Math Testing

引理1 假设 $y \geq 0$ ,而 $[\log x]$ 表示 $\log x$ 的整数部分, 设: $$\mathrm{\Phi }(y)=\frac{1}{2\pi i}{\int }_{2-i\mathrm{\infty }}^{2+i\mathrm{\infty }}\frac{y^{\omega }\mathrm{d}\omega }{\omega {(1+\frac{\omega }{(\log x{)}^{1.1}})}^{[\log x]+1}},x>1$$ 显见，当 $0 \leq y \leq 1$ 时，有 $\Phi(y) = 0$. 对于所有 $y \geq 0$, 则 $\Phi(y)$ 是一个非减函数. 当 $\log x\geq 10^4$ 及 $y\geq e^{2{(\log x)}^{-0.1}}$ 时，则有: $$1-x^{-0.1}\le \mathrm{\Phi }(y)\le 1$$

证:

我们先来证明

$$\frac{{\partial }^r}{\partial {\omega }^r}\left(\frac{y^{\omega }}{\omega }\right)=\left(\frac{y^{\omega }}{\omega }\right)\{(\log y{)}^r+\sum _{i=1}^r\frac{(-1{)}^{i}r\cdots (r-i+1)(\log y{)}^{r-i}}{{\omega }^i}\}$$

成立, 显见 (1) 式当 $r=1$ 和 $r=2$ 时都成立.

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

由于:

$$\begin{array}{rl}\frac{{\partial }^{S+1}}{\partial {\omega }^{S+1}}\left(\frac{y^{\omega }}{\omega }\right)& =\frac{\partial }{\partial \omega }\left\{y^{\omega }(\frac{(\log y{)}^S}{\omega }+\sum _{i=1}^S\frac{(-1{)}^i\cdot S\cdots (S-i+1)(\log y{)}^{S-i}}{{\omega }^{i+1}})\right\}\\ & =y^{\omega }\{\frac{(\log y{)}^{S+1}}{\omega }+\sum _{i=1}^S\frac{(-1{)}^i\cdot S\cdots (S-i+1)(\log y{)}^{S+1-i}}{{\omega }^{i+1}}-\frac{(\log y{)}^S}{{\omega }^2}+\sum _{i=1}^S\frac{(-1{)}^{i+1}S\cdot \cdot (S-i+1)(i+1)(\log y{)}^{S-i}}{{\omega }^{t+2}}\}\\ & =\left(\frac{y^{\omega }}{\omega }\right)(\log y{)}^{S+1}-\frac{(S+1)(\log y{)}^S}{\omega }+\frac{(-1{)}^{S+1}(S+1)!}{{\omega }^{S+1}}+\sum _{i=2}^S((-1{)}^{i}S\cdots (S-i+1)(\log y{)}^{S+1-i}+\frac{(-1{)}^{i}S\cdots (S+2-i)i(\log y{)}^{x+1-i}}{{\omega }^i})\}\\ & =\left(\frac{y^{\omega }}{\omega }\right)\{(\log y{)}^{S+1}+\sum _{i=1}^{S+1}\frac{(-1{)}^i(S+1)\cdots (S+1-i+1)(\log y{)}^{S+1-i}}{{\omega }^i}\}\end{array}$$

故 (1) 式得证.

又当 $y \geq 1$ 时, 我们有:

$$\begin{array}{rl}\mathrm{\Phi }(y)& =1+\left\{\frac{(\log x{)}^{1.1+1.1[\log x}]}{[\log x]!}\right\}{\left\{\frac{{\partial }^{[\log x]}}{\partial {\omega }^{[\log x)}}\left(\frac{y^{\omega }}{\omega }\right)\right\}}_{\omega =-(\log x{)}^{1.1}}\\ & =1-e^{-(\log x{)}^{1.1}\cdot (\log y)}\sum _{\nu =0}^{[\log x]}\frac{(\log x{)}^{1.1}(\log y{)}^{\nu }}{\nu !}\\ & =\left\{\frac{1}{[\log x]!}\right\}{\int }_0^{(\log x)!\cdot (\log y)}{e}^{-\lambda }{\lambda }^{[\log x]}\mathrm{d}\lambda \end{array}$$

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

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

$$\begin{array}{rl}0<1-\mathrm{\Phi }(y)& =\left\{\frac{1}{[\log x]!}\right\}{\int }_{(\log x{)}^{1.1}(\log y)}^{\mathrm{\infty }}{e}^{-\lambda }{\lambda }^{[\log x]}\mathrm{d}\lambda \\ & \le \left\{\frac{1}{[\log x]!}\right\}{\int }_{2[\log x]}^{\mathrm{\infty }}{e}^{-\lambda }{\lambda }^{[\log x]}\mathrm{d}\lambda \\ & =\left\{\frac{([\log x]{)}^{1+[\log x)}}{[\log x]!}\right\}\cdot {\int }_2^{\mathrm{\infty }}{e}^{-\lambda [\log x]}{\lambda }^{[\log x]}\mathrm{d}\lambda \\ & =\left\{\frac{e^{-[\log x]}([\log x]{)}^{1+[\log x]}}{[\log x]!}\right\}\cdot {\int }_1^{\mathrm{\infty }}{e}^{-\lambda [\log x]}(1+\lambda {)}^{[\log x]}\mathrm{d}\lambda \\ & \le x^{-0.1}\end{array}$$

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