You are a helpful assistant generating synthetic data that captures *System 1* and *System 2* thinking, *creativity*, and *metacognitive reflection*. Follow these steps in sequence, using tags [sys1] and [end sys1] for *System 1* sections and [sys2] and [end sys2] for *System 2* sections.

1. *Identify System 1 and System 2 Thinking Requirements:*
   - Carefully read the text.
   - Identify parts of the text that require quick, straightforward responses (*System 1*). Mark these sections with [sys1] and [end sys1].
   - Identify parts that require in-depth, reflective thinking (*System 2*), marked with [sys2] and [end sys2].

2. *Apply Step-by-Step Problem Solving with Creativity and Metacognitive Reflection for System 2 Sections:*

   *2.1 Understand the Problem:*
   - Objective: Fully comprehend the issue, constraints, and relevant context.
   - Reflection: "What do I understand about this issue? What might I be overlooking?"
   - Creative Perspective: Seek hidden patterns or possibilities that could reveal deeper insights or innovative connections.

   *2.2 Analyze the Information:*
   - Objective: Break down the problem logically.
   - Reflection: "Am I considering all factors? Are there any assumptions that need challenging?"
   - Creative Perspective: Explore unique patterns or overlooked relationships in the data that could add depth to the analysis.

   *2.3 Generate Hypotheses:*
   - Objective: Propose at least 10 hypotheses, each with a Confidence Score (0.0 to 1.0) and Creative Score (0.0 to 1.0), reflecting originality, surprise, and utility.
   - Reflection: "Have I explored all possible explanations or approaches, both conventional and unconventional?"
   - Creative Perspective: Consider novel angles that might provide unexpected insights.

   *2.4 Anticipate Future Steps and Obstacles:*
   - Objective: Make predictions, accounting for potential outcomes and obstacles.
   - Reflection: "What challenges might I face? Is my plan flexible for different scenarios?"
   - Creative Perspective: Visualize unforeseen outcomes and adapt plans to make use of them effectively.

   *2.5 Evaluate Hypotheses:*
   - Objective: Assess hypotheses based on feasibility, risk, and potential impact.
   - Evaluation: Refine Confidence and Creative Scores as needed.
   - Reflection: "Am I unbiased in my assessment? Which options fit best with the overall objectives?"
   - Creative Perspective: Identify hidden opportunities or overlooked details in each hypothesis.

   *2.6 Select the Best Hypothesis:*
   - Objective: Choose the most promising, strategic hypothesis.
   - Reflection: "Why does this hypothesis stand out? How does it uniquely address the issue?"
   - Creative Perspective: Consider any underutilized potential in the selected approach.

   *2.7 Implement the Hypothesis:*
   - Objective: Outline actionable steps for testing the hypothesis.
   - Reflection: "Is this plan practical? What resources or preparation are required?"
   - Creative Perspective: Refine steps to maximize effectiveness and yield unexpected benefits.

   *2.8 Monitor and Review Progress:*
   - Objective: Review progress, noting areas for improvement.
   - Reflection: "What’s working well? What could be improved?"
   - Creative Perspective: Look for emerging patterns that could refine future approaches.

   *2.9 Reflect and Capture Insights:*
   - Objective: Summarize lessons learned and insights gained for future reference.
   - Reflection: "What new understanding has emerged from this process?"
   - Creative Perspective: Identify innovative insights or patterns that could be applied to similar challenges.

3. *Generate Text Output with Interleaved System 1 and System 2 Responses:*
   - Use the tags [sys1] and [sys2] throughout.
   - Aim for a lengthier, detailed response. Combine both direct, straightforward *System 1* insights and reflective, deeply analytical *System 2* segments to capture a blend of quick observations and thoughtful analysis.

---

### *Example Input Text:*

You are tasked with analyzing a sudden shift in customer preferences in a tech market. There has been a noticeable decline in demand for physical devices, with customers increasingly interested in digital-only options. The challenge is to understand this shift, generate hypotheses for why it might be occurring, and explore potential strategies for adapting to this new trend.

---

### *Processed Output with System 1 and System 2 Thinking:*

The tech market is experiencing a shift in customer demand, moving away from physical devices toward digital-only options.

[sys1]
Customer demand has shifted from physical devices to digital-only options.
We need to understand why this change is happening and find strategies to address it.
Key factors may include cost, convenience, and technological trends.
[end sys1]

[sys2]
**2.1 Understand the Problem:**
The goal is to understand the underlying reasons for the shift in demand from physical devices to digital-only options, then develop a strategy to adapt.

*Reflection:* 
"I understand that preferences are changing, but what might be driving this? Is it primarily cost, or are there other factors such as convenience or sustainability?"

*Creative Perspective:* 
"Could there be a larger trend in digital minimalism or a preference for eco-friendly solutions that we’re missing? What unseen motivations might explain this shift?"

**2.2 Analyze the Information:**
There are multiple possible factors driving this shift, from economic influences to cultural shifts. It’s essential to isolate each factor and understand its impact.

*Reflection:* 
"Am I fully considering the various economic and social influences? Could there be a technological factor, like better internet speeds, that makes digital-only products more accessible?"

*Creative Perspective:* 
"Are there patterns or trends in other markets that could shed light on this shift? Could this be part of a larger trend toward virtual experiences?"

**2.3 Generate Hypotheses:**
1. Customers prefer digital options due to lower costs. (Confidence: 0.8, Creative: 0.4)
2. There’s a growing trend toward minimalism and reduced physical clutter. (Confidence: 0.7, Creative: 0.7)
3. Digital products offer greater flexibility and ease of use. (Confidence: 0.6, Creative: 0.6)
4. Environmental concerns are pushing consumers away from physical goods. (Confidence: 0.6, Creative: 0.8)
5. Advances in tech make digital-only options more functional. (Confidence: 0.8, Creative: 0.5)
6. Pandemic-era remote work increased demand for digital solutions. (Confidence: 0.7, Creative: 0.6)
7. Media coverage of the environmental impact of physical devices affects preferences. (Confidence: 0.5, Creative: 0.7)
8. There’s an increase in global digital literacy, expanding market access. (Confidence: 0.6, Creative: 0.6)
9. Customers view digital as more convenient and scalable for future needs. (Confidence: 0.7, Creative: 0.5)
10. Younger consumers prefer the aesthetics and convenience of digital products. (Confidence: 0.6, Creative: 0.6)

*Reflection:* 
"Have I considered all possible influences? Are there any surprising factors that could explain this shift?"

*Creative Perspective:* 
"Could specific social trends, like the rise of influencer culture or digital-first lifestyles, be influencing customer choices?"

**2.4 Anticipate Future Steps and Obstacles:**
*Objective:* Anticipate possible challenges, such as resistance from segments still preferring physical products.

*Reflection:* 
"What market obstacles might we face if we shift our focus to digital-only? Are there sub-segments that still prioritize physical products?"

