2019年考研数二第19题解析：波纹函数、定积分累加求和、等比数列

一、题目

难度评级：

二、解析

根据「荒原之梦考研数学」的文章《周期函数的兄弟：波纹函数》和《单路径约束法：快速判断一个函数是否为周期函数》可知，题目中的函数 $y = \mathrm{e}^{-x} \sin x$ 是一个波纹函数。

接着，当 $k$ $=$ $0$, $1$, $2$, $\cdots$ 时，有：

$$
\textcolor{yellow}{
\sin \left( k \pi \right) = 0
}
$$

于是：

$$
\textcolor{violet}{
\mathrm{e}^{-k \pi} \sin \left( k \pi \right) = 0
}
$$

进一步分析可知，当 $k$ 为奇数时，$k + 1$ 为偶数，此时：

$$
\begin{aligned}
& \int_{k \pi}^{(k+1) \pi} \mathrm{e}^{-x} \sin x \mathrm{~d} x < 0 \\ \\ \textcolor{lightgreen}{ \leadsto } \ & \textcolor{yellow}{ \left( -1 \right)^{k} < 0 } \\ \\ \textcolor{lightgreen}{ \leadsto } \ \textcolor{lightgreen}{ \left( -1 \right)^{k} } & \textcolor{lightgreen}{ \int_{k \pi}^{(k+1) \pi} \mathrm{e}^{-x} \sin x \mathrm{~d} x > 0 }
\end{aligned}
$$

当 $k$ 为偶数时，$k + 1$ 为奇数，此时：

$$
\begin{aligned}
& \int_{k \pi}^{(k+1) \pi} \mathrm{e}^{-x} \sin x \mathrm{~d} x > 0 \\ \\
\textcolor{lightgreen}{ \leadsto } \ & \textcolor{yellow}{ \left( -1 \right)^{k} > 0 } \\ \\
\textcolor{lightgreen}{ \leadsto } \ \textcolor{lightgreen}{ \left( -1 \right)^{k} } & \textcolor{lightgreen}{ \int_{k \pi}^{(k+1) \pi} \mathrm{e}^{-x} \sin x \mathrm{~d} x > 0 }
\end{aligned}
$$

又因为，当 $\textcolor{orange}{ k = 0 }$ 时，$\textcolor{magenta}{ k \pi = 0 }$, 当 $\textcolor{orange}{ k = n-1 }$ 时，$\textcolor{magenta}{ (k + 1) \pi = n \pi }$.

于是，在区间 $\left( 0, n \pi \right)$ 上，有：

$$
S_{n} = \sum_{\textcolor{orange}{ k = 0 }}^{\textcolor{orange}{ n-1 }}(-1)^{k} \int_{\textcolor{magenta}{ k \pi }}^{\textcolor{magenta}{ (k+1) \pi }} \mathrm{e}^{-x} \sin x \mathrm{~d} x
$$

接着，根据「荒原之梦考研数学」的《考研数学解题思路积累：和 $\mathrm{e}^{x}$ 有关的那些式子》这篇文章可知：

$$
\textcolor{yellow}{
\int \mathrm{e}^{-x} \sin x \mathrm{~d} x = \frac{-1}{2} \mathrm{e}^{-x} ( \sin x + \cos x )
}
$$

