用定积分的定义计算数列和时怎么确定积分上下限？

一、题目

$$
\begin{aligned}
I_{1} & = \lim_{n \rightarrow \infty} \ln \sqrt[n]{\left(1 + \frac{1}{n} \right)^{2} \left(1 + \frac{2}{n} \right)^{2} \cdots \left(1 + \frac{n}{n} \right)^{2}} \\ \\
I_{2} & = \lim_{n \rightarrow \infty} \ln \sqrt[n]{\left(1 + \frac{1}{n} \right)^{2} \left(1 + \frac{2}{n} \right)^{2} \cdots \left(1 + \frac{n}{n} \right)^{2} \cdots \textcolor{orange}{ \left(1 + \frac{2n}{n} \right)^{2} } } \\ \\
\end{aligned}
$$

二、解析

$I_{1}$

$$
\begin{aligned}
I_{1} & = \lim_{n \rightarrow \infty} \ln \sqrt[n]{\left(1 + \frac{1}{n} \right)^{2} \left(1 + \frac{2}{n} \right)^{2} \cdots \left(1 + \frac{n}{n} \right)^{2} } \\ \\
& = \lim_{n \rightarrow \infty} \frac{2}{n} \ln \left(1 + \frac{1}{n} \right) \left(1 + \frac{2}{n} \right) \cdots \left(1 + \frac{n}{n} \right) \\ \\
\textcolor{yellow}{\Large{\boldsymbol{\star}}} & = 2 \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{\textcolor{lightgreen}{i = 1}}^{\textcolor{lightgreen}{n}} \ln \left(1 + \textcolor{lightgreen}{\frac{i}{n}} \right) \\ \\
\textcolor{yellow}{\Large{\boldsymbol{\star}}} & = 2 \int_{\textcolor{pink}{0}}^{\textcolor{pink}{1}} \ln(1 + x) \mathrm{~d} x \\ \\
& = 2 \int_{0}^{1} \ln(1 + x) \mathrm{~d}(x+1) \\ \\
& \textcolor{lightgreen}{ \leadsto } \textcolor{gray}{t = x + 1} \textcolor{lightgreen}{ \leadsto } \textcolor{gray}{t \in (0, 1+1)} \\ \\
& = 2 \int_{1}^{2} \ln t \mathrm{~d} t \\ \\
& \textcolor{lightgreen}{ \leadsto } \textcolor{gray}{x = t} \\ \\
& = \textcolor{springgreen}{\boldsymbol{ 2 \int_{1}^{2} \ln x \mathrm{~d} x }}
\end{aligned}
$$

在上面的计算中，$2 \int_{\textcolor{pink}{0}}^{\textcolor{pink}{1}} \ln(1 + x) \mathrm{~d} x$ 中积分上下限的确定，是通过 $2 \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{\textcolor{lightgreen}{i = 1}}^{\textcolor{lightgreen}{n}} \ln \left(1 + \textcolor{lightgreen}{\frac{i}{n}} \right)$ 得到的：

由于 $\textcolor{lightgreen}{i}$ 的取值是从 $\textcolor{lightgreen}{1}$ 到 $\textcolor{lightgreen}{n}$, 因此，当 $n \rightarrow \infty$ 的时候，$\textcolor{lightgreen}{\frac{i}{n}}$ 的取值就是从 $\lim_{n \rightarrow \infty} \frac{1}{n} = \textcolor{pink}{0}$ 到 $\lim_{n \rightarrow \infty} \frac{n}{n} = \textcolor{pink}{1}$.

$I_{2}$

$$
\begin{aligned}
I_{1} & = \lim_{n \rightarrow \infty} \ln \sqrt[n]{\left(1 + \frac{1}{n} \right)^{2} \left(1 + \frac{2}{n} \right)^{2} \cdots \left(1 + \frac{n}{n} \right)^{2} \cdots \textcolor{orange}{ \left(1 + \frac{2n}{n} \right)^{2} } } \\ \\
& = \lim_{n \rightarrow \infty} \frac{2}{n} \ln \left(1 + \frac{1}{n} \right) \left(1 + \frac{2}{n} \right) \cdots \left(1 + \frac{n}{n} \right) \cdots \textcolor{orange}{ \left(1 + \frac{2n}{n} \right) } \\ \\
\textcolor{yellow}{\Large{\boldsymbol{\star}}} & = 2 \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{\textcolor{lightgreen}{i = 1}}^{\textcolor{lightgreen}{2n}} \ln \left(1 + \textcolor{lightgreen}{\frac{i}{n}} \right) \\ \\
\textcolor{yellow}{\Large{\boldsymbol{\star}}} & = 2 \int_{\textcolor{pink}{0}}^{\textcolor{pink}{2}} \ln(1 + x) \mathrm{~d} x \\ \\
& = 2 \int_{0}^{2} \ln(1 + x) \mathrm{~d}(x+1) \\ \\
& \textcolor{lightgreen}{ \leadsto } \textcolor{gray}{t = x + 1} \textcolor{lightgreen}{ \leadsto } \textcolor{gray}{t \in (0, 2+1)} \\ \\
& = 2 \int_{1}^{3} \ln t \mathrm{~d} t \\ \\
& \textcolor{lightgreen}{ \leadsto } \textcolor{gray}{x = t} \\ \\
& = \textcolor{springgreen}{\boldsymbol{ 2 \int_{1}^{3} \ln x \mathrm{~d} x }}
\end{aligned}
$$

在上面的计算中，$2 \int_{\textcolor{pink}{0}}^{\textcolor{pink}{2}} \ln(1 + x) \mathrm{~d} x$ 中积分上下限的确定，是通过 $2 \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{\textcolor{lightgreen}{i = 1}}^{\textcolor{lightgreen}{2n}} \ln \left(1 + \textcolor{lightgreen}{\frac{i}{n}} \right)$ 得到的：

由于 $\textcolor{lightgreen}{i}$ 的取值是从 $\textcolor{lightgreen}{1}$ 到 $\textcolor{lightgreen}{2n}$, 因此，当 $n \rightarrow \infty$ 的时候，$\textcolor{lightgreen}{\frac{i}{n}}$ 的取值就是从 $\lim_{n \rightarrow \infty} \frac{1}{n} = \textcolor{pink}{0}$ 到 $\lim_{n \rightarrow \infty} \frac{2n}{n} = \textcolor{pink}{2}$.

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