华里士公式的推导及其推广

基础知识

华里士公式

In=∫0π2sin⁡nxdx=∫0π2cos⁡nxdx={n−1nn−3n−2⋯23nisodd,n−1nn−3n−2⋯12π2niseven\Large \begin{aligned} I_n = \int_{0}^{\frac{\pi}{2}} \sin^n{x} \mathrm{d}x = \int_{0}^{\frac{\pi}{2}} \cos^n{x} \mathrm{d}x = \begin{cases} \frac{n-1}{n} \frac{n-3}{n-2} \cdots \frac{2}{3} &&n\ is\ odd,\\ \\ \frac{n-1}{n} \frac{n-3}{n-2} \cdots \frac{1}{2} \frac{\pi}{2} &&n\ is\ even \end{cases} \end{aligned} In=∫02πsinnxdx=∫02πcosnxdx=⎩⎪⎪⎪⎪⎨⎪⎪⎪⎪⎧nn−1n−2n−3⋯32nn−1n−2n−3⋯212πn is odd,n is even

基础公式的推导

仅以sin⁡nx\sin^n{x}sinnx为例，有
In=−∫0π2sin⁡n−1xdcos⁡x=−sin⁡n−1x⋅cos⁡x∣0π2+∫0π2(n−1)⋅cos⁡2x⋅sin⁡n−2xdx=(n−1)∫0π2(1−sin⁡2x)⋅sin⁡n−2xdx=(n−1)In−2−(n−1)In\large \begin{aligned} I_n &= -\int_{0}^{\frac{\pi}{2}} \sin^{n-1}{x} \mathrm{d}{\cos{x}} \\[2ex] &= -\left. \sin^{n-1}x\cdot\cos{x} \right|_{0}^{\frac{\pi}{2}} + \int_{0}^{\frac{\pi}{2}} (n - 1)\cdot\cos^2{x} \cdot \sin^{n-2}{x} \mathrm{d}{x} \\[2ex] &= (n-1) \int_{0}^{\frac{\pi}{2}} (1-\sin^2{x}) \cdot \sin^{n-2}{x} \mathrm{d}x \\[2ex] &= (n-1) I_{n-2} - (n-1) I_n \end{aligned} In=−∫02πsinn−1xdcosx=−sinn−1x⋅cosx∣∣∣02π+∫02π(n−1)⋅cos2x⋅sinn−2xdx=(n−1)∫02π(1−sin2x)⋅sinn−2xdx=(n−1)In−2−(n−1)In
即
In=n−1nIn−2\Large I_n = \frac{n-1}{n}I_{n-2} In=nn−1In−2
分情况讨论，求出通项公式，即可得原式成立。

华里士公式的推广

推广公式

∫0m2πsin⁡nxdx={mInniseven,Innisodd,m=2k+1,2Innisodd,m=4k+2,0nisodd,m=4k\Large \begin{aligned} &\int_{0}^{\frac{m}{2}{\pi}} \sin^n{x} \mathrm{d}x = \begin{cases} m I_n &n\ is\ even,\\ \\ I_n &n\ is\ odd,\ \ m=2k+1,\\ \\ 2I_n &n\ is\ odd,\ \ m=4k+2,\\ \\ 0 &n\ is\ odd,\ \ m=4k \end{cases} \end{aligned} ∫02mπsinnxdx=⎩⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎧mInIn2In0n is even,n is odd, m=2k+1,n is odd, m=4k+2,n is odd, m=4k

∫0m2πcos⁡nxdx={mInniseven,Innisodd,m=4k+1,−Innisodd,m=4k+3,0nisodd,m=2k\Large \begin{aligned} &\int_{0}^{\frac{m}{2}{\pi}} \cos^n{x} \mathrm{d}x = \begin{cases} m I_n &n\ is\ even,\\ \\ I_n &n\ is\ odd,\ \ m=4k+1,\\ \\ -I_n &n\ is\ odd,\ \ m=4k+3,\\ \\ 0 &n\ is\ odd, \ \ m=2k \end{cases} \end{aligned} ∫02mπcosnxdx=⎩⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎧mInIn−In0n is even,n is odd, m=4k+1,n is odd, m=4k+3,n is odd, m=2k

当 m=1m=1m=1 时，该公式退化为原华里士公式。

推广公式的图像理解

这是 f(x)=sin⁡3xf(x)=\sin^3xf(x)=sin3x 的图像（对应n为奇数时的情况）：

由图可知，f(x)f(x)f(x) 是以 2π2\pi2π 为周期的周期函数，且函数在[0,π2][0, \frac{\pi}{2}][0,2π] 与 [π2,π][\frac{\pi}{2}, \pi][2π,π] 的部分各自与x轴围成的面积相等，函数在[0,π][0, \pi][0,π] 与 [π,2π][\pi, 2\pi][π,2π] 的部分各自与x轴围成的面积也相等。

这是 f(x)=sin⁡4xf(x)=\sin^4xf(x)=sin4x 的图像（对应n为偶数时的情况）：

由图可知，f(x)f(x)f(x) 是以 π\piπ 为周期的周期函数，且函数在[0,π2][0, \frac{\pi}{2}][0,2π] 与 [π2,π][\frac{\pi}{2}, \pi][2π,π] 的部分各自与x轴围成的面积相等。

故无论m的值为多少，积分值都是 [0,π2][0, \frac{\pi}{2}][0,2π] 的 mmm 倍。

推广公式的特例

m=2

∫0πsin⁡nxdx=2In∫0πcos⁡nxdx={2Inniseven0nisodd\Large \begin{aligned} &\int_{0}^{\pi} \sin^n{x} \mathrm{d}x = 2I_n \\[4ex] \Large &\int_{0}^{\pi} \cos^n{x} \mathrm{d}x = \begin{cases} 2I_n \quad &n\ is\ even \\[2ex] 0 \quad &n\ is\ odd \end{cases} \end{aligned} ∫0πsinnxdx=2In∫0πcosnxdx=⎩⎪⎪⎪⎨⎪⎪⎪⎧2In0n is evenn is odd

m=4

∫02πsin⁡nxdx=∫02πcos⁡nxdx={4Inniseven0nisodd\Large \begin{aligned} \int_{0}^{2\pi} \sin^n{x} \mathrm{d}x = \int_{0}^{2\pi} \cos^n{x} \mathrm{d}x = \begin{cases} 4I_n \quad &n\ is\ even \\[2ex] 0 \quad &n\ is\ odd \end{cases} \end{aligned} ∫02πsinnxdx=∫02πcosnxdx=⎩⎪⎪⎪⎨⎪⎪⎪⎧4In0n is evenn is odd