于是：

$$
\begin{aligned}
\textcolor{springgreen}{\boldsymbol{ S_{n} }} & = \sum_{k = 0}^{n-1} (-1)^{k} \int_{k \pi }^{(k+1) \pi} \textcolor{yellow}{ \mathrm{e}^{-x} \sin x } \mathrm{~d} x \\ \\
& = \sum_{k = 0}^{n−1} ( −1 )^{k} \left[ \textcolor{yellow}{ \frac{-1}{2} \mathrm{e}^{−x}\left(\sin x + \cos x\right) } \right] \Bigg|_{k ⁢⁢⁢\pi }^{ (k + 1)⁢ ⁢\pi } \\ \\
& = \frac{1}{2} \sum_{k = 0}^{n – 1} \textcolor{orangered}{ (-1)^{k+1} } \left[ \mathrm{e}^{-(k + 1) \pi} \textcolor{orangered}{ (-1)^{k+1} } – \mathrm{e}^{- k \pi} \textcolor{orangered}{ (-1)^{k} } \right] \\ \\
& = \frac{1}{2} \sum_{k = 0}^{n – 1} \textcolor{orangered}{ (-1)^{k+1+k+1 + k} } \left[ \mathrm{e}^{-(k + 1) \pi} – \mathrm{e}^{- k \pi} \right] \\ \\
& = \frac{1}{2} \sum_{k = 0}^{n – 1} \textcolor{orangered}{ (-1)^{3k+2} } \left[ \mathrm{e}^{-(k + 1) \pi} – \mathrm{e}^{- k \pi} \right] \\ \\
& = \frac{1}{2} \sum_{k = 0}^{n – 1} \textcolor{orangered}{ 1 } \left[ \mathrm{e}^{-(k + 1) \pi} – \mathrm{e}^{- k \pi} \right] \\ \\
& = \frac{1}{2} \textcolor{#c4cbcf}{ \sum_{k = 0}^{n – 1} } \left[ \mathrm{e}^{-(k+1) \pi } + \mathrm{e}^{- k \pi} \right] \\ \\
& = \frac{1}{2} \left[ \textcolor{#c4cbcf}{ \sum_{k = 0}^{n – 1} } \textcolor{#57c3c2}{ \mathrm{e}^{-(k+1) \pi } } + \textcolor{#c4cbcf}{ \sum_{k = 0}^{n – 1} } \mathrm{e}^{- k \pi} \right] \\ \\
& \textcolor{lightgreen}{ \leadsto } \textcolor{gray}{f = k + 1} \textcolor{lightgreen}{ \leadsto } \textcolor{gray}{ k \in \left( 0, n-1 \right) } \textcolor{lightgreen}{ \leadsto } \textcolor{gray}{ f \in \left( 1, n \right) } \\ \\
& = \frac{1}{2} \left[ \textcolor{#f0f5e5}{ \sum_{f = 1}^{n} } \textcolor{#57c3c2}{ \mathrm{e}^{- f \pi } } + \textcolor{#c4cbcf}{ \sum_{k = 0}^{n – 1} } \textcolor{#fed71a}{ \mathrm{e}^{- k \pi} } \right] \\ \\
& = \frac{1}{2} \left[ \textcolor{#f0f5e5}{ \sum_{f = 1}^{n} } \mathrm{e}^{- f \pi } + \textcolor{#fed71a}{\mathrm{e}^{0}} + \textcolor{#f0f5e5}{ \sum_{k = 1}^{n – 1} } \textcolor{#fed71a}{ \mathrm{e}^{- k \pi} } \right] \\ \\
& \textcolor{lightgreen}{ \leadsto } \textcolor{gray}{k = f} \\ \\
& = \frac{1}{2} \left[ \textcolor{#b2cf87}{ \sum_{k = 1}^{n} \mathrm{e}^{- k \pi } } + \textcolor{#fed71a}{1} + \textcolor{#f0f5e5}{ \sum_{k = 1}^{n – 1} } \mathrm{e}^{- k \pi} \right] \\ \\
& = \frac{1}{2} \left[ \textcolor{#b2cf87}{ \sum_{k = 1}^{n-1} \mathrm{e}^{- k \pi } } + \textcolor{#b2cf87}{\mathrm{e}^{- n \pi}} + \textcolor{#fed71a}{1} + \textcolor{#f0f5e5}{ \sum_{k = 1}^{n – 1} } \mathrm{e}^{- k \pi} \right] \\ \\
& = \frac{1}{2} \left[ 1 + \textcolor{violet}{ 2 \sum_{k = 1}^{n – 1} \mathrm{e}^{- k \pi} } \ \textcolor{white}{\colorbox{red}{+}} \ \mathrm{e}^{- n \pi} \right] \\ \\
& = \frac{1}{2} \left[ 1 + \textcolor{violet}{ 2 \sum_{k = 1}^{n} \mathrm{e}^{- k \pi} – 2 \mathrm{e}^{- n \pi} } \ \textcolor{white}{\colorbox{red}{+}} \ \mathrm{e}^{- n \pi} \right] \\ \\
& = \frac{1}{2} \left[ 1 + 2 \textcolor{pink}{ \sum_{k = 1}^{n} \mathrm{e}^{- k \pi} } \ \textcolor{black}{\colorbox{lightgreen}{-}} \ \mathrm{e}^{- n \pi} \right]
\end{aligned}
$$

又因为 $\textcolor{pink}{ \sum_{k = 1}^{n} \mathrm{e}^{- k \pi} }$ 展开之后其实是一个首项 $a_{1}$ $=$ $\mathrm{e}^{- \pi}$, 公比 $q$ $=$ $\mathrm{e}^{- \pi}$ 的等比数列的求和，即：

$$
\textcolor{pink}{ \sum_{k = 1}^{n} \mathrm{e}^{- k \pi} } = \mathrm{e}^{- \pi} + \mathrm{e}^{-2 \pi} + \mathrm{e}^{-3 \pi} + \cdots + \mathrm{e}^{- n \pi}
$$

所以，根据等比数列的求和公式可知：

$$
\textcolor{pink}{ \sum_{k = 1}^{n} \mathrm{e}^{- k \pi} } = \frac{a_{1} \cdot \left( 1 – q^{n} \right)}{1 – q} = \textcolor{pink}{ \frac{ \mathrm{e}^{- \pi} \left( 1 – \mathrm{e}^{- n \pi} \right) }{1 – \mathrm{e}^{- \pi}} }
$$

于是：

$$
\begin{aligned}
\textcolor{springgreen}{\boldsymbol{ S_{n} }} & = \frac{1}{2} \left[ 1 + 2 \textcolor{pink}{ \sum_{k = 1}^{n} \mathrm{e}^{- k \pi} } \ \textcolor{black}{\colorbox{lightgreen}{-}} \ \mathrm{e}^{- n \pi} \right] \\ \\
& = \frac{1}{2} \left[ 1 + 2 \cdot \textcolor{pink}{ \frac{ \mathrm{e}^{- \pi} \left( 1 – \mathrm{e}^{- n \pi} \right) }{1 – \mathrm{e}^{- \pi}} } \ – \ \mathrm{e}^{- n \pi} \right] \\ \\
\end{aligned}
$$

综上：

$$
\begin{aligned}
\lim_{n \rightarrow \infty} \boldsymbol{\textcolor{springgreen}{ S_{n} }} & = \lim_{n \rightarrow \infty} \frac{1}{2} \left[ 1 + 2 \cdot \frac{ \mathrm{e}^{- \pi} \left(1 – \mathrm{e}^{- n \pi} \right)}{1 – \mathrm{e}^{- \pi}} – \mathrm{e}^{- n \pi} \right] \\ \\
& = \frac{1}{2} + \lim_{n \rightarrow \infty} \frac{ \mathrm{e}^{- \pi} \left(1 – \mathrm{e}^{- n \pi} \right)}{1 – \mathrm{e}^{- \pi}} – \lim_{n \rightarrow \infty} \mathrm{e}^{- n \pi} \\ \\
& = \frac{1}{2} + \frac{ \mathrm{e}^{- \pi} \left(1 – \textcolor{magenta}{ \lim_{n \rightarrow \infty} \mathrm{e}^{- n \pi} } \right)}{1 – \mathrm{e}^{- \pi}} – \textcolor{magenta}{ \lim_{n \rightarrow \infty} \mathrm{e}^{- n \pi} } \\ \\
& = \frac{1}{2} + \frac{ \mathrm{e}^{- \pi} \left(1 – \textcolor{magenta}{ 0 } \right)}{1 – \mathrm{e}^{- \pi}} – \textcolor{magenta}{ 0 } \\ \\
& = \frac{1}{2} + \frac{ \mathrm{e}^{- \pi} }{1 – \mathrm{e}^{- \pi}} \\ \\
& = \frac{1}{2} + \frac{ \frac{1}{\mathrm{e}^{\pi}} }{1 – \frac{1}{\mathrm{e}^{\pi}}} \\ \\
& = \frac{1}{2} + \frac{1}{\textcolor{yellow}{ \mathrm{e}^{\pi} }} \cdot \frac{\textcolor{yellow}{ \mathrm{e}^{\pi} }}{\mathrm{e}^{\pi} – 1} \\ \\
& = \boldsymbol{\textcolor{springgreen}{ \frac{1}{2} + \frac{1}{\mathrm{e}^{\pi} – 1} }}
\end{aligned}
$$

即：

$$
\boldsymbol{\textcolor{springgreen}{ S_{n} = \frac{1}{2} + \frac{1}{\mathrm{e}^{\pi} – 1} }}
$$

拓展资料

事实上，本题中的曲线函数 $y = \mathrm{e}^{-x} \sin x$ 的函数图像示意图如图 01 所示，其在区间 $\left( 0, n \pi \right)$ 上的函数图像看上去就像是一条“直线”：

即便我们将 $y = \mathrm{e}^{-x} \sin x$ 在区间 $\left( 0, n \pi \right)$ 上的函数图象再放大一些，其仍然看上去像是一条直线，如图 02 所示：

不过，当我们继续放大，就会看到波纹函数 $y = \mathrm{e}^{-x} \sin x$ “浅浅”的“波纹”了，如图 03 所示：

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