引理1 假设 y0y \geq 0 ,而 [logx][\log x] 表示 logx\log x 的整数部分, 设: Φ(y)=12πi2i2+iyωdωω(1+ω(logx)1.1)[logx]+1,x>1 显见,当 0y10 \leq y \leq 1 时,有 Φ(y)=0\Phi(y) = 0. 对于所有 y0y \geq 0, 则 Φ(y)\Phi(y) 是一个非减函数. 当 logx104\log x\geq 10^4ye2(logx)0.1y\geq e^{2{(\log x)}^{-0.1}} 时,则有: 1x0.1Φ(y)1

证:

我们先来证明

rωr(yωω)=(yωω){(logy)r+i=1r(1)ir(ri+1)(logy)riωi}

成立, 显见 (1) 式当 r=1r=1r=2r=2 时都成立.

现假定 (1) 式对于 r=2,,Sr = 2 , \cdots , S 时都成立, 而证明对于 S+1S+1 也成立.

由于:

S+1ωS+1(yωω)=ω{yω((logy)Sω+i=1S(1)iS(Si+1)(logy)Siωi+1)}=yω{(logy)S+1ω+i=1S(1)iS(Si+1)(logy)S+1iωi+1(logy)Sω2+i=1S(1)i+1S(Si+1)(i+1)(logy)Siωt+2}=(yωω)(logy)S+1(S+1)(logy)Sω+(1)S+1(S+1)!ωS+1+i=2S((1)iS(Si+1)(logy)S+1i+(1)iS(S+2i)i(logy)x+1iωi)}=(yωω){(logy)S+1+i=1S+1(1)i(S+1)(S+1i+1)(logy)S+1iωi}

故 (1) 式得证.

又当 y1y \geq 1 时, 我们有:

Φ(y)=1+{(logx)1.1+1.1[logx][logx]!}{[logx]ω[logx)(yωω)}ω=(logx)1.1=1e(logx)1.1(logy)ν=0[logx](logx)1.1(logy)νν!={1[logx]!}0(logx)!(logy)eλλ[logx]dλ

因为 0y10\leq y \leq1 时, Φ(y)=0\Phi(y)=0 . 故由上式得到: 当 y0y\geq0 时, 则 Φ(y)\Phi(y) 是一个非减函数.

又当 ye2(logx)1.0y \geq e^{2 (\log x) - 1.0} 时, 有:

0<1Φ(y)={1[logx]!}(logx)1.1(logy)eλλ[logx]dλ{1[logx]!}2[logx]eλλ[logx]dλ={([logx])1+[logx)[logx]!}2eλ[logx]λ[logx]dλ={e[logx]([logx])1+[logx][logx]!}1eλ[logx](1+λ)[logx]dλx0.1

其中用到 logx104\log x\geq10^4 及当 λ1\lambda\geq1 时, 有 elog(1+λ)eλlog2e^{\log (1 + \lambda)} \leq e^{\lambda \log 2} .