推导过程参考自陈希孺《数理统计学教程》1.4节，在原文基础上补充了一些细节。
文章目录
预备知识
标准正态分布
- 概率密度函数： φ ( x ) = 1 2 π e − x 2 2 \varphi(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}} φ(x)=2π1e−2x2
- 分布函数： ϕ ( x ) = ∫ − ∞ x 1 2 π e − t 2 2 d t \phi(x)=\int_{-\infty}^{x}\frac{1}{\sqrt{2\pi}}e^{-\frac{t^2}{2}}\text{d} t ϕ(x)=∫−∞x2π1e−2t2dt
Γ \Gamma Γ函数（伽马函数）
- 定义： Γ ( s ) = ∫ 0 + ∞ e − t t s − 1 d t \Gamma(s)=\int_{0}^{+\infty}e^{-t}t^{s-1}\text{d} t Γ(s)=∫0+∞e−tts−1dt
- 递推公式： Γ ( s + 1 ) = s Γ ( s ) , ( s > 0 ) \Gamma(s+1)=s\Gamma(s),\space(s>0) Γ(s+1)=sΓ(s), (s>0)
- 几个重要的值： Γ ( 1 ) = 1 \Gamma(1)=1 Γ(1)=1， Γ ( 1 2 ) = π \Gamma(\frac{1}{2})=\sqrt{\pi} Γ(21)=π
推导目标
已知有 n n n个独立同分布的随机变量 Y 1 , Y 2 , ⋯ , Y n Y_1,Y_2,\cdots,Y_n Y1,Y2,⋯,Yn，它们均服从标准正态分布 N ( 0 , 1 ) N(0,1) N(0,1)。
求 X = ∑ i = 1 n Y i 2 X=\sum\limits_{i=1}^{n}Y_i^2 X=i=1∑nYi2的概率密度函数。
推导过程
当 x ⩽ 0 x \leqslant 0 x⩽0时，显然有 X = ∑ i = 1 n Y i 2 ⩾ 0 ⩾ x X=\sum\limits_{i=1}^{n}Y_i^2 \geqslant 0 \geqslant x X=i=1∑nYi2⩾0⩾x。于是 P ( X < x ) = 0 P(X < x) = 0 P(X<x)=0， p ( x ) = 0 p(x) = 0 p(x)=0。下面针对 x > 0 x > 0 x>0的情况进行推导。
我们使用分布函数法，先计算分布函数 F X ( x ) = P ( X < x ) F_{X}(x) = P(X<x) FX(x)=P(X<x)，再对它求导得到概率密度函数。
简单情况
我们先推导 n = 1 n=1 n=1和 n = 2 n=2 n=2时的情况。从简单情况出发，可以更好地把握整体思路。
n = 1 n = 1 n=1时，有
F X ( x ) = P ( X < x ) = P ( Y 1 2 < x ) = ∫ − x x 1 2 π e − t 2 2 d t ， (1) F_{X}(x) = P(X<x) = P(Y_1^2 < x) = \int_{-\sqrt{x}}^{\sqrt{x}} \frac{1}{\sqrt{2\pi}}e^{-\frac{t^2}{2}}\text{d} t \thinspace，\tag{1} FX(x)=P(X<x)=P(Y12<x)=∫−xx2π1e−2t2dt，(1)求导得
p ( x ) = F X ′ ( x ) = 1 2 π e − x 2 ( 1 2 x − ( − 1 2 x ) ) = 1 2 π e − x 2 x − 1 2 。 (2) p(x) = F_{X}'(x) = \frac{1}{\sqrt{2\pi}} e^{-\frac{x}{2}}(\frac{1}{2\sqrt{x}}-(-\frac{1}{2\sqrt{x}})) = \frac{1}{\sqrt{2\pi}} e^{-\frac{x}{2}}x^{-\frac{1}{2}} \thinspace。\tag{2} p(x)=FX′(x)=2π1e−2x(2x1−(−2x1))=2π1e−2xx−21。(2)
n = 2 n = 2 n=2时，有
F X ( x ) = P ( Y 1 2 + Y 2 2 < x ) = ∬ D 1 2 π e − x 1 2 + x 2 2 2 d x 1 d x 2 = 1 2 π ∫ 0 2 π d θ ∫ 0 x e − r 2 2 r d r = − e − r 2 2 ∣ 0 x = 1 − e − x 2 ， (3) \begin{aligned} F_{X}(x) &= P(Y_1^2+Y_2^2 < x) \\ &= \begin{array}{c}\iint \\ D\end{array} \frac{1}{2\pi}e^{-\frac{x_1^2+x_2^2}{2}}\text{d} x_1 \text{d} x_2 \\ &= \frac{1}{2\pi}\int_{0}^{2\pi}\text{d}\theta\int_{0}^{\sqrt{x}}e^{-\frac{r^2}{2}}r \text{d} r \\ &= -e^{-\frac{r^2}{2}} {\Bigg|}_0^{\sqrt{x}} \\ &= 1-e^{-\frac{x}{2}} \thinspace， \end{aligned} \tag{3} FX(x)=P(Y12+Y22<x)=∬D2π1e−2x12+x22dx1dx2=2π1∫02πdθ∫0xe−2r2rdr=−e−2r2∣∣∣∣∣0x=1−e−2x，(3)其中 D D D表示圆 { ( x 1 , x 2 ) ∣ x 1 2 + x 2 2 < x } \{(x_1,x_2)|x_1^2+x_2^2<x\} {
