升级版的点火公式

一、题目

$$
K_{m} = \int_{0}^{\pi} x \sin ^{m} x \mathrm{~d} x = ?
$$

其中，$m$ 为自然数。

难度评级：

二、解析

为了解答本题，我们还需要知道什么是“自然数”——自然数的另一个名字是“非负整数”，也就是 $0$, $1$, $2$, $3$, $\cdots$

$$
\begin{aligned}
K_{m} = & \ \int_{0}^{\pi} x \sin ^{m} x \mathrm{~d} x \\ \\
= & \ \int_{0}^{\pi} x \sin ^{m-1} x (\textcolor{pink}{\sin x}) \mathrm{~d} x \\ \\
= & \ \textcolor{pink}{-} \int_{0}^{\pi} x \sin ^{m-1} x \mathrm{~d} (\textcolor{pink}{\cos x}) \\ \\
= & \ \textcolor{magenta}{ – x \sin ^{m-1} x (\cos x) \Big|_{0}^{\pi} } + \int_{0}^{\pi} \cos x \mathrm{~d} (x \sin ^{m-1} x) \\ \\
= & \ \textcolor{magenta}{ 0 } + \int_{0}^{\pi} \cos x \mathrm{~d} (x \sin ^{m-1} x) \\ \\
= & \ \int_{0}^{\pi} \cos x \mathrm{~d} (x \sin ^{m-1} x) \\ \\
= & \ \int_{0}^{\pi} \cos x \left[ \sin ^{m-1} x + x(m-1) \sin ^{m-2} x (\cos x) \right] \mathrm{~d} x \\ \\
= & \ \int_{0}^{\pi} \textcolor{orangered}{\cos x} (\sin ^{m-1} x) \mathrm{~d} x + \int_{0}^{\pi} x (m – 1) \sin ^{m-2} x (\textcolor{springgreen}{ \cos ^{2} x }) \mathrm{~d} x \\ \\
= & \ \int_{0}^{\pi} (\sin ^{m-1} x) \mathrm{~d} (\textcolor{orangered}{\sin x}) + \int_{0}^{\pi} x (\textcolor{tan}{ m – 1 }) \sin ^{m-2} x (\textcolor{springgreen}{ 1 – \sin ^{2} x }) \mathrm{~d} x \\ \\
= & \ \int_{0}^{\pi} (\sin ^{m-1} x) \mathrm{~d} (\sin x) + (\textcolor{tan}{ m – 1 }) \int_{0}^{\pi} x \sin ^{m-2} x (1 – \sin ^{2} x) \mathrm{~d} x \\ \\
= & \ \textcolor{yellow}{ \frac{1}{m} \sin ^{m} x \Big|_{0}^{\pi} } + (m-1) \int_{0}^{\pi} x \sin ^{m-2} x \mathrm{~d} x – (m-1) \int_{0}^{\pi} x \sin ^{m} x \mathrm{~d} x \\ \\
= & \ \int_{0}^{\pi} (\sin ^{m-1} x) \mathrm{~d} (\sin x) + (\textcolor{tan}{ m – 1 }) \int_{0}^{\pi} x \sin ^{m-2} x (1 – \sin ^{2} x) \mathrm{~d} x \\ \\
= & \ \textcolor{yellow}{ 0 } + (m-1) \int_{0}^{\pi} x \sin ^{m-2} x \mathrm{~d} x – (m-1) \int_{0}^{\pi} x \sin ^{m} x \mathrm{~d} x \\ \\
= & \ (m-1) \textcolor{orange}{ \int_{0}^{\pi} x \sin ^{m-2} x \mathrm{~d} x } – (m-1) \textcolor{purple}{ \int_{0}^{\pi} x \sin ^{m} x \mathrm{~d} x } \\ \\
= & \ (m-1) \textcolor{orange}{ K_{m-2} } – (m-1) \textcolor{purple}{ K_{m} }
\end{aligned}
$$

即：

$$
\textcolor{springgreen}{
K_{m} = (m-1) K_{m-2} – (m-1) K_{m}
}
$$

接着：

$$
\begin{aligned}
& K_{m} = (m-1) K_{m-2} – (m-1) K_{m} \\ \\
\Rightarrow & \ K_{m} + (m-1)K_{m} = (m-1) K_{m-2} \\ \\
\Rightarrow & \ m K_{m} = (m-1) K_{m-2} \\ \\
\Rightarrow & \ \textcolor{springgreen}{ K_{m} = \frac{m-1}{m} K_{m-2} }
\end{aligned}
$$

递推可知：

$$
\begin{aligned}
K_{m-2} & = \frac{(m-2) – 1}{m-2} K_{(m-2)-2} \\ \\
& = \frac{m-3}{m-2} K_{m-4} \\ \\
\textcolor{gray}{\downarrow} \\ \\
K_{m-4} & = \frac{(m-2) – 3}{(m-2) – 2} K_{(m-2)-4} \\ \\
& = \frac{m-5}{m-4} K_{m-6} \\ \\
\textcolor{gray}{\downarrow} \\ \\
K_{m-6} & = \frac{m-7}{m-6} K_{m-8}
\end{aligned}
$$

观察上面的式子可知，从 $K_{m-2}$ 到 $K_{m-4}$, 以及到 $K_{m-6}$, 每次都是减去 $2$, 又因为偶数减去 $2$ 一定得偶数，奇数减去 $2$ 一定得奇数，而 $0$ 是一个偶数，说明 $K_{0}$ 的下标 $0$ 只能通过偶数减 $2$ 得到；$1$ 是一个奇数，说明 $K_{1}$ 的下标只能通过奇数减 $2$ 得到——于是可知，对于 $K_{m}$ 的表达式，我们应该分成 $m$ 为奇数和 $m$ 为偶数两部分表述。

此外需要注意的是，由于 $m$ 是自然数，也就是非负整数，所以，下标最小的 $K_{m}$ 是 $K_{0}$, 之后是 $K_{1}$, 并不存在 $K_{-1}$ 或者 $K_{-2}$ 等下标为负数的 $K_{m}$.

于是可知：

$$
\begin{aligned}
K_{m} & = \frac{m-1}{m} K_{m-2} \\ \\
& = \begin{cases}
\frac{m-1}{m} \cdot \frac{m-3}{m-2} \cdot \frac{m-5}{m-4} \cdot \frac{m-7}{m-6} \cdots \frac{3}{4} \cdot \frac{1}{2} \cdot K_{0}, & m \text{ 为偶数} \\ \\
\frac{m-1}{m} \cdot \frac{m-3}{m-2} \cdot \frac{m-5}{m-4} \cdot \frac{m-7}{m-6} \cdots \frac{4}{5} \cdot \frac{2}{3} \cdot K_{1}, & m \text{ 为奇数}
\end{cases}
\end{aligned}
$$

又因为：

$$
\textcolor{yellow}{
\begin{aligned}
K_{0} & = \int_{0}^{\pi} x \mathrm{~d} x \\ \\
& = \frac{1}{2} x^{2} \Big|_{0}^{\pi} \\ \\
& = \textcolor{springgreen}{ \frac{\pi ^{2}}{2} } \\ \\ \\
K_{1} & = \int_{0}^{\pi} x \sin x \mathrm{~d} x \\ \\
& = – \int_{0}^{\pi} x \mathrm{~d} (\cos x) \\ \\
& = -x \cos x \Big|_{0}^{\pi} + \int_{0}^{\pi} \cos x \mathrm{~d} x \\ \\
& = 0 + (- \sin x) \Big|_{0}^{\pi} \\ \\
& = \textcolor{springgreen}{ \pi }
\end{aligned}
}
$$

综上可知：

$$
\textcolor{springgreen}{
\boldsymbol{
\begin{aligned}
K_{m} & = \int_{0}^{\pi} x \sin ^{m} x \mathrm{~d} x\\ \\
& = \begin{cases}
\frac{m-1}{m} \cdot \frac{m-3}{m-2} \cdot \frac{m-5}{m-4} \cdot \frac{m-7}{m-6} \cdots \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi^{2}}{2}, & m \text{ 为偶数} \\ \\
\frac{m-1}{m} \cdot \frac{m-3}{m-2} \cdot \frac{m-5}{m-4} \cdot \frac{m-7}{m-6} \cdots \frac{4}{5} \cdot \frac{2}{3} \cdot \pi, & m \text{ 为奇数}
\end{cases}
\end{aligned}
}
}
$$

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