{Bi}^2 x}\mathrm dx,$$ where $\operatorname{Ai}$, $\operatorname{Bi}$ are Airy functions: $$\operatorname{Ai}\,x=\frac{1}{\pi}\int_0^\infty\cos\left(x\,z+\frac{z^3}{3}\right)\,\mathrm dz,$$ $$\operatorname{Bi}\,x=\frac{1}{\pi}\int_0^\infty\left(\sin\left(x\,z+\frac{z^3}{3}\right)+\exp\left(x\,z-\frac{z^3}{3}\right)\right)\,\mathrm dz.$$ 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: $$\mathcal{K}(3)\stackrel?=\frac{5\,\pi^2}{32},\ \ \mathcal{K}(6)\stackrel?=\frac{565\,\pi^2}{512}.$$

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    $\begingroup$ Your conjectures hold with very high precision. Next several values of the form $\mathcal{K}(3\,n),\, n\in\mathbb{N}$ also seem to be rational multiples of $\pi^2$, where denominators are powers of $2$, but numerators show no obvious pattern. But I have no idea how to find a general form either. $\endgroup$ – Liu Jin Tsai Sep 30 '13 at 17:20
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    $\begingroup$ Interestingly, the graph of $Ai^2(x)+Bi^2(x) is rid of any oscillation (wolframalpha.com/input/?i=Ai%28x%29%5E2%2BBi%28x%29%5E2) $\endgroup$ – AstroSharp Oct 2 '13 at 15:48
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    $\begingroup$ @LailaPodlesny for $x>0$, yes: $\frac{1}{2}\frac{d^2}{dx^2}((A^2+B^2))=(A')^2+(B')^2+(A^2+B^2)x >0$ for $x>0$ since $A$ and $B$ solve $y''(x)=xy(x)$. $\endgroup$ – AstroSharp Oct 3 '13 at 22:13
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    $\begingroup$ A starting point may be the observation that $Ai^2,Bi^2$ and $Ai^2+Bi^2$ satisfy the DE $f'''=2f+4xf'$. Does integration by parts lead somewhere? $\endgroup$ – Jack D'Aurizio Oct 6 '13 at 17:05
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    $\begingroup$ The numerators of those rational numbers appear to be the even-indexed numbers in the sequence A226260. $\endgroup$ – Kirill Oct 7 '13 at 16:31
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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.

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  • $\begingroup$ Excellent. Can you comment on how the numbers in the expansion $\theta$ in my post is related to the residues? Is my value for M(10) correct? $\endgroup$ – Bennett Gardiner Oct 12 '13 at 1:37
  • $\begingroup$ Oh, I meant to write down $M(10)$ but maybe divided by too many 2s. $\endgroup$ – Bennett Gardiner Oct 12 '13 at 2:16
  • $\begingroup$ My first question, restated, is - can your theory account for whether or not the values 1105, 82825, ... eg. the "other" numerators from the mass formula sequence, occur in for $\mathcal{K}(3n+1)$ or $\mathcal{K}(3n+2)$? $\endgroup$ – Bennett Gardiner Oct 12 '13 at 2:22
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    $\begingroup$ @BennettGardiner No, it gives directly the values. You must have missed some brackets. Try this: M[n_] := -Pi^2 SeriesCoefficient[ Series[(3 s/4)^(2 n) D[-Log[s]/6 + Log[Sum[(-1)^k Pochhammer[1/6, k] Pochhammer[5/6, k]/(k! s^k), {k, 0, 2 n}]], s], {s, Infinity, 1}], 1] $\endgroup$ – Start wearing purple Oct 12 '13 at 14:11
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    $\begingroup$ @O.L. Yes, it feels very similar. Indeed, one can ask many similar questions, involving a monotonic sum of squares of a pair of oscillating functions (e.g. Bessel $J_\nu(x)$, $Y_\nu(x)$, Fresnel $S(x)$, $C(x)$, trig integrals $\operatorname{Si}(x)$, $\operatorname{Ci}(x)$, Airy prime functions $\operatorname{Ai}'(x)$, $\operatorname{Bi}'(x)$ etc). It would be nice to find a general approach to all these problems. $\endgroup$ – Vladimir Reshetnikov Oct 13 '13 at 18:46
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I can give a partial answer. Using DLMF, Chapter 9 "Airy and Related Functions", Airy Functions:

