New estimate in Vinogradov's mean-value theorem (Q650419)

Here the author refines the upper bound for the Vinogradov integral defined by \[ J_{s,k}(P)=\int_{[0,1]^k}\left|\sum_{1\leq x\leq P}e(\alpha_1x+\ldots+\alpha_kx^k)\right|^{2s}\,d\alpha. \] He proves the following mean-value theorems, which improve results due to \textit{K. Ford} [Proc. Lond. Math. Soc. (3) 85, No. 3, 565--633 (2002; Zbl 1034.11044)]. Theorem 1. Suppose that \(k\) and \(s\) are integers, \(k\geq 2000\), \(D\), \(1\leq D\leq (k-1)/2\), is a real number, \[ H = H(D, k) =\frac{(7/8)D - 7D(D + 1)/(16k)}{1 + (2Dk - D(D + 1))/(2k^2)},\qquad H\geq 1, \] and \[ 2k^2\leq s\leq \frac 12 k^2 \left(\frac 12 + \log\left(\frac{3k}{8D}\right)-\left(\frac{1 - D/k}{2 - D/k}\right)\frac Dk+\frac 5{6k}+\frac{0.67}{Dk}\left(1 +\frac 1{2^{H-2}} +\frac 3{2k-1}\right)\right). \] Then the following estimate holds: \[ J_{s,k}(P) \leq k^{2.055k^3-5.91k^2+3s}1,06^{sk+2s^2/k-9.7278k^3} P^{2s-k(k+1)/2+\Delta_s}, \quad P\geq 1, \] where \[ \Delta_s =\frac 38 k^2 \exp\left(\frac 12 -\frac{2s}{k^2}-\left(\frac{1-D/k}{2- D/k}\right)\frac Dk+\frac 5{6k}+ 0.67\frac{1 + 1/2^{H-2} + 3/(2k-1)}{Dk }+\frac{2(s \bmod k)}{k^3}\right). \] Theorem 2. Suppose that \(k\) and \(s\) are integers. Then the estimate \[ J_{s,k}(P) \leq k^{\theta k^3}P^{2s-k(k+1)/2+0.001k^2}, \qquad P\geq 1, \] holds for \[ 129 \leq k < 16000,\quad s\geq 3.20354k^2,\quad \theta= 2.4183; \] \[ 4k \geq 16000,\quad s\geq 3.213303k^2,\quad \theta= 2.3291. \]