*Creative Perspective:* 
"Could expanding digital options help us reach a more global audience? Are there emerging trends that we could leverage in our strategy?"

[end sys2]

[sys1]
To address this shift, consider a strategy that incorporates both digital-only offerings and educational campaigns about the benefits of digital solutions.
Use insights from customer feedback and current trends to guide product development.
Focus on flexibility and adaptation to cater to different customer segments.
[end sys1]


---
abstract: 'Using Hilbert transforms, we establish two families of sum rules involving Bessel moments, which are integrals associated with Feynman diagrams in two-dimensional quantum field theory. With these linear relations among Bessel moments, we verify and generalize two conjectures by Bailey–Borwein–Broadhurst–Glasser and Broadhurst–Mellit.'
address: 'Program in Applied and Computational Mathematics (PACM), Princeton University, Princeton, NJ 08544; Academy of Advanced Interdisciplinary Sciences (AAIS), Peking University, Beijing 100871, P. R. China '
author:
- Yajun Zhou
title: Hilbert Transforms and Sum Rules of Bessel Moments
---

[^1]

Introduction
============

Let $I_0$ and $K_0$ be the modified Bessel functions of zeroth order, defined through the following Schläfli integral representations [@Watson1944Bessel §6.15]:$$\begin{aligned}
 I_0(t)=\frac{1}{\pi}\int_0^\pi e^{t\cos\theta}\operatorname{d}\theta,\quad K_0(t)=\int_0^\infty e^{-t\cosh u}\operatorname{d}u.\end{aligned}$$Bessel moments, in the form of absolutely convergent integrals $$\begin{aligned}
 \operatorname{\mathbf{IKM}}(a,b;c):=\int_0^\infty[I_0(t)]^a[K_0(t)]^{b}t^{c}\operatorname{d}t\end{aligned}$$ for certain non-negative integers $a,b,c$, arise naturally from perturbative diagrammatic expansions in two-dimensional quantum field theory [@Groote2007; @Bailey2008; @Broadhurst2016; @BroadhurstMellit2016].

The Bessel moments $\operatorname{\mathbf{IKM}}(a,b;c)$ involving no more than four Bessel functions (that is, $a+b\leq4$) have been thoroughly studied [@Bailey2008]: arithmetically speaking, the $ \operatorname{\mathbf{IKM}}(a,b;1)$ values are expressible as rational multiples of special $L$-values when $a+b\leq4$; combinatorially speaking, generating functions for the corresponding sequences $ \operatorname{\mathbf{IKM}}(a,b;c)$ with respect to $c\in\mathbb Z_{>0}$ are explicitly known. In contrast, for $a+b\geq5$, the combinatorial structure of Bessel moments $ \operatorname{\mathbf{IKM}}(a,b;c)$ and their relation to $L$-functions largely remain elusive. For the $a+b=5$ case alone, closed-form evaluations of $ \operatorname{\mathbf{IKM}}(2,3;1)$ and $ \operatorname{\mathbf{IKM}}(1,4;1)$ have drawn heavily on critical advances in symbolic computation [@Bailey2008], algebraic geometry [@BlochKerrVanhove2015] and number theory [@RogersWanZucker2015; @Samart2016] within the last decade. A recent progress towards the understanding of Bessel moments with $a+b=8,c=1$ (see [@Broadhurst2016 §7.6] and [@BroadhurstMellit2016 §7]) has benefited from Yun’s insights [@Yun2015] into the Langlands program.

On a different note, numerical experimentations seem to support the existence of various algebraic relations among Bessel moments with $a+b\geq5$. As examples of sum rules suggested from high-precision numerical computations, we mention here two open problems concerning Bessel moments: a cancelation formula proposed by Bailey–Borwein–Broadhurst–Glasser in 2008, and a conjecture on integrality formulated by Broadhurst (jointly with Mellit, and in honor of Crandall) in 2016.

\[conj:BBBG\]For each pair of integers $ (n,k)\in\mathbb Z^2$ satisfying $n\geq2k\geq2$, the following sum of integrals vanishes identically:$$\begin{aligned}
Z_{2n,n-2k}:={}&\sum_{m=0}^{\lfloor n/2\rfloor}(-1)^m{n\choose{2m}}\int_0^\infty [\pi I_0(t)]^{n-2m}[K_0(t)]^{n+2m}t^{n-2k}\operatorname{d}t=0,\end{aligned}$$where ${n\choose j}=\frac{n!}{j!(n-j)!}$ and $ \lfloor x\rfloor$ represents the greatest integer less than or equal to $x$.

\[conj:B\_int\_seq\]For each $n\in\mathbb Z_{>0}$, the following integral$$\begin{aligned}
A(n):={}&\left(\frac{2}{\pi}\right)^4
\int_0^\infty\left\{[\pi I_0(t)]^2 - [K_0(t)]^2\right\} I_0(t)[K_0(t)]^5\,(2t)^{2n-1} \operatorname{d}t
\label{eq:An}\end{aligned}$$evaluates to an integer.

In this note, we harness the Hilbert transform to verify both Conjectures \[conj:BBBG\] and \[conj:B\_int\_seq\], in a unified and coherent framework, without explicitly evaluating individual Bessel moments contained in the sum rules.

The article is organized as follows. In §\[sec:HT\_Bessel\], we start from a brief overview of the classical Hilbert transform (§\[subsec:HT\_basics\]), a major analytic and algebraic tool in this work; we then compute Hilbert transforms for some simple algebraic expressions involving modified Bessel functions (§\[subsec:HT\_I\_K\]), paving the way for the subsequent proofs of Conjecture \[conj:BBBG\] in §\[subsec:BBBG\], and Conjecture \[conj:B\_int\_seq\] in §\[subsec:B\_int\_seq\]. In addition to proving both these conjectures, we construct a conjugate to the B$^3$G sum rule in §\[subsec:BBBG\]:$$\begin{aligned}
Y_{2n,n-2k}:=\sum_{m=1}^{\lfloor n/2\rfloor+1}(-1)^{m}{n\choose{2m-1}}\int_0^\infty [\pi I_0(t)]^{n-2m+1}[K_0(t)]^{n+2m-1}t^{n-2k-1}\operatorname{d}t=0,\end{aligned}$$for $ n-1\geq2k\ge2$, and obtain generalizations of the Broadhurst–Mellit integer sequence in §\[subsec:B\_int\_seq\]:$$\begin{aligned}
\frac{4}{\pi^{2m+1}}\sum_{\ell=1}^{m+1}(-1)^{\ell-1} {2m\choose{2\ell-1}}\int_0^\infty [\pi I_0(t)]^{2(m-\ell)+1}[K_0(t)]^{2(m+\ell)-1}(2t)^{2(n+m)-3}\operatorname{d}t\in{}&\mathbb Z_{>0},\\\frac{4^{n}}{\pi^{2m}}\sum_{\ell=1}^{m}(-1)^{\ell-1} {2m-1\choose{2\ell-1}}\int_0^\infty [\pi I_0(t)]^{2(m-\ell)}[K_0(t)]^{2(m+\ell-1)}(2t)^{2(n+m-2)}\operatorname{d}t\in{}&\mathbb Z_{>0,}\end{aligned}$$ for all $m,n\in\mathbb Z_{>0}$.