we can see that $$\frac{d}{dx}\arctan\frac{\operatorname{Ai}x}{\operatorname{Bi}x}=-\frac1\pi\frac1{\operatorname{Ai}^2x+\operatorname{Bi}^2x}$$ and $$\int_0^\infty\frac{dx}{\operatorname{Ai}^2x+\operatorname{Bi}^2x}=\frac{\pi^2}6.$$

This settles the question for the particular case $p=0$.

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    $\begingroup$ nice work. $\pi^2/6$ shows up again! $\endgroup$ – Bennett Gardiner Oct 7 '13 at 7:34
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Here is an explicit formula based on O.L.'s answer: $$\mathcal{K}(3n)=\frac{\pi^2}{6\cdot64^n}a_{2n},$$ where $a_n$ is the sequence defined recursively as follows: $$a_0=1,\ \ a_{n+1}=(6\,n+4)\,a_n+\sum\limits_{i=0}^n a_i\,a_{n-i}$$

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    $\begingroup$ It appears that this formula works for half-integer values of $n$ too, e.g. for $n=\frac12$, $\mathcal{K}\left(\frac32\right)\stackrel?=\frac{5\,\pi^2}{48}$. $\endgroup$ – Vladimir Reshetnikov Oct 13 '13 at 21:32
  • $\begingroup$ Did you ever get a proof of this conjecture? Seems like a worthy question. $\endgroup$ – Brevan Ellefsen Jan 13 '17 at 15:57
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This might have to be a team effort guys. I will give a partial answer but hopefully someone will take the ball and run with it. Please note that the following post contains large swathes of experimental and unproven mathematics. It is also a sort of stream of consciousness of my solution attempt. If that offends your sensibilities, please look away!

Let's use @TauMu's excellent answer for the case where $p=0$ to our advantage. Multiply both sides of his identity by $x^p$, $$x^p\frac{d}{dx}\arctan\frac{\operatorname{Ai}x}{\operatorname{Bi}x}=-\frac1\pi\frac{x^p}{\operatorname{Ai}^2x+\operatorname{Bi}^2x},$$ then I believe integrating by parts gives $$ \mathcal{K}(p) = \pi p \int^{\infty}_0 \! x^{p-1} \theta(x) \ \mathrm{d}x, \qquad p\ge 1, $$ where $$ \theta(x) = \arctan{\frac{\operatorname{Ai} x}{\operatorname {Bi}x}}. $$ I am hoping someone can evaluate THIS integral instead for any value of $p$.

However, for now let's look at what @TauMu's reference gives as the asymptotic expansion of $\theta(x)$ as $x \rightarrow -\infty$: $$ \theta\left(x\right)\sim\frac{\pi}{4}+\frac{2}{3}(-x)^{{% 3/2}}\left(1+\frac{5}{32}\frac{1}{x^{3}}+\frac{1105}{6144}\frac{1}{x^{6}}+% \frac{82825}{65536}\frac{1}{x^{9}}+\frac{1282031525}{58720256}\frac{1}{x^{% {12}}}+\cdots\right), $$ Between looking at the numerators of the bracketed terms and browsing the OEIS link given by @Kirill (http://oeis.org/A226260) I noticed that the conjecture for $\mathcal{K}(3)$ and $\mathcal{K}(6)$ fit this sequence not just in numerator but with a slightly modified denominator (Edit - I just saw that @Liu Jin Tsai mentioned this).

Edit - note that all the terms other than the first term do not feature in the following work for determining values for $\mathcal{K}(3n)$, but still appear (modified by a factor of $3/2^k$) in the given sequence $M$. Therefore I simply MUST believe they have something to do either with the $3n+1$ or $3n+2$ case - simply not along with the factor $\pi^2$.

