用户: TravorLZH/一个二重素数倒数和的估计

本文中, 我们将估计下列和式的上界: $$S=\sum _{y<p\le x}\sum _{\begin{array}{c}y<p^{\prime }\le x/y\\ p^{\prime }\equiv 1(p)\end{array}}\frac{1}{p{p}^{\prime }}.$$

1推导过程

为了估计内侧求和, 我们直接计算: $$S_1=\sum _{\begin{array}{c}{p}^{\prime }\le x\\ p^{\prime }\equiv 1(p)\end{array}}\frac{1}{p^{\prime }}$$的上界. 拆分区间可知: $$\begin{array}{rl}{S}_1& \le \sum _{\begin{array}{c}n\le x_0\\ n\equiv 1(p)\end{array}}\frac{1}{n}+\sum _{\begin{array}{c}{x}_0<p^{\prime }\le x\\ p^{\prime }\equiv 1(p)\end{array}}\frac{1}{p^{\prime }}\\ & =O\left(\frac{\log x_0}{p}\right)+{\int }_{x_0}^x\frac{\mathrm{d}\pi (t;p,1)}{t}\\ & \le \frac{\pi (x;p,1)}{x}+{\int }_{x_0}^x\frac{\pi (t;p,1)}{t^2}\mathrm{d}t+O\left(\frac{\log x_0}{p}\right).\end{array}$$为了估计右侧积分, 我们引入 Brun–Titchmarsh 不等式 [2], 即存在固定常数 $M>0$ 使 $x\ge q\ge 2$ 时: $$\pi (x;q,a)\le \frac{Mx}{\phi (q)\log (x/q)},$$则有: $$\begin{array}{rl}{S}_1& \ll \frac{1}{p}{\int }_{x_0}^x\frac{\mathrm{d}t}{t\log t/p}+\frac{\log x_0}{p}+\frac{1}{p\log (x_0/p)}\\ & =\frac{1}{p}\left[\frac{1}{\log x_0/p}+{\int }_{x_0/p}^{x/p}\frac{\mathrm{d}u}{u\log u}+\log x_0\right].\end{array}$$现在设置 $x_0=p^2+1$, 即得: $$S_1\ll \frac{\log p+\log \log x}{p}.$$(1)将 (1) 回代至 $S$ 中, 便有: $$\begin{array}{rl}S& \ll \sum _{y<p\le x}\frac{\log p+\log \log x}{p^2}\le \sum _{n>y}\frac{\log n+\log \log x}{n^2}\\ & \ll \frac{\log \log x}{y}+{\int }_y^{\infty }\frac{\log t}{t^2}\mathrm{d}t\ll \frac{\log \log x}{y}+\frac{\log y}{y}.\end{array}$$这个上界被 Erdös[1] 用来证明:

$$\sum _{\begin{array}{c}n\le x\\ (n,\phi (n))=1\end{array}}1\sim \frac{e^{-\gamma }x}{\log \log \log x}.$$

参考文献

[1]	Erdös, P. (1948). Some asymptotic formulas in number theory. J. Indian Math. Soc., 12, 75–78.
[2]	Titchmarsh, E. C. (1930). A divisor problem. Rend. Circ. Mat. Palermo, 54, 414–429.