(x1,x2)∣x12+x22<x}。求导得
p ( x ) = F X ′ ( x ) = 1 2 e − x 2 。 (4) p(x) = F_{X}'(x) = \frac{1}{2} e^{-\frac{x}{2}} \thinspace。\tag{4} p(x)=FX′(x)=21e−2x。(4)
正式推导
同样地，我们先写出 P ( X < x ) P(X<x) P(X<x)的表达式
P ( X < x ) = P ( ∑ i = 1 n Y i 2 < x ) = ( 2 π ) − n 2 ∫ ⋯ ∫ B e − 1 2 ∑ i = 1 n x i 2 d x 1 ⋯ d x n ， (5) P(X<x) = P(\sum\limits_{i=1}^{n}Y_i^2 < x) = (2\pi)^{-\frac{n}{2}}\begin{array}{c}\int\cdots\int \\ B\end{array}e^{-\frac{1}{2}\sum\limits_{i=1}^{n} x_i^2} \text{d} x_1\cdots \text{d} x_n \thinspace，\tag{5} P(X<x)=P(i=1∑nYi2<x)=(2π)−2n∫⋯∫Be−21i=1∑nxi2dx1⋯dxn，(5)其中 B B B为球体 { ( x 1 , ⋯ , x n ) ∣ ∑ i = 1 n x i 2 < x } \{(x_1,\cdots,x_n)|\sum\limits_{i=1}^{n}x_i^2 < x\} {
(x1,⋯,xn)∣i=1∑nxi2<x}。
根据 n n n维球坐标的雅可比行列式（详细计算过程见后文）
J = ∂ ( x 1 , x 2 , ⋯ x n ) ∂ ( r , θ 1 , ⋯ θ n − 1 ) = r n − 1 sin n − 2 θ 1 sin n − 3 θ 2 ⋯ sin θ n − 2 ， (6) \mathcal{J} = \frac{\partial(x_1,x_2,\cdots x_n)}{\partial(r,\theta_1,\cdots \theta_{n-1})} = r^{n-1}\sin^{n-2}\theta_1 \sin^{n-3}\theta_2\cdots \sin\theta_{n-2} \thinspace，\tag{6} J=∂(r,θ1,⋯θn−1)∂(x1,x2,⋯xn)=rn−1sinn−2θ1sinn−3θ2⋯sinθn−2，(6)并且令
c n = ( 2 π ) − n 2 ∫ 0 π sin n − 2 θ 1 d θ 1 ⋯ ∫ 0 π sin θ n − 2 d θ n − 2 ∫ 0 2 π d θ n − 1 (7) c_n = (2\pi)^{-\frac{n}{2}} \int_0^{\pi}\sin^{n-2}\theta_1\text{d}\theta_1\cdots \int_0^{\pi}\sin\theta_{n-2}\text{d}\theta_{n-2} \int_0^{2\pi}\text{d}\theta_{n-1} \tag{7} cn=(2π)−2n∫0πsinn−2θ1dθ1⋯∫0πsinθn−2dθn−2∫02πdθn−1(7)是某个与 n n n有关的常数，从而公式 ( 5 ) (5) (5)可化简为
P ( X < x ) = c n ∫ 0 x e − r 2 2 r n − 1 d r 。 (8) P(X<x) = c_n \int_0^{\sqrt{x}}e^{-\frac{r^2}{2}}r^{n-1}\text{d} r \thinspace。\tag{8} P(X<x)=cn∫0xe−2r2rn−1dr。(8)
接下来我们需要计算 c n c_n cn。其实 c n c_n cn可以直接使用公式 ( 7 ) (7) (7)来计算（会稍微复杂一点，详细计算过程在后文中给出），但是《数理统计学教程》中给出了一种更为巧妙和精简的方法，摘录如下：
