On Laporta's 4-loop sunrise formulae

Understand

We prove Laporta's conjecture\begin{align*}&\int_0^\infty\frac{\mathrm d\, x_1}{x_1}\int_0^\infty\frac{\mathrm d\, x_2}{x_2}\int_0^\infty\frac{\mathrm d\, x_3}{x_3}\int_0^\infty\frac{\mathrm d\, x_4}{x_4}\frac{1}{\left(1+\sum^4_{k=1}x_k\right)\left(1+\sum^4_{k=1}\frac{1}{x_{k}} \right)-1}\\={}&\frac43 \int_{0}^\pi\mathrm d\, \phi_1 \int_{0}^\pi\mathrm d\, \phi_2\int_{0}^\pi\mathrm d\, \phi_3 \int_{0}^\pi\mathrm d\, \phi_4\frac{1}{4-\sum_{k=1}^4\cos \phi_k}, \end{align*} which relates the 4-loop sunrise diagram in 2-dimensional quantum field theory to Watson's integral for 4-dimensional hypercubic lattice.

• We also establish several related integral identities proposed by Laporta, including a reduction of the 4-loop sunrise diagram to special values of Euler's gamma function and generalized hypergeometric series:\begin{align*} \frac{4 \pi ^{5/2}}{\sqrt{3}}\left\{ \frac{\sqrt{3} }{2^6 }\left[\frac{\Gamma \left(\frac{1}{3}\right)}{\sqrt{\pi}}\right]^9\, _4F_3\left(\left.
• \begin{array}{c}\frac{1}{6},\frac{1}{3},\frac{1}{3},\frac{1}{2}\\[4pt]\frac{2}{3},\frac{5}{6},\frac{5}{6}\end{array} \right|1\right)-\frac{2^{4}}{3}\left[\frac{\sqrt{\pi}}{\Gamma \left(\frac{1}{3}\right)}\right]^9\, _4F_3\left(\left.
• \begin{array}{c}\frac{1}{2},\frac{2}{3},\frac{2}{3},\frac{5}{6}\\[4pt]\frac{7}{6},\frac{7}{6},\frac{4}{3}\end{array} \right|1\right) \right\}.