先假设你有一只兔子。

假设有人又给了你另一只兔子。

现在，数一下你所拥有的兔子数量，你会得到结果是两只。也就是说一只兔子加一只兔子等于两只兔子，也就是一加一等于二。

$$1+1=2$$

这就是算数的运算方法了。

那么，现在你已经对算数的基本原理有了一定了解，就让我们来看一看下面这个例子，把刚刚学到的知识运用到实践中去。

试试看

$$\begin{array}{r}\begin{array}{c}\log \mathrm{\Pi }(N)=(N+{\displaystyle \frac{1}{2}})\log N-N+A-{\int }_N^{\mathrm{\infty }}{\displaystyle \frac{{\bar{B}}_1(x)\mathrm{d}x}{x}},A=1+{\int }_1^{\mathrm{\infty }}{\displaystyle \frac{{\bar{B}}_1(x)\mathrm{d}x}{x}}\\ \log \mathrm{\Pi }(s)=(s+{\displaystyle \frac{1}{2}})\log s-s+A-{\int }_0^{\mathrm{\infty }}{\displaystyle \frac{{\bar{B}}_1(t)\mathrm{d}t}{t+s}}\end{array}\\ \\ \begin{array}{rl}\log \mathrm{\Pi }(s)=& \underset{n\to \mathrm{\infty }}{lim}[s\log (N+1)+\sum _{n=1}^N\log n-\sum _{n=1}^N\log (s+n)]\\ =& \underset{n\to \mathrm{\infty }}{lim}[s\log (N+1)+{\int }_1^N\log x\mathrm{d}x-{\displaystyle \frac{1}{2}}\log N+{\int }_1^N{\displaystyle \frac{{\bar{B}}_1\mathrm{d}x}{x}}\\ & -{\int }_1^N\log (s+x)\mathrm{d}x-{\displaystyle \frac{1}{2}}[\log (s+1)+\log (s+N)]\\ & -{\int }_1^N{\displaystyle \frac{{\bar{B}}_1(x)\mathrm{d}x}{s+x}}]\\ =& \underset{n\to \mathrm{\infty }}{lim}[s\log (N+1)+N\log N-N+1+{\displaystyle \frac{1}{2}}\log N+{\int }_1^N{\displaystyle \frac{{\bar{B}}_1(x)\mathrm{d}x}{x}}\\ & -(s+N)\log (s+N)+(s+N)+(s+1)\log (s+1)\\ & -(s+1)-{\displaystyle \frac{1}{2}}\log (s+1)-{\displaystyle \frac{1}{2}}\log (s+N)-{\int }_1^N{\displaystyle \frac{{\bar{B}}_1(x)\mathrm{d}x}{s+x}}]\\ =& (s+{\displaystyle \frac{1}{2}})\log (s+1)+{\int }_1^{\mathrm{\infty }}{\displaystyle \frac{{\bar{B}}_1(x)\mathrm{d}x}{x}}-{\int }_1^N{\displaystyle \frac{{\bar{B}}_1(x)\mathrm{d}x}{s+x}}\\ & +\underset{n\to \mathrm{\infty }}{lim}[s\log (N+1)+(N+{\displaystyle \frac{1}{2}})\log N\\ & -(s+N+{\displaystyle \frac{1}{2}})\log (s+N)]\\ =& (s+{\displaystyle \frac{1}{2}})\log (s+1)+(A-1)-{\int }_1^{\mathrm{\infty }}{\displaystyle \frac{{\bar{B}}_1(x)\mathrm{d}x}{s+x}}\\ & +lim[s\log {\displaystyle \frac{N+1}{2}}-(N+{\displaystyle \frac{1}{2}})\log (1+{\displaystyle \frac{s\phantom{1}}{2}})]\end{array}\end{array}$$