试计算：
\[\int_{0}^{1}\int_{0}^{1}\cdots\int_{0}^{1}\dfrac{1}{1-x_1x_2\cdots x_n}\text{d}x_1\text{d}x_2\cdots\text{d}x_n
\]
解法
$$\begin{aligned}I=&\int_{0}^{1}\int_{0}^{1}\cdots\int_{0}^{1}\sum\limits_{i=0}(x_1x_2\cdots x_n)^i\text{d}x_1\text{d}x_2\cdots\text{d}x_n\\
=&\sum\limits_{i=0}\int_{0}^{1}\int_{0}^{1}\cdots\int_{0}^{1}(x_1x_2\cdots x_n)^i\text{d}x_1\text{d}x_2\cdots\text{d}x_n\\
=&\sum\limits_{i=0}\left(\int_0^1x^i\text{d}x\right)^n\\
=&\sum\limits_{i=0}\left(\dfrac{1}{i+1}\right)^{n}=\zeta(n)\end{aligned}$$
试计算：
\[\int_{0}^{1}\left\{\dfrac{1}{1-x}\right\}\text{d}x
\]
解法
$$\begin{aligned}I=&\int_{0}^1\left(\dfrac{1}{x}-\left\lfloor\dfrac{1}{x}\right\rfloor\right)\text{d}x\\
=&\lim\limits_{n\rightarrow+\infty}\int_{\frac{1}{n}}^1\left(\dfrac{1}{x}-\left\lfloor\dfrac{1}{x}\right\rfloor\right)\text{d}x\\
=&\lim\limits_{n\rightarrow+\infty}\left(\ln1-\ln\dfrac{1}{n}-\int_{\frac{1}{n}}^{1}\sum\limits_{i=1}\left[\left\lfloor\dfrac{1}{x}\right\rfloor\ge i\right]\right)\\
=&\lim\limits_{n\rightarrow+\infty}\left(\ln1-\ln\dfrac{1}{n}-\sum\limits_{i=1}^{n}\left(\dfrac{1}{i}-\dfrac{1}{n}\right)\right)\\
=&\lim\limits_{n\rightarrow+\infty}\left(\ln n-H(n)+1\right)\\
=&1-\gamma\end{aligned}$$