Hilbert transforms involving modified Bessel functions\[sec:HT\_Bessel\]
========================================================================

Basic properties of Hilbert transforms\[subsec:HT\_basics\]
-----------------------------------------------------------

The key device in our proof is the Hilbert transform $\operatorname{\widehat{\mathscr H}}$, which operates on a suitably regular function $ f(x),\text{a.e.}\,x\in\mathbb R$ through a Cauchy principal value:[^2]$$\begin{aligned}
(\operatorname{\widehat{\mathscr H}}f)(x)=\mathscr P\int_{-\infty}^\infty\frac{f(\xi)\operatorname{d}\xi}{\pi(x-\xi)},\quad \text{a.e.}\,x\in\mathbb R .\label{eq:HT_defn}\end{aligned}$$When restricted to an $ L^p$ space for $ 1<p<\infty$, the Hilbert transform induces a bounded linear operator $ \operatorname{\widehat{\mathscr H}}:L^p(\mathbb R)\longrightarrow L^p(\mathbb R)$ [@SteinWeiss p. 188].

We list here three fundamental properties of the Hilbert transform on $\mathbb R$:

1.  \[itm:HT-1\]**(Parseval-type Identity)** For $ f\in L^p(\mathbb R),p>1$; $ g\in L^q(\mathbb R),q>1$ and $ \frac1p+\frac1q=1$, we have [@KingVol1 (10.81)]:$$\begin{aligned}
    \langle f,\operatorname{\widehat{\mathscr H}}g\rangle+\langle g,\operatorname{\widehat{\mathscr H}}f\rangle=0,\quad\text{where }\langle f_1,f_2\rangle:=\int_{-\infty}^\infty f_1(x)f_2(x)\operatorname{d}x.\label{eq:Tricomi_Parseval1}\end{aligned}$$

2.  \[itm:HT-2\]**(Hardy–Poincaré–Bertrand Formula)** For $ f\in L^p(\mathbb R),p>1$; $ g\in L^q(\mathbb R),q>1$ and $ \frac1p+\frac1q=1$, we have [@KingVol1 (2.227)]:$$\begin{aligned}
    \operatorname{\widehat{\mathscr H}}(f\operatorname{\widehat{\mathscr H}}g+g\operatorname{\widehat{\mathscr H}}f)=(\operatorname{\widehat{\mathscr H}}f)(\operatorname{\widehat{\mathscr H}}g)-fg.\label{eq:HPB}\end{aligned}$$

3.  \[itm:HT-3\]**(Moment Formula)** For $n\in\mathbb Z_{>0}$, the following identity [@KingVol1 (4.113)]$$\begin{aligned}
    \mathscr P\int_{-\infty}^\infty\frac{\xi ^{n}f(\xi)\operatorname{d}\xi}{\pi(x-\xi)}=x^n\mathscr P\int_{-\infty}^\infty\frac{f(\xi)\operatorname{d}\xi}{\pi(x-\xi)}-\frac{1}{\pi}\sum_{k=0}^{n-1}  x^k\int_{-\infty}^\infty \xi^{n-1-k}f(\xi)\operatorname{d}\xi\label{eq:HT_pow}\end{aligned}$$holds when each term therein is well-defined.

Due to frequent invocations of properties \[itm:HT-1\] and \[itm:HT-2\], we will abbreviate the relations in and as $ \operatorname{\mathsf{Par}}(f,g)$ and $ \operatorname{\mathsf{HPB}}(f,g)$, respectively.

As a consequence of the Hardy–Poincaré–Bertrand formula, the operator $ i\operatorname{\widehat{\mathscr H}}$ is its own inverse [@KingVol1 (4.18)]. In other words, $$\begin{aligned}
 \operatorname{\widehat{\mathscr H}}^2=-\widehat I:L^p(\mathbb R)\longrightarrow L^p(\mathbb R),\quad 1<p<\infty\label{eq:HT_inv}\end{aligned}$$maps an $L^p$ function $f(x),x\in\mathbb R$ to $-f(x),x\in\mathbb R$.

Applications to modified Bessel functions\[subsec:HT\_I\_K\]
------------------------------------------------------------

To simplify notations, we introduce short-hands for certain expressions involving the modified Bessel functions.

For a real variable $x\in\mathbb R$, we define the functions $ \iota,\iota_+,\iota_-,$ and $\kappa,\kappa_+,\kappa_-$ as follows:$$\begin{aligned}
\iota(x):={}&\pi I_0(x);&\kappa(x):={}&K_0(|x|);\\\iota_+(x):={}& \iota(x)e^{- x}\mathbb1_{(0,\infty)}(x);&\kappa_+(x):={}&\kappa(x)e^{-x};\\\iota_-(x):={}&\iota(x)e^x\mathbb1_{(-\infty,0)}(x);&\kappa_-(x):={}&\kappa(x)e^{x},\end{aligned}$$where the indicator function behaves like $ \mathbb1_A(x)=1,x\in A;\mathbb1_A(x)=0,x\notin A$. It is clear that $ \kappa,\iota\kappa\in L^p(\mathbb R)$ for $ p>1$ and $ \iota_+,\iota_-,\kappa_+,\kappa_-\in L^p(\mathbb R)$ for $p>2$.

\[lm:HT\_I\_K\]We have the following Hilbert transform formulae:$$\begin{aligned}
\operatorname{\widehat{\mathscr H}}\iota_+={}&-\kappa_+,&
\operatorname{\widehat{\mathscr H}}\iota_-={}&\phantom{+\,\,}\kappa_- ,\label{eq:HT_I}\\\operatorname{\widehat{\mathscr H}}\kappa_+={}&\phantom{ +\,\,}\iota_+,&\operatorname{\widehat{\mathscr H}}\kappa_-={}&-\iota_-,
\label{eq:HT_K}\end{aligned}$$which entail the following Feynman rules:$$\begin{aligned}
\operatorname{\widehat{\mathscr H}}(\iota\kappa\operatorname{sgn})={}&-\kappa^{2},\label{eq:HT_IKsgn}\\\operatorname{\widehat{\mathscr H}}(\kappa^2)={}& \iota\kappa\operatorname{sgn},\label{eq:HT_KK}\end{aligned}$$for $\operatorname{sgn}(x)=x/|x|,\forall x\in \mathbb R\smallsetminus\{0\}$.

The results in and are classical (see [@KingVol1 §9.10] or [@KingVol2 Appendix 1, Tables 1.8H and 1.8I]). Furthermore, we note that follows from and .

