Arithmeticity conjecture for fifth-moment Berndt-type integrals

At least 1 year old · documented by

Let Γ=Γ(1/4)\Gamma=\Gamma(1/4) and, for n≥1n\geq1, define

J2n−1= ⁣∫0∞ ⁣x5 dx(cos⁡x+cosh⁡x)2n−1,J2n= ⁣∫0∞ ⁣x5 dx(cos⁡x+cosh⁡x)2n.J_{2n-1}=\displaystyle\!\int_0^\infty\displaystyle\!\frac{x^5\,dx}{(\cos x+\cosh x)^{2n-1}},\qquad J_{2n}=\displaystyle\!\int_0^\infty\displaystyle\!\frac{x^5\,dx}{(\cos x+\cosh x)^{2n}}.

Here Q\mathbb{Q} denotes the rational numbers. Fifth-moment arithmeticity conjecture. For every integer n≥1n\geq1,

J2n−1∈ ⁣Γ4πQ+ ⁣Γ4π2Q+ ⁣∑j=2n−3 ⁣∑i=16 ⁣Γ8j−4π6j−10+iQ+ ⁣∑j=13 ⁣∑i=17−2j ⁣Γ8j+8n−20π6j+6n−22+iQ,J_{2n-1}\in\displaystyle\!\frac{\Gamma^4}{\pi}\mathbb{Q}+\displaystyle\!\frac{\Gamma^4}{\pi^2}\mathbb{Q}+\displaystyle\!\sum_{j=2}^{n-3}\displaystyle\!\sum_{i=1}^{6}\displaystyle\!\frac{\Gamma^{8j-4}}{\pi^{6j-10+i}}\mathbb{Q}+\displaystyle\!\sum_{j=1}^{3}\displaystyle\!\sum_{i=1}^{7-2j}\displaystyle\!\frac{\Gamma^{8j+8n-20}}{\pi^{6j+6n-22+i}}\mathbb{Q},

and

J2n∈Q+ ⁣∑i=14 ⁣Γ8πi+1Q+ ⁣∑i=1\i≠26 ⁣Γ16πi+5Q+ ⁣∑j=3n−2 ⁣∑i=16 ⁣Γ8jπ6j−7+iQ+ ⁣∑j=13 ⁣∑i=17−2j ⁣Γ8j+8n−16π6j+6n−19+iQ.J_{2n}\in\mathbb{Q}+\displaystyle\!\sum_{i=1}^{4}\displaystyle\!\frac{\Gamma^8}{\pi^{i+1}}\mathbb{Q}+\displaystyle\!\sum_{\substack{i=1\i\ne2}}^{6}\displaystyle\!\frac{\Gamma^{16}}{\pi^{i+5}}\mathbb{Q}+\displaystyle\!\sum_{j=3}^{n-2}\displaystyle\!\sum_{i=1}^{6}\displaystyle\!\frac{\Gamma^{8j}}{\pi^{6j-7+i}}\mathbb{Q}+\displaystyle\!\sum_{j=1}^{3}\displaystyle\!\sum_{i=1}^{7-2j}\displaystyle\!\frac{\Gamma^{8j+8n-16}}{\pi^{6j+6n-19+i}}\mathbb{Q}.

These membership assertions refine the paper's proven degree bounds for related integrals and are supported by extensive numerical evidence, but remain conjectural.

References

Primary source

Ce Xu and Jianqiang Zhao, “General Berndt-Type Integrals and Series Associated with Jacobi Elliptic Functions”, arXiv:2401.01385 (2024).

Progress summary

Never refreshed

Nothing recorded yet. Refresh searches the literature and the public web for attempts on this problem, and writes the first summary here.

Solutions 0

No solutions have been posted yet.