1 = lim x → + ∞ P ( X < x ) = c n ∫ 0 + ∞ e − r 2 2 r n − 1 d r = r = 2 t c n ∫ 0 + ∞ e − t ( 2 t ) n − 1 d 2 t = c n ⋅ ( 2 ) n ∫ 0 + ∞ e − t ( t ) n − 1 1 2 t d t = c n ⋅ 2 n 2 − 1 Γ ( n 2 ) ， (9) \begin{aligned} 1 &\space\space\,=\space\space\, \lim\limits_{x\rightarrow+\infty}P(X<x) \\ &\space\space\,=\space\space\, c_n \int_0^{+\infty}e^{-\frac{r^2}{2}}r^{n-1}\text{d} r \\ &\overset{r=\sqrt{2t}}{=} c_n \int_0^{+\infty}e^{-t}(\sqrt{2t})^{n-1}\text{d}\sqrt{2t} \\ &\space\space\,=\space\space\, c_n\cdot (\sqrt{2})^n\int_0^{+\infty}e^{-t}(\sqrt{t})^{n-1}\frac{1}{2\sqrt{t}}\text{d} t \\ &\space\space\,=\space\space\, c_n\cdot 2^{\frac{n}{2}-1}\Gamma(\frac{n}{2}) \thinspace， \end{aligned} \tag{9} 1 = x→+∞limP(X<x) = cn∫0+∞e−2r2rn−1dr=r=2tcn∫0+∞e−t(2t)n−1d2t = cn⋅(2)n∫0+∞e−t(t)n−12t1dt = cn⋅22n−1Γ(2n)，(9)于是解得
c n = 2 1 − n 2 ⋅ 1 Γ ( n / 2 ) 。 (10) c_n = 2^{1-\frac{n}{2}}\cdot \frac{1}{\Gamma(n/2)} \thinspace。\tag{10} cn=21−2n⋅Γ(n/2)1。(10)将 c n c_n cn代入公式 ( 8 ) (8) (8)并求导，即可得到
p ( x ) = d d x P ( X < x ) = c n ( e − x 2 x n − 1 2 ⋅ 1 2 x − 0 ) = 1 2 n / 2 Γ ( n / 2 ) e − x 2 x n 2 − 1 。 (11) p(x) = \frac{\text{d}}{\text{d}x} P(X<x) = c_n(e^{-\frac{x}{2}}x^{\frac{n-1}{2}}\cdot\frac{1}{2\sqrt{x}} - 0) = \frac{1}{2^{n/2}\Gamma(n/2)}e^{-\frac{x}{2}}x^{\frac{n}{2}-1} \thinspace。\tag{11} p(x)=dxdP(X<x)=cn(e−2xx2n−1⋅2x1−0)=2n/2Γ(n/2)1e−2xx2n−1。(11)
综上， X X X的概率密度函数为
p ( x ) = { 1 2 n / 2 Γ ( n / 2 ) e − x 2 x n 2 − 1 , x > 0 0 , x ⩽ 0 。 (12) p(x) = \left\{\begin{array}{cc} \frac{1}{2^{n/2}\Gamma(n/2)}e^{-\frac{x}{2}}x^{\frac{n}{2}-1}, & x>0 \\ 0, & x\leqslant 0 \end{array}\right. \thinspace。\tag{12} p(x)={
2n/2Γ(n/2)1e−2xx2n−1,0,x>0x⩽0。(12)
补充说明
对于公式 ( 6 ) (6) (6)和公式 ( 7 ) (7) (7)， n n n要满足 n ⩾ 3 n\geqslant 3 n⩾3是一个比较直观的想法。为了避免麻烦，我们可以直接说明 n = 1 n=1 n=1和 n = 2 n=2 n=2时的情况（即公式 ( 2 ) (2) (2)和公式 ( 4 ) (4) (4)）满足公式 ( 12 ) (12) (12)。
n n n维球坐标雅可比行列式的计算
以下内容摘录自https://www.zhihu.com/question/332530250/answer/781120941
n n n维球坐标变换如下：
{ x 1 = r cos θ 1 x 2 = r sin θ 1 cos θ 2 x 3 = r sin θ 1 sin θ 2 cos θ 3 ⋯ x n − 1 = r sin θ 1 sin θ 2 ⋯ sin θ n − 2 cos θ n − 1 x n = r sin θ 1 sin θ 2 ⋯ sin θ n − 2 sin θ n − 1 ， (13) \left\{\begin{aligned} x_1 &= r\cos\theta_1 \\ x_2 &= r\sin\theta_1\cos\theta_2 \\ x_3 &= r\sin\theta_1\sin\theta_2\cos\theta_3 \\ &\cdots \\ x_{n-1} &= r\sin\theta_1\sin\theta_2\cdots \sin\theta_{n-2}\cos\theta_{n-1} \\ x_n &= r\sin\theta_1\sin\theta_2\cdots \sin\theta_{n-2}{\color{red}\;\sin\,}\theta_{n-1} \\ \end{aligned}\right. \thinspace，\tag{13} ⎩⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎧x1x2x3xn−1xn=rcosθ1=rsinθ1cosθ2=rsinθ1sinθ2cosθ3⋯=rsinθ1sinθ2⋯sinθn−2cosθn−1=rsinθ1sinθ2⋯sinθn−2sinθn−1，(13)其中