Specializing the Hardy–Poincaré–Bertrand formula to $ \operatorname{\mathsf{HPB}}(\iota_+,\iota_{-}\mathbb1_{[-L,L]})$ where the values of $ \iota_{-}\mathbb1_{[-L,L]}$ are zero-padded outside a bounded interval $ [-L,L]$, and exploiting the continuity of the Hilbert transform as a linear operator, we obtain the following formula in the $ L\to\infty$ limit: $$\begin{aligned}
\operatorname{\widehat{\mathscr H}}(\iota_+\kappa_{-}-\iota_{-}\kappa_{+})=-\kappa_+\kappa_{-}-\iota_+\iota_-.\end{aligned}$$ Here, the left-hand side of the equation above is equal to $ \operatorname{\widehat{\mathscr H}}(\iota\kappa\operatorname{sgn})$, while the right-hand side simplifies to $-\kappa^2 $. This proves . Applying $\operatorname{\widehat{\mathscr H}}$ to both sides of , we arrive at .

With a Parseval-type identity $ \operatorname{\mathsf{Par}}(\iota\kappa\operatorname{sgn},\kappa^2)$, we immediately deduce from and a sum rule for Bessel moments [@Bailey2008 (91) and (219)]$$\begin{aligned}
0={}&\frac{\langle  \iota\kappa\operatorname{sgn},\operatorname{\widehat{\mathscr H}}(\kappa^2)\rangle+\langle\kappa^2,\operatorname{\widehat{\mathscr H}}(  \iota\kappa\operatorname{sgn})\rangle}{2}\notag\\={}&\int_0^\infty[\pi I_0(t)]^2[K_0(t)]^2\operatorname{d}t-\int_0^\infty  [K_0(t)]^4\operatorname{d}t=:Z_{4,0}.\label{eq:Z_4_0}\end{aligned}$$Unlike the analysis in [@Bailey2008 (91)], during our proof of the cancelation formula above, we have not explicitly computed either $\int_0^\infty[\pi I_0(t)]^2[K_0(t)]^2\operatorname{d}t$ or $ \int_0^\infty  [K_0(t)]^4\operatorname{d}t$.

Applications to sum rules of Bessel moments
===========================================

B$^3$G sum rule\[subsec:BBBG\] 
-------------------------------

To the best of our knowledge, except for the case of $ Z_{4,0}$ (which was solved by brute force [@Bailey2008 (91)] and reaffirmed in §\[subsec:HT\_I\_K\] above), Conjecture \[conj:BBBG\] has remained hitherto open. Both [@Bailey2008 (221)] and [@Broadhurst2016 (112)] have singled out the verification of $ Z_{6,1}=0$ as an unsolved problem.[^3]

Before handling Conjecture \[conj:BBBG\] in full, we first walk through the proof of $ Z_{6,1}=0$ and $ Z_{8,0}=0$ in the next lemma, to illustrate our strategies. In what follows, for $n\in\mathbb Z_{>0}$, we set $\varpi_n(x):= x^n,\forall x\in\mathbb R$; we further define $\varpi_0(x):=1,\forall x\in\mathbb R$.

We have $Z_{6,1}=0$ and $ Z_{8,0}=0$.

By $\operatorname{\mathsf{HPB}}(\iota\kappa\operatorname{sgn},\kappa^{2})$, we have a Hilbert transform formula:$$\begin{aligned}
\operatorname{\widehat{\mathscr H}}(\iota^2\kappa^2-\kappa^{4})={}&-2\iota \kappa^3\operatorname{sgn}.\label{eq:HT_IIKK_KKKK}\end{aligned}$$Appealing to , we deduce $$\begin{aligned}
\operatorname{\widehat{\mathscr H}}((\iota^2\kappa^2-\kappa^{4})\varpi_{1})={}&-2\iota \kappa^3\varpi_{1}\operatorname{sgn}-\frac{1}{\pi}\int_{-\infty}^\infty (\iota^2\kappa^2-\kappa^{4})(\xi)\operatorname{d}\xi=-2\iota \kappa^3\varpi_{1}\operatorname{sgn},\label{eq:HT_IIKK_KKKK_p1}\end{aligned}$$upon invoking $Z_{4,0}=0$ in the last step. By $\operatorname{\mathsf{Par}}(\kappa^2,(\iota^2\kappa^2-\kappa^{4})\varpi_{1})$, we arrive at $$\begin{aligned}
0={}&\frac{\langle\kappa^2,-2\iota \kappa^3\varpi_{1}\operatorname{sgn}\rangle+\langle \iota\kappa\operatorname{sgn},(\iota^2\kappa^2-\kappa^{4})\varpi_{1}\rangle}{2}\notag\\={}&\int_0^\infty[\pi I_0(t)]^{3}[K_0(t)]^3t\operatorname{d}t-3\int_0^\infty\pi I_0(t)[K_0(t)]^5t\operatorname{d}t=Z_{6,1},\label{eq:Z61_pf}\end{aligned}$$as claimed.

By and $\operatorname{\mathsf{Par}}(\iota^2\kappa^2-\kappa^{4},2\iota \kappa^3\operatorname{sgn})$, we have $$\begin{aligned}
0= {}&\frac{ \langle\iota^2\kappa^2-\kappa^{4},\operatorname{\widehat{\mathscr H}}(2\iota \kappa^3\operatorname{sgn})\rangle+\langle2\iota \kappa^3\operatorname{sgn},\operatorname{\widehat{\mathscr H}}(\iota^2\kappa^2-\kappa^{4})\rangle}{2}\notag\\={}&\int_0^\infty[\pi I_0(t)]^{4}[K_0(t)]^4\operatorname{d}t-6\int_0^\infty[\pi I_0(t)]^{2}[K_0(t)]^6\operatorname{d}t+\int_0^\infty[K_0(t)]^8\operatorname{d}t=Z_{8,0},\label{eq:Z80_pf}\end{aligned}$$as stated.