Let $M$ be the sequence given by the "mass formula[e] for connected vacuum graphs on 2$n$ nodes for a $\phi^3$ field theory", whatever the hell that means! The sequence is as follows, $$ M = \left\{{5\over24}, {5\over16}, {1105\over1152}, {565\over128}, {82825\over3072}, {19675\over96}, {1282031525\over688128}, {80727925\over4096},\cdots\right\} $$ Then our conjecture as a group seems to be that $$ \mathcal{K}(3n) = 2^{-k}\pi^2M(2n), \qquad n \ge 1 $$ where $k$ depends on $n$ in some way. When $n=3$, $k=3$ didn't seem to work, so it isn't as simple as $k=n$. I tried $$ \mathcal{K}(9) = 2^{-4}\pi^2M(6) = \frac{19675}{16 \times 96}\pi^2 = \frac{19675}{768}\pi^2 $$ I spent a long time trying to validate this with Mathematica. I thought my conjecture must've been wrong so I moved on in desperation to $n=4$, for which I found $$ \mathcal{K}(12) = 2^{-5}\pi^2M(8) = {80727925\over32\times4096}\pi^2 = {80727925\over131072}\pi^2 $$ which agrees with the evaluation of the integral to 100 digits. I figured out that there is a missing factor of $3/2$ in the term $M(6)$! The corrected value is $$ M(6) = {19675\over64}. $$ This makes more sense, as 64 is a power of 2, like the rest of the denominators of the even terms in $M$.

I suspect that $k$ skips multiples of 3. So $k=1,2,4,5,7,8,10,.\ldots$ I have not thought of any way to show this conclusively however. Note that the OEIS sequence for the mass formulae actually has only been calculated up to 8 terms. The original paper is here - http://arxiv.org/pdf/hep-th/9304052v1.pdf. Equation 15 is the relevant sequence.

However, if the theory for the values of $k$ is correct, we can now achieve something really, really cool. I have NO idea what $\phi^3$ theory is, but through our conjecture and the power of Mathematica we can guess at the even terms of this sequence by trying to convert our decimal approximations for $\mathcal{K}(3n)$ into fractions!

For example, by finding the value of $\mathcal{K}(15)$ I suspect that $$ M(10) ={66049225625\over2097152}, $$ or something like that. Note that $2097152=2^{12}$. If we say the sequence $M_{2}$ is the even terms of $M$, rewritten slightly, with the corrected value of $M(6)$ (red), the conjectured value of $M(10)$ (blue) and factors of 2 expressed explicitly, $$ M_2 = \left\{{5\over2^4}, {565\over2^7}, \color{red}{19675\over2^6}, {80727925\over2^{12}},\color{blue}{66049225625\over2^{21}}\cdots\right\} $$ I have no idea what sort of pattern the powers on the denominators follow.

I realise that conjectures based off one piece of evidence (that $k=3$ was skipped) and supposing an error in the original paper is a very tenuous prospect. It's not even worthy of the name conjecture. But it's just too pretty to not suggest.

If anyone would like to extend to two or four more terms the sequence given in Eqn 15 of this paper I will be forever gratefully. My curiosity has been well piqued by this problem and its rather obscure and beautiful connection to this problem of connected vacuum graphs. It's the unexpected connections between branches of maths that make it so beautiful. The question remains - what is the value of the integral when $p$ is not a multiple of 3, is it a rational multiple of $\pi^2$ and does it have anything to do with $\phi$ theory?

Edit - I'm sorry to continue gushing words, but I just realised that Carl Bender coauthored the paper twenty years ago, he wrote my favourite text (http://www.amazon.com/Advanced-Mathematical-Methods-Scientists-Engineers/dp/0387989315) - I wish I could tell him one of his numbers was wrong, even though I don't understand the work! :)

Second Edit - I reread what I wrote and what everyone else had already written and now I feel silly, it seems everyone knew this already. I'm hoping someone at least enjoyed seeing it written out in order.

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