{ 0 ⩽ r ⩽ R 0 ⩽ θ 1 ⩽ π 0 ⩽ θ 2 ⩽ π ⋯ 0 ⩽ θ n − 2 ⩽ π 0 ⩽ θ n − 1 ⩽ 2 π 。 (14) \left\{\begin{aligned} &0 \leqslant r \leqslant R \\ &0 \leqslant \theta_1 \leqslant \pi \\ &0 \leqslant \theta_2 \leqslant \pi \\ &\cdots \\ &0 \leqslant \theta_{n-2} \leqslant \pi \\ &0 \leqslant \theta_{n-1} \leqslant {\color{red} 2}\pi \\ \end{aligned}\right. \thinspace。\tag{14} ⎩⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎧0⩽r⩽R0⩽θ1⩽π0⩽θ2⩽π⋯0⩽θn−2⩽π0⩽θn−1⩽2π。(14)由雅可比行列式的定义和矩阵转置的行列式不变，有
J = ∂ ( x 1 , x 2 , ⋯ x n ) ∂ ( r , θ 1 , ⋯ θ n − 1 ) = ∣ ∂ x 1 ∂ r ∂ x 1 ∂ θ 1 ⋯ ∂ x 1 ∂ θ n − 1 ∂ x 2 ∂ r ∂ x 2 ∂ θ 1 ⋯ ∂ x 2 ∂ θ n − 1 ⋮ ⋮ ⋱ ⋮ ∂ x n ∂ r ∂ x n ∂ θ 1 ⋯ ∂ x n ∂ θ n − 1 ∣ = ∣ ∂ x 1 ∂ r ∂ x 2 ∂ r ⋯ ∂ x n ∂ r ∂ x 1 ∂ θ 1 ∂ x 2 ∂ θ 1 ⋯ ∂ x n ∂ θ 1 ⋮ ⋮ ⋱ ⋮ ∂ x 1 ∂ θ n − 1 ∂ x 2 ∂ θ n − 1 ⋯ ∂ x n ∂ θ n − 1 ∣ 。 (15) \mathcal{J} = \frac{\partial(x_1,x_2,\cdots x_n)}{\partial(r,\theta_1,\cdots \theta_{n-1})} = \left|\begin{array}{cccc} \frac{\partial x_1}{\partial r} & \frac{\partial x_1}{\partial \theta_1} & \cdots & \frac{\partial x_1}{\partial \theta_{n-1}} \\ \frac{\partial x_2}{\partial r} & \frac{\partial x_2}{\partial \theta_1} & \cdots & \frac{\partial x_2}{\partial \theta_{n-1}} \\ \vdots & \vdots & \ddots & \vdots \\ \frac{\partial x_n}{\partial r} & \frac{\partial x_n}{\partial \theta_1} & \cdots & \frac{\partial x_n}{\partial \theta_{n-1}} \\ \end{array}\right| = \left|\begin{array}{cccc} \frac{\partial x_1}{\partial r} & \frac{\partial x_2}{\partial r} & \cdots & \frac{\partial x_n}{\partial r} \\ \frac{\partial x_1}{\partial \theta_1} & \frac{\partial x_2}{\partial \theta_1} & \cdots & \frac{\partial x_n}{\partial \theta_1} \\ \vdots & \vdots & \ddots & \vdots \\ \frac{\partial x_1}{\partial \theta_{n-1}} & \frac{\partial x_2}{\partial \theta_{n-1}} & \cdots & \frac{\partial x_n}{\partial \theta_{n-1}} \\ \end{array}\right| \thinspace。\tag{15} J=∂(r,θ1,⋯θn−1)∂(x1,x2,⋯xn)=∣∣∣∣∣∣∣∣∣∣∂r∂x1∂r∂x2⋮∂r∂xn∂θ1∂x1∂θ1∂x2⋮∂θ1∂xn⋯⋯⋱⋯∂θn−1∂x1∂θn−1∂x2⋮∂θn−1∂xn∣∣∣∣∣∣∣∣∣∣=∣∣∣∣∣∣∣∣∣∂r∂x1∂θ1∂x1⋮∂θn−1∂x1∂r∂x2∂θ1∂x2⋮∂θn−1∂x2⋯⋯⋱⋯∂r∂xn∂θ1∂xn⋮∂θn−1∂xn∣∣∣∣∣∣∣∣∣。(15)分别计算出各项导数