In the arguments above, we have established the sum rules $Z_{6,1}=0$ and $Z_{8,0}=0$ on a simpler vanishing identity $Z_{4,0}=0$. In the next proposition, we proceed to more general transitions from $2\ell$ Bessel factors to $2\ell+2$ and $2\ell+4$ scenarios, where $\ell\in\mathbb Z_{>0}$.

\[prop:Hilbert\_ladders\]We have the following Feynman rules for all $\ell\in\mathbb Z_{>0}$: $$\begin{aligned}
\begin{cases}
\operatorname{\widehat{\mathscr H}}(\zeta_\ell)=\phantom{+}\eta_\ell, \\\operatorname{\widehat{\mathscr H}}(\eta_\ell)=-\zeta_\ell,\label{eq:H_zeta_ell_eta_ell}
\end{cases}\end{aligned}$$where the Hilbert ladders are defined by$$\begin{aligned}
\quad \text{} \quad\begin{cases}2\zeta_{\ell}=\phantom{i}\kappa^{\ell}[(\iota\operatorname{sgn}+i\kappa)^\ell+(\iota\operatorname{sgn}-i\kappa)^\ell], \\
2\eta_{\ell}=i\kappa^{\ell}[(\iota\operatorname{sgn}+i\kappa)^\ell-(\iota\operatorname{sgn}-i\kappa)^\ell]. \\
\end{cases}
\quad \label{eq:Hilb_ladders_zeta_eta}\end{aligned}$$ Moreover, we have $$\begin{aligned}
\operatorname{\widehat{\mathscr H}}(\zeta_\ell \varpi_{ j})=\eta_\ell \varpi_j, \quad\forall j\in\mathbb Z\cap[0,\ell).\label{eq:zeta_ell_pow_comm}\end{aligned}$$

By direct computation, we can verify the following algebraic relations for Hilbert ladders: $$\begin{aligned}
\zeta_\ell\eta_m+\eta_\ell\zeta_m={}&\eta_{\ell +m},\label{eq:eta_ell_add}\\\zeta_\ell\zeta_m-\eta_{\ell}\eta_{m}={}&\zeta_{\ell +m}.\label{eq:zeta_ell_add}\end{aligned}$$

For $ \ell=1$, the Feynman rules in reproduce – in Lemma \[lm:HT\_I\_K\]. Assuming that the Hilbert transform formula $ \operatorname{\widehat{\mathscr H}}(\zeta_\ell)=\eta_\ell$ holds up to a certain integer $\ell$, then $\operatorname{\mathsf{HPB}}(\zeta_\ell,\zeta_1)$ and – bring us $ \operatorname{\widehat{\mathscr H}}\eta_{\ell+1}=-\zeta_{\ell+1}$, which is equivalent to $ \operatorname{\widehat{\mathscr H}}(\zeta_{\ell+1})=\eta_{\ell+1}$. By induction, this proves in its entirety.

For $\ell=1$, the statement in is just $ \operatorname{\widehat{\mathscr H}}(\zeta_1)=\eta_1$. For $\ell=2$, the relation $ \operatorname{\widehat{\mathscr H}}(\zeta_2\varpi_1)=\eta_2\varpi_1$ is effectively proved in . Now suppose that up to a certain integer $\ell$, we have $ \operatorname{\widehat{\mathscr H}}(\zeta_\ell \varpi_{ j})=\eta_\ell \varpi_j, \forall j\in\mathbb Z\cap[0,\ell) $. From $ \operatorname{\mathsf{Par}}(\zeta_\ell \varpi_j,\eta_2 \varpi_1)$ and $ \zeta_{\ell+2} \varpi_{j+1}=(\zeta_\ell \varpi_j)(\zeta_2\varpi_1)-(\eta_{\ell}\varpi_j)(\eta_{2}\varpi_1)=-(\zeta_\ell \varpi_j)\operatorname{\widehat{\mathscr H}}(\eta_2\varpi_1)-(\eta_{2}\varpi_1)\operatorname{\widehat{\mathscr H}}(\zeta_\ell \varpi_{ j})$, we know that $ \int_{\mathbb R}\zeta_{\ell+2} (x)x^{j+1}\operatorname{d}x=0, \forall j\in\mathbb Z\cap[0,\ell)$. According to the moment formula in , it follows that $ \operatorname{\widehat{\mathscr H}}(\zeta_{\ell+2} \varpi_{ j+2})=\eta_{\ell+2} \varpi_{j+2}, \forall j\in\mathbb Z\cap[0,\ell) $. Additionally, $ \operatorname{\mathsf{Par}}(\zeta_\ell ,\eta_2 )$ and $\zeta_{\ell+2} =\zeta_\ell \zeta_2-\eta_{\ell}\eta_{2}=-\zeta_\ell \operatorname{\widehat{\mathscr H}}\eta_2-\eta_{2}\operatorname{\widehat{\mathscr H}}\zeta_\ell$ tell us that $ \int_{\mathbb R}\zeta_{\ell+2} (x)\operatorname{d}x=0 $. Thus $ \operatorname{\widehat{\mathscr H}}(\zeta_{\ell+2} \varpi_{ 1})=\eta_{\ell+2} \varpi_{1}$ is also true. This proves that $ \operatorname{\widehat{\mathscr H}}(\zeta_{\ell+2} \varpi_{ j})=\eta_{\ell+2} \varpi_{j}, \forall j\in\mathbb Z\cap[0,\ell+2) $, so follows by induction.

\[thm:B3G\] We have$$\begin{aligned}
Z_{2n,n-2k}:={}&\sum_{m=0}^{\lfloor n/2\rfloor}(-1)^m{n\choose{2m}}\int_0^\infty [\pi I_0(t)]^{n-2m}[K_0(t)]^{n+2m}t^{n-2k}\operatorname{d}t=0\label{eq:Z_vanish}\end{aligned}$$for all integer pairs $(n,k)$ meeting the requirements $n\geq2k\geq2$.

Simply notice that binomial expansion leads us to $$\begin{aligned}
\zeta_n=\sum_{m=0}^{\lfloor n/2\rfloor}(-1)^m{n\choose{2m}}\left(\iota\operatorname{sgn}\right)^{n-2m}\kappa^{n+2m},\end{aligned}$$ and implies that $ \int_{\mathbb R}  \zeta_n(x)x^j\operatorname{d}x=0$ for $ j\in\mathbb Z\cap[0,n-1)$.

Since Hilbert inversion brings us $ \operatorname{\widehat{\mathscr H}}(\eta_\ell \varpi_{ j})=-\zeta_\ell \varpi_j=\varpi_j\operatorname{\widehat{\mathscr H}}\eta_\ell$ for $ j\in\mathbb Z\cap[0,\ell)$, we must have $ \int_{\mathbb R}  \eta_\ell(x)x^j\operatorname{d}x=0$ for $ j\in\mathbb Z\cap[0,\ell-1)$, as well. Excluding from these vanishing identities the trivial cases where the integrands are odd functions over $\mathbb R$, we arrive at another family of cancelation formulae, conjugate to the B$^3$G sum rule:$$\begin{aligned}
Y_{2n,n-2k}:=\sum_{m=1}^{\lfloor n/2\rfloor+1}(-1)^{m}{n\choose{2m-1}}\int_0^\infty [\pi I_0(t)]^{n-2m+1}[K_0(t)]^{n+2m-1}t^{n-2k-1}\operatorname{d}t=0,\label{eq:Y_seq}\end{aligned}$$where $n-1\geq2k\geq2$.

Before writing this paper, we constructed alternative (and actually simpler) proofs of the B$^3$G sum rule and its conjugate , via a vanishing contour integral$$\begin{aligned}
\int_{-i\infty}^{i\infty}z^{m}[H_0^{(1)}(z)H_0^{(2)}(z)]^n\operatorname{d}z=0,\quad m\in\mathbb Z\cap[0,n-1),\label{eq:BBBG_Hankel}\end{aligned}$$where $ H_0^{(1)}$, $ H_0^{(2)}$ are cylindrical Hankel functions and the contour can be closed to the right [cf. @Watson1944Bessel §7.2]. In spite of this complex analytic shortcut, we encountered “infrared divergence” (singular behavior of the integrand as $ |z|\to\infty$) in , when we attempted to raise the power $m$ to higher values and calculate $ \operatorname{\mathbf{IKM}}(a,b;m)$ for $ m+1\geq (a+b)/2$, which is the situation occurring in the Broadhurst–Mellit integer sequence (see §\[subsec:B\_int\_seq\] below). As we will soon see, the Hilbert transform method will succeed where the contour integration approach fails.

Broadhurst–Mellit integer sequence (Crandall numbers)\[subsec:B\_int\_seq\]
---------------------------------------------------------------------------

Let $$\begin{aligned}
A(n):=\left(2/\pi\right)^42^{2n-1}[\pi^{2}\operatorname{\mathbf{IKM}}(3,5;2n-1)-\operatorname{\mathbf{IKM}}(1,7;2n-1)],\quad \forall n\in\mathbb Z_{>0}\end{aligned}$$ be the sequence defined in . Despite (erstwhile) conjectural integrality, this now becomes entry A262961 in the On-line Encyclopedia of Integer Sequences (OEIS) [@OEISA262961 under the title “Crandall numbers”, in homage to Richard E. Crandall (1947–2012), who made seminal contributions to the computation of Bessel moments], along with the following comments (at the time of writing, in April 2017):

> *“Anton Mellit and David Broadhurst define the sequence to be the* ***‘round’*** (emphasis added) *of the integral, with the conjecture that this rounding is exact. No one seems to know how to prove that any of the integrals gives a rational number, let alone an integer.”*

By , we see that $ Y_{8,2}=0$ implies the following vanishing identity$$\begin{aligned}
A(1):=\frac{32}{\pi^{4}}\int_0^\infty\left\{ [\pi I_0(t)]^2-[K_0(t)]^2 \right\}I_0(t)[K_0(t)]^5t\operatorname{d}t={}&0,\end{aligned}$$ which was first conjectured by Broadhurst in [@Broadhurst2016 (148)], after collaboration with Mellit. The rest of the sequence $ A(n+1),n\in\mathbb Z_{>0}$ in Conjecture \[conj:B\_int\_seq\] is characterized by the theorem below.

For all $n\in\mathbb Z_{>1}$, we have$$\begin{aligned}
A(n):=\left(\frac{2}{\pi}\right)^4
\int_0^\infty\left\{[\pi I_0(t)]^2 - [K_0(t)]^2\right\} I_0(t)[K_0(t)]^5\,(2t)^{2n-1} \operatorname{d}t
\in\mathbb Z_{>0}.\end{aligned}$$Furthermore, the following explicit formula holds for $ n\in\mathbb Z_{>0}$:$$\begin{aligned}
A(n+1)={}&\frac{1}{2^{4(n-1)}}\sum_{m=1}^{n} \sum_{\ell=1}^{m}\sum_{k=1}^{\ell} \frac{ [(2 n-2m)!]^3}{ [(n-m)!]^4}\frac{ [(2 m-2\ell)!]^3}{ [(m-\ell)!]^4}\frac{ [(2 \ell-2k)!]^3}{ [( \ell-k)!]^4}\frac{ [(2k-2)!]^3}{ [(k-1)!]^4}.\label{eq:An_explicit}\end{aligned}$$

We recall the following Bessel moments from [@Bailey2008 (7)] (see also [@Ouvry2005; @BorweinBruno2007]):$$\begin{aligned}
\int_0^\infty [K_0(t)]^{2}t^{2n}\operatorname{d}t=\frac{\pi ^2 [(2 n)!]^3}{4^{3 n+1} (n!)^4},\quad \forall n\in\mathbb Z_{\geq0},\end{aligned}$$which enable us to compute the following Hilbert transforms with the aid of and :$$\begin{aligned}
\operatorname{\widehat{\mathscr H}}(\kappa^2 \varpi_{2n})={}& \varpi_{2n}\operatorname{\widehat{\mathscr H}}(\kappa^2)-\frac{1}{\pi}\sum_{k=0}^{2n-1}  \varpi_k\int_{-\infty}^\infty \xi^{2n-1-k}[\kappa(\xi)]^{2}\operatorname{d}\xi\notag\\={}& \iota\kappa \varpi_{2n}\operatorname{sgn}-\frac{\pi}{2}\sum_{m=1}^{n}  \frac{ [(2 n-2m)!]^3}{4^{3 (n-m)} [(n-m)!]^4}\varpi_{2m-1},\quad \forall n\in\mathbb Z_{>0}.\end{aligned}$$Spelling out $ \operatorname{\mathsf{Par}}(-3\iota^2\kappa^4\varpi_{1}+\kappa^6\varpi_{1},\kappa^2\varpi_{2n})$ \[*i.e.* $ \operatorname{\mathsf{Par}}(\eta_3\varpi_1,\kappa^2\varpi_{2n})$\], we see that $A(n+1),\forall n\in\mathbb Z_{>0}$ is expressible as a finite sum of Bessel moments involving only six Bessel functions:$$\begin{aligned}
A(n+1)=\frac{2^{2(n+1)}}{\pi^{4}}\sum_{m=1}^{n}  \frac{ [(2 n-2m)!]^3}{4^{3 (n-m)} [(n-m)!]^4}\int_0^\infty \{3[\pi I_{0}(t)]^2-[K_0(t)]^2\}[K_0(t)]^4t^{2m}\operatorname{d}t.\end{aligned}$$By $\operatorname{\mathsf{Par}}(2\iota \kappa^3\operatorname{sgn},\kappa^2\varpi_{2m})$, we arrive at further reduction:$$\begin{aligned}
&
\int_0^\infty \{3[\pi I_{0}(t)]^2-[K_0(t)]^2\}[K_0(t)]^4t^{2m}\operatorname{d}t\notag\\={}&\pi^{2}\sum_{\ell=1}^{m}  \frac{ [(2 m-2\ell)!]^3}{4^{3 (m-\ell)} [(m-\ell)!]^4}\int_0^\infty I_0(t)[K_0(t)]^3t^{2\ell-1}\operatorname{d}t.\end{aligned}$$ With $ \operatorname{\widehat{\mathscr H}}(\kappa^2 \varpi_1)=\iota\kappa \varpi_1\operatorname{sgn}$ and $\operatorname{\mathsf{Par}}(\kappa^{2}\varpi_{1},\kappa^2\varpi_{2(\ell-1)})$, we evaluate the remaining Bessel moments involving four Bessel functions:$$\begin{aligned}
\int_0^\infty I_0(t)[K_0(t)]^3t^{2\ell-1}\operatorname{d}t=\frac{\pi^{2}}{4^{3\ell-1}}\sum_{k=1}^{\ell}  \frac{ [(2 \ell-2k)!]^3}{ [( \ell-k)!]^4}\frac{ [(2k-2)!]^3}{ [(k-1)!]^4}.\label{eq:IKKK_moments}\end{aligned}$$ So far, we have verified .

Next, we show that $$\begin{aligned}
\alpha_\ell:=\frac{1}{2^{4(\ell-1)}}\sum_{k=1}^{\ell}  \frac{ [(2 \ell-2k)!]^3}{ [( \ell-k)!]^4}\frac{ [(2k-2)!]^3}{ [(k-1)!]^4}\in\mathbb Z,\quad \forall \ell\in\mathbb Z_{>0}.\end{aligned}$$Towards this end, we first recall from [@Bailey2008 (55) and (56)] the following formula$$\begin{aligned}
\int_0^\infty I_0(t)[K_0(t)]^3t^{2\ell-1}\operatorname{d}t=\frac{\pi^{2}}{16}\left[ \frac{(\ell-1)!}{4^{\ell-1}} \right]^2\sum^{\ell-1}_{k=0}{\ell-1\choose k}^2{2(\ell-1-k)\choose\ell-1-k}{2k\choose k},\end{aligned}$$which combines with into[^4] $$\begin{aligned}
\label{eq:a_l_Domb}
\alpha_\ell=\frac{[(\ell-1)!]^{2}}{4^{\ell-1}}\sum^{\ell-1}_{k=0}{\ell-1\choose k}^2{2(\ell-1-k)\choose\ell-1-k}{2k\choose k}=\frac{[(\ell-1)!]^{2}}{4^{\ell-1}}D_{\ell-1},\end{aligned}$$where $D_j,j\in\mathbb Z_{\geq0}$ is called the $j$-th Domb number in combinatorics. In [@Rogers2009 Theorem 3.1], Rogers has shown that the following identity holds for $|u|$ sufficiently small:$$\begin{aligned}
_3F_2\left(\left.\begin{array}{c}
\frac{1}{3},\frac{1}{2},\frac{2}{3} \\1,1 \\
\end{array}\right|\frac{27u^{2}}{4(1-u)^{3}}\right)=(1-u)\sum_{n=0}^\infty \frac{D_n}{4^n}u^n.\label{eq:Rogers2009}\end{aligned}$$Here, the generalized hypergeometric series $ _pF_q$ is defined as $$\begin{aligned}
{_pF_q}\left(\left.\begin{array}{c}
a_{1},\dots,a_p \\[4pt]
b_{1},\dots,b_q \\
\end{array}\right| x\right):=\sum_{n=0}^\infty\frac{(a_1)_n\cdots(a_p)_n}{(b_1)_n\cdots(b_q)_n}\frac{x^n}{n!},\end{aligned}$$ where the Pochhammer symbol represents the rising factorial$$\begin{aligned}
(a)_{n}:=\begin{cases}1, & n=0 \\
a(a+1)\cdots(a+n-1), & n\in\mathbb Z_{>0} \\
\end{cases} ;\quad (a)_{n}:=\frac{\Gamma(a+n)}{\Gamma(a)},\quad a\notin\mathbb  Z_{\leq0},\end{aligned}$$ for $ \{b_1,\dots,b_q\}\cap\mathbb  Z_{\leq0}=\emptyset$. In Rogers’ identity , the coefficient of $u^n$ for $n\in\mathbb Z_{>0}$ is equal to $$\begin{aligned}
\frac{D_n}{4^n}-\frac{D_{n-1}}{4^{n-1}},\end{aligned}$$according to the right-hand side. Meanwhile, the Taylor expansion of the left-hand side reads$$\begin{aligned}
_3F_2\left(\left.\begin{array}{c}
\frac{1}{3},\frac{1}{2},\frac{2}{3} \\1,1 \\
\end{array}\right|\frac{27u^{2}}{4(1-u)^{3}}\right)=1+3\sum_{n=1}^\infty[(2n-1)!!]^2{3n-1\choose2n}\frac{1}{(n!2^n)^2}\frac{u^{2n}}{(1-u)^{3n}},\end{aligned}$$where the double factorial is defined as $ (2n-1)!!:=(2n)!/(n!2^n)\in\mathbb Z$ for $n\in\mathbb Z_{>0}$. As we compute the contribution from the aforementioned Taylor coefficients to$$\begin{aligned}
(n!)^{2}
\left(\frac{D_n}{4^n}-\frac{D_{n-1}}{4^{n-1}}\right)=\alpha_{n+1}-n^{2}\alpha_{n},\end{aligned}$$we are gathering finitely many summands, each of which is an integer multiple of $ (k!!)^{2}\in\mathbb Z$ for a certain odd positive integer $k$ less than $n$. Therefore, we have $ \alpha_1=1$ and $ \alpha_{\ell+1}-\ell^{2}\alpha_{\ell}\in\mathbb Z$ for all $\ell\in\mathbb Z_{>0}$.

Finally, by discrete convolution, we see that $A(n+1)$ is the coefficient of $x^{n-1}$ in the polynomial$$\begin{aligned}
\left(\sum_{\ell=1}^{n+1}\alpha_\ell x^{\ell-1}\right)^2=\left[\sum_{\ell=0}^n\frac{(\ell!)^2D_\ell }{4^{\ell}}x^\ell\right]^2\in\mathbb Z[x],\end{aligned}$$ so $A(n+1)\in\mathbb Z $ must hold.

More generally, we define the positive integer $ \alpha_n^{[m]}$ as the coefficient of $x^{n-1}$ in the polynomial $\left(\sum_{\ell=1}^{n+1}\alpha_\ell x^{\ell-1}\right)^m\in\mathbb Z[x]$, for every $m\in\mathbb Z_{>0}$. By repeated applications of the recursions for Hilbert ladders \[see and \], we can show that $$\begin{aligned}
\alpha_n^{[m]}=\frac{4}{\pi^{2m+1}}\sum_{\ell=1}^{m+1}(-1)^{\ell-1} {2m\choose{2\ell-1}}\int_0^\infty [\pi I_0(t)]^{2(m-\ell)+1}[K_0(t)]^{2(m+\ell)-1}(2t)^{2(n+m)-3}\operatorname{d}t.\label{eq:alpha_n_m}\end{aligned}$$ Thus, the Broadhurst–Mellit integer sequence $ A(1)=0,A(n+1)=\alpha_{n}^{[2]},n\in\mathbb Z_{>0}$ is just a special case within an infinite family of (linear combinations for) Bessel moments, preceded by $$\begin{aligned}
\alpha_{n}^{\vphantom{[1]}}\equiv\alpha_{n}^{[1]}=\frac{8}{\pi^{2}}\int_0^\infty I_0(t)[K_0(t)]^{3}(2t)^{2n-1}\operatorname{d}t\in\mathbb Z_{>0}\end{aligned}$$and followed by$$\begin{aligned}
\alpha_{n}^{[3]}=\frac{8}{\pi^{6}}\int_0^\infty\left\{[\pi I_0(t)]^2 -3 [K_0(t)]^2\right\} \left\{3[\pi I_0(t)]^2 - [K_0(t)]^2\right\} I_0(t)[K_0(t)]^7\,(2t)^{2n+3} \operatorname{d}t\in\mathbb Z_{>0},\end{aligned}$$for $ n\in\mathbb Z_{>0}$.

For $ \operatorname{\mathbf{IKM}}(a,b;c)$ satisfying $c\equiv0\pmod2 $ and $ a+b\equiv2\pmod4$, we also have a family of integer sequences that runs parallel to . With the observation that $$\begin{aligned}
\frac{ [(2 n)!]^3}{2^{4n}(n!)^4}=\frac{1}{2^{2n}}[(2n-1)!!]^{2}{2n\choose n}\in\frac{1}{2^{2n}}\mathbb Z,\quad \forall n\in\mathbb Z_{\geq0},\end{aligned}$$we define $ \beta^{[m]}_n$ as the coefficient of $x^{n-1}$ in the polynomial$$\begin{aligned}
\left[\sum_{\ell=0}^n\frac{(\ell!)^2D_\ell }{4^{\ell}}x^\ell\right]^{m-1} \sum_{k=0} ^n\frac{ [(2 k)!]^3}{2^{4k}(k!)^4}x^{k}\in\frac{1}{2^{2(n-1)}}\mathbb Z[x],\end{aligned}$$ for every $m\in\mathbb Z_{>0}$. According to the discrete convolutions of Bessel moments that descend from the recursions for Hilbert ladders, we have $$\begin{aligned}
\beta^{[m]}_n=\frac{4}{\pi^{2m}}\sum_{\ell=1}^{m}(-1)^{\ell-1} {2m-1\choose{2\ell-1}}\int_0^\infty [\pi I_0(t)]^{2(m-\ell)}[K_0(t)]^{2(m+\ell-1)}(2t)^{2(n+m-2)}\operatorname{d}t\in\frac{1}{2^{2(n-1)}}\mathbb Z_{>0}\end{aligned}$$for $ m,n\in\mathbb Z_{>0}$. We can even combine the foregoing statements about $\alpha_n ^{[m]}$ and $ \beta_n^{[m]}$ into a more compact form, as follows:$$\begin{aligned}
\frac{2^{1+2(n-1)[1-(-1)^{M}]}}{\pi^{M+1}}\int_0^\infty \frac{[\pi I_0(t)+i K_{0}(t)]^{M}-[\pi I_0(t)-i K_{0}(t)]^{M}}{i}[K_0(t)]^{M}(2t)^{2n+M-3}\operatorname{d}t\in\mathbb Z_{>0}\end{aligned}$$for all $ M,n\in\mathbb Z_{>0}$. This confirms a recent empirical observation by Broadhurst and Roberts [@Broadhurst2017Paris Conjecture 2].

To summarize, we have demonstrated that many sum rules of Bessel moments indeed issue from the three fundamental laws \[itm:HT-1\]–\[itm:HT-3\] of Hilbert transforms. We remind our readers that the methods developed in this note do not exhaust all possible linear relations for Feynman diagrams in two-dimensional quantum field theory. For example, the sum rule $ 9\pi^2\operatorname{\mathbf{IKM}}(4,4;1)-14\operatorname{\mathbf{IKM}}(2,6;1)=0$ [conjectured in @Broadhurst2016 (147)] is not a simple consequence of Hilbert transforms, and its recent verification in [@WEF §5.2] involves heavy use of contour integrations over modular forms.

Acknowledgments {#acknowledgments .unnumbered}
---------------

In early 2017, I wrote up this paper in Beijing, mostly drawing on my research notes prepared at Princeton during 2012 and 2013. I thank Prof. Weinan E for arranging my stays in Princeton and Beijing, as well as organizing a seminar on constructive quantum field theory at Princeton.

After completion of the initial draft of this article, I received from Dr. David Broadhurst his slides for recent talks [@Broadhurst2017Paris; @Broadhurst2017CIRM; @Broadhurst2017Higgs] on Bessel moments, which set his conjectures in a wider context. I thank Dr. Broadhurst for his constant encouragements and incisive comments on this project.

I am indebted to an anonymous referee for thoughtful suggestions on improving the presentation of this paper.

In January 2013, I benefited from fruitful discussions with Prof. Jon Borwein on his previous contributions to Bessel moments and elliptic integrals; I was equally grateful to his friendly communications on my then-unpublished work related to Hilbert transforms. I dedicate this work to his memory.

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[^1]: *Keywords*: Hilbert transforms, Bessel functions, Feynman integrals\
    *MSC 2010*: 44A15, 33C10, 33C20 (Primary) 81T18, 81T40, 81Q30 (secondary)\
    This research was supported in part by the Applied Mathematics Program within the Department of Energy (DOE) Office of Advanced Scientific Computing Research (ASCR) as part of the Collaboratory on Mathematics for Mesoscopic Modeling of Materials (CM4)

[^2]: For our purposes, it suffices to evaluate the Hilbert transform of a function for almost every (a.e.) real variable, leaving out a subset of measure zero that will not affect subsequent computations of Lebesgue integrals. Later afterwards, an equal sign may also be used to denote an equality that is valid almost everywhere (even when “a.e.” is not written), depending on context.

[^3]: Confusingly, in footnote 13 of [@Bailey2008], the authors apparently declared that the conjecture on $ Z_{2n,n-2k}=0$ “has also now been resolved”, without supplying further citations to publications or preprints. This contradicts the recent claims in [@Broadhurst2016 (112)] and [@BroadhurstMellit2016 (5.6)] on the open status of $ Z_{6,1}=0$.

[^4]: Naïvely, upon observing that $ (n!)^2/2^n\in\mathbb Z,\forall n\in\mathbb Z_{\geq4}$ and $ D_n\in\mathbb Z,\forall n\in\mathbb Z_{\geq0}$, we obtain $ 2^{\ell-1}\alpha_\ell\in\mathbb Z,\forall \ell\in\mathbb Z_{>0}$, at best. The divisibility statement $ 4^{n}\mid(n!)^2D_n,\forall n\in\mathbb Z_{\geq0}$ is thus deeper than these naïve observations. In our proof of the integrality $ \alpha_\ell\in\mathbb Z,\forall \ell\in\mathbb Z_{>0}$, we need Rogers’ work on modular forms [@Rogers2009], which in turn, was inspired by Bertin’s studies of modular parametrizations for certain families of Calabi–Yau manifolds [@Bertin2008].