J = ∣ cos θ 1 sin θ 1 cos θ 2 sin θ 1 sin θ 2 cos θ 3 ⋯ ∏ i = 1 n − 2 sin θ i cos θ n − 1 ∏ i = 1 n − 1 sin θ i − r sin θ 1 r cos θ 1 cos θ 2 r cos θ 1 sin θ 2 cos θ 3 ⋯ r cos θ 1 ∏ i = 2 n − 2 sin θ i cos θ n − 1 r cos θ 1 ∏ i = 2 n − 1 sin θ i 0 − r sin θ 1 sin θ 2 r sin θ 1 cos θ 2 cos θ 3 ⋯ r cos θ 2 ∏ i = 1 i ≠ 2 n − 2 sin θ i cos θ n − 1 r cos θ 2 ∏ i = 1 i ≠ 2 n − 1 sin θ i ⋮ ⋮ ⋮ ⋱ ⋮ ⋮ 0 0 0 ⋯ r cos θ n − 2 ∏ i = 1 n − 3 sin θ i cos θ n − 1 r cos θ n − 2 ∏ i = 1 i ≠ n − 2 n − 1 sin θ i 0 0 0 ⋯ − r ∏ i = 1 n − 1 sin θ i r cos θ n − 1 ∏ i = 1 n − 2 sin θ i ∣ 。 (16) \mathcal{J} = \left|\begin{array}{cccccc} \cos\theta_1 & \sin\theta_1\cos\theta_2 & \sin\theta_1\sin\theta_2\cos\theta_3 & \cdots & \prod\limits_{i=1}^{n-2}\sin\theta_i\cos\theta_{n-1} & \prod\limits_{i=1}^{n-1}\sin\theta_i \\ -r\sin\theta_1 & r\cos\theta_1\cos\theta_2 & r\cos\theta_1\sin\theta_2\cos\theta_3 & \cdots & r\cos\theta_1\prod\limits_{i=2}^{n-2}\sin\theta_i\cos\theta_{n-1} & r\cos\theta_1\prod\limits_{i=2}^{n-1}\sin\theta_i \\ 0 & -r\sin\theta_1\sin\theta_2 & r\sin\theta_1\cos\theta_2\cos\theta_3 & \cdots & r\cos\theta_2\prod\limits_{i=1\atop i\neq2}^{n-2}\sin\theta_i\cos\theta_{n-1} & r\cos\theta_2\prod\limits_{i=1\atop i\neq2}^{n-1}\sin\theta_i \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & r\cos\theta_{n-2}\prod\limits_{i=1}^{n-3}\sin\theta_i\cos\theta_{n-1} & r\cos\theta_{n-2}\prod\limits_{i=1\atop i\neq n-2}^{n-1}\sin\theta_i \\ 0 & 0 & 0 & \cdots & -r\prod\limits_{i=1}^{n-1}\sin\theta_i & r\cos\theta_{n-1}\prod\limits_{i=1}^{n-2}\sin\theta_i \\ \end{array}\right| \thinspace。\tag{16} J=∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣cosθ1−rsinθ10⋮00sinθ1cosθ2rcosθ1cosθ2−rsinθ1sinθ2⋮00sinθ1sinθ2cosθ3rcosθ1sinθ2cosθ3rsinθ1cosθ2cosθ3⋮00⋯⋯⋯⋱⋯⋯i=1∏n−2sinθicosθn−1rcosθ1i=2∏n−2sinθicosθn−1rcosθ2i=2i=1∏n−2sinθicosθn−1⋮rcosθn−2i=1∏n−3sinθicosθn−1−ri=1∏n−1sinθii=1∏n−1sinθircosθ1i=2∏n−1sinθircosθ2i=2i=1∏n−1sinθi⋮rcosθn−2i=n−2i=1∏n−1sinθircosθn−1i=1∏n−2sinθi∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣。(16)从第二行到最后一行提出公因式 r r r，得到
J = r n − 1 ∣ cos θ 1 sin θ 1 cos θ 2 sin θ 1 sin θ 2 cos θ 3 ⋯ ∏ i = 1 n − 2 sin θ i cos θ n − 1 ∏ i = 1 n − 1 sin θ i − sin θ 1 cos θ 1 cos θ 2 cos θ 1 sin θ 2 cos θ 3 ⋯ cos θ 1 ∏ i = 2 n − 2 sin θ i cos θ n − 1 cos θ 1 ∏ i = 2 n − 1 sin θ i 0 − sin θ 1 sin θ 2 sin θ 1 cos θ 2 cos θ 3 ⋯ cos θ 2 ∏ i = 1 i ≠ 2 n − 2 sin θ i cos θ n − 1 cos θ 2 ∏ i = 1 i ≠ 2 n − 1 sin θ i ⋮ ⋮ ⋮ ⋱ ⋮ ⋮ 0 0 0 ⋯ cos θ n − 2 ∏ i = 1 n − 3 sin θ i cos θ n − 1 cos θ n − 2 ∏ i = 1 i ≠ n − 2 n − 1 sin θ i 0 0 0 ⋯ − ∏ i = 1 n − 1 sin θ i cos θ n − 1 ∏ i = 1 n − 2 sin θ i ∣ 。 (17) \mathcal{J} = r^{n-1}\left|\begin{array}{cccccc} \cos\theta_1 & \sin\theta_1\cos\theta_2 & \sin\theta_1\sin\theta_2\cos\theta_3 & \cdots & \prod\limits_{i=1}^{n-2}\sin\theta_i\cos\theta_{n-1} & \prod\limits_{i=1}^{n-1}\sin\theta_i \\ -\sin\theta_1 & \cos\theta_1\cos\theta_2 & \cos\theta_1\sin\theta_2\cos\theta_3 & \cdots & \cos\theta_1\prod\limits_{i=2}^{n-2}\sin\theta_i\cos\theta_{n-1} & \cos\theta_1\prod\limits_{i=2}^{n-1}\sin\theta_i \\ 0 & -\sin\theta_1\sin\theta_2 & \sin\theta_1\cos\theta_2\cos\theta_3 & \cdots & \cos\theta_2\prod\limits_{i=1\atop i\neq2}^{n-2}\sin\theta_i\cos\theta_{n-1} & \cos\theta_2\prod\limits_{i=1\atop i\neq2}^{n-1}\sin\theta_i \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & \cos\theta_{n-2}\prod\limits_{i=1}^{n-3}\sin\theta_i\cos\theta_{n-1} & \cos\theta_{n-2}\prod\limits_{i=1\atop i\neq n-2}^{n-1}\sin\theta_i \\ 0 & 0 & 0 & \cdots & -\prod\limits_{i=1}^{n-1}\sin\theta_i & \cos\theta_{n-1}\prod\limits_{i=1}^{n-2}\sin\theta_i \\ \end{array}\right| \thinspace。\tag{17} J=rn−1∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣cosθ1−sinθ10⋮00sinθ1cosθ2cosθ1cosθ2−sinθ1sinθ2⋮00sinθ1sinθ2cosθ3cosθ1sinθ2cosθ3sinθ1cosθ2cosθ3⋮00⋯⋯⋯⋱⋯⋯i=1∏n−2sinθicosθn−1cosθ1i=2∏n−2sinθicosθn−1cosθ2i=2i=1∏n−2sinθicosθn−1⋮cosθn−2i=1∏n−3sinθicosθn−1−i=1∏n−1sinθii=1∏n−1sinθicosθ1i=2∏n−1sinθicosθ2i=2i=1∏n−1sinθi⋮cosθn−2i=n−2i=1∏n−1sinθicosθn−1i=1∏n−2sinθi∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣。(17)接着用第一行的 sin θ 1 cos θ 1 \frac{\sin\theta_1}{\cos\theta_1} cosθ1sinθ1倍加到第二行
J = r n − 1 ∣ cos θ 1 sin θ 1 cos θ 2 sin θ 1 sin θ 2 cos θ 3 ⋯ ∏ i = 1 n − 2 sin θ i cos θ n − 1 ∏ i = 1 n − 1 sin θ i 0 1 cos θ 1 cos θ 2 1 cos θ 1 sin θ 2 cos θ 3 ⋯ ∏ i = 2 n − 2 sin θ i cos θ n − 1 cos θ 1 1 cos θ 1 ∏ i = 2 n − 1 sin θ i 0 − sin θ 1 sin θ 2 sin θ 1 cos θ 2 cos θ 3 ⋯ cos θ 2 ∏ i = 1 i ≠ 2 n − 2 sin θ i cos θ n − 1 cos θ 2 ∏ i = 1 i ≠ 2 n − 1 sin θ i ⋮ ⋮ ⋮ ⋱ ⋮ ⋮ 0 0 0 ⋯ cos θ n − 2 ∏ i = 1 n − 3 sin θ i cos θ n − 1 cos θ n − 2 ∏ i = 1 i ≠ n − 2 n − 1 sin θ i 0 0 0 ⋯ − ∏ i = 1 n − 1 sin θ i cos θ n − 1 ∏ i = 1 n − 2 sin θ i ∣ 。 (18) \mathcal{J} = r^{n-1}\left|\begin{array}{cccccc} \cos\theta_1 & \sin\theta_1\cos\theta_2 & \sin\theta_1\sin\theta_2\cos\theta_3 & \cdots & \prod\limits_{i=1}^{n-2}\sin\theta_i\cos\theta_{n-1} & \prod\limits_{i=1}^{n-1}\sin\theta_i \\ 0 & \frac{1}{\cos\theta_1}\cos\theta_2 & \frac{1}{\cos\theta_1}\sin\theta_2\cos\theta_3 & \cdots & \prod\limits_{i=2}^{n-2}\sin\theta_i\frac{\cos\theta_{n-1}}{\cos\theta_1} & \frac{1}{\cos\theta_1}\prod\limits_{i=2}^{n-1}\sin\theta_i \\ 0 & -\sin\theta_1\sin\theta_2 & \sin\theta_1\cos\theta_2\cos\theta_3 & \cdots & \cos\theta_2\prod\limits_{i=1\atop i\neq2}^{n-2}\sin\theta_i\cos\theta_{n-1} & \cos\theta_2\prod\limits_{i=1\atop i\neq2}^{n-1}\sin\theta_i \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & \cos\theta_{n-2}\prod\limits_{i=1}^{n-3}\sin\theta_i\cos\theta_{n-1} & \cos\theta_{n-2}\prod\limits_{i=1\atop i\neq n-2}^{n-1}\sin\theta_i \\ 0 & 0 & 0 & \cdots & -\prod\limits_{i=1}^{n-1}\sin\theta_i & \cos\theta_{n-1}\prod\limits_{i=1}^{n-2}\sin\theta_i \\ \end{array}\right| \thinspace。\tag{18} J=rn−1∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣cosθ100⋮00sinθ1cosθ2cosθ11cosθ2−sinθ1sinθ2⋮00sinθ1sinθ2cosθ3cosθ11sinθ2cosθ3sinθ1cosθ2cosθ3⋮00⋯⋯⋯⋱⋯⋯i=1∏n−2sinθicosθn−1i=2∏n−2sinθicosθ1cosθn−1cosθ2i=2i=1∏n−2sinθicosθn−1⋮cosθn−2i=1∏n−3sinθicosθn−1−i=1∏n−1sinθii=1∏n−1sinθicosθ11i=2∏n−1sinθicosθ2i=2i=1∏n−1sinθi⋮cosθn−2i=n−2i=1∏n−1sinθicosθn−1i=1∏n−2sinθi∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣。(18)按照第一行展开，有
J = r n − 1 cos θ 1 ∣ 1 cos θ 1 cos θ 2 1 cos θ 1 sin θ 2 cos θ 3 ⋯ 1 cos θ 1 ∏ i = 2 n − 2 sin θ i cos θ n − 1 1 cos θ 1 ∏ i = 2 n − 1 sin θ i − sin θ 1 sin θ 2 sin θ 1 cos θ 2 cos θ 3 ⋯ cos θ 2 ∏ i = 1 i ≠ 2 n − 2 sin θ i cos θ n − 1 cos θ 2 ∏ i = 1 i ≠ 2 n − 1 sin θ i ⋮ ⋮ ⋱ ⋮ ⋮ 0 0 ⋯ cos θ n − 2 ∏ i = 1 n − 3 sin θ i cos θ n − 1 cos θ n − 2 ∏ i = 1 i ≠ n − 2 n − 1 sin θ i 0 0 ⋯ − ∏ i = 1 n − 1 sin θ i cos θ n − 1 ∏ i = 1 n − 2 sin θ i ∣ 。 (19) \mathcal{J} = r^{n-1}\cos\theta_1\left|\begin{array}{ccccc} \frac{1}{\cos\theta_1}\cos\theta_2 & \frac{1}{\cos\theta_1}\sin\theta_2\cos\theta_3 & \cdots & \frac{1}{\cos\theta_1}\prod\limits_{i=2}^{n-2}\sin\theta_i\cos\theta_{n-1} & \frac{1}{\cos\theta_1}\prod\limits_{i=2}^{n-1}\sin\theta_i \\ -\sin\theta_1\sin\theta_2 & \sin\theta_1\cos\theta_2\cos\theta_3 & \cdots & \cos\theta_2\prod\limits_{i=1\atop i\neq2}^{n-2}\sin\theta_i\cos\theta_{n-1} & \cos\theta_2\prod\limits_{i=1\atop i\neq2}^{n-1}\sin\theta_i \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & \cos\theta_{n-2}\prod\limits_{i=1}^{n-3}\sin\theta_i\cos\theta_{n-1} & \cos\theta_{n-2}\prod\limits_{i=1\atop i\neq n-2}^{n-1}\sin\theta_i \\ 0 & 0 & \cdots & -\prod\limits_{i=1}^{n-1}\sin\theta_i & \cos\theta_{n-1}\prod\limits_{i=1}^{n-2}\sin\theta_i \\ \end{array}\right| \thinspace。\tag{19} J=rn−1cosθ1∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣cosθ11cosθ2−sinθ1sinθ2⋮00cosθ11sinθ2cosθ3sinθ1cosθ2cosθ3⋮00⋯⋯⋱⋯⋯cosθ11i=2∏n−2sinθicosθn−1cosθ2i=2i=1∏n−2sinθicosθn−1⋮cosθn−2i=1∏n−3sinθicosθn−1−i=1∏n−1sinθicosθ11i=2∏n−1sinθicosθ2i=2i=1∏n−1sinθi⋮cosθn−2i=n−2i=1∏n−1sinθicosθn−1i=1∏n−2sinθi∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣。(19)对于公式 ( 19 ) (19) (19)中的行列式，从第一行提出公因式 1 cos θ 1 \frac{1}{\cos\theta_1} cosθ11，从第二行到最后一行提出公因式 sin θ 1 \sin\theta_1 sinθ1，得到
J = r n − 1 cos θ 1 sin n − 2 θ 1 cos θ 1 ∣ cos θ 2 sin θ 2 cos θ 3 ⋯ ∏ i = 2 n − 2 sin θ i cos θ n − 1 ∏ i = 2 n − 1 sin θ i − sin θ 2 cos θ 2 cos θ 3 ⋯ cos θ 2 ∏ i = 3 n − 2 sin θ i cos θ n − 1 cos θ 2 ∏ i = 3 n − 1 sin θ i ⋮ ⋮ ⋱ ⋮ ⋮ 0 0 ⋯ cos θ n − 2 ∏ i = 2 n − 3 sin θ i cos θ n − 1 cos θ n − 2 ∏ i = 2 i ≠ n − 2 n − 1 sin θ i 0 0 ⋯ − ∏ i = 2 n − 1 sin θ i cos θ n − 1 ∏ i = 2 n − 2 sin θ i ∣ 。 (20) \mathcal{J} = r^{n-1}\cos\theta_1\frac{\sin^{n-2}\theta_1}{\cos\theta_1}\left|\begin{array}{ccccc} \cos\theta_2 & \sin\theta_2\cos\theta_3 & \cdots & \prod\limits_{i=2}^{n-2}\sin\theta_i\cos\theta_{n-1} & \prod\limits_{i=2}^{n-1}\sin\theta_i \\ -\sin\theta_2 & \cos\theta_2\cos\theta_3 & \cdots & \cos\theta_2\prod\limits_{i=3}^{n-2}\sin\theta_i\cos\theta_{n-1} & \cos\theta_2\prod\limits_{i=3}^{n-1}\sin\theta_i \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & \cos\theta_{n-2}\prod\limits_{i=2}^{n-3}\sin\theta_i\cos\theta_{n-1} & \cos\theta_{n-2}\prod\limits_{i=2\atop i\neq n-2}^{n-1}\sin\theta_i \\ 0 & 0 & \cdots & -\prod\limits_{i=2}^{n-1}\sin\theta_i & \cos\theta_{n-1}\prod\limits_{i=2}^{n-2}\sin\theta_i \\ \end{array}\right| \thinspace。\tag{20} J=rn−1cosθ1cosθ1sinn−2θ1∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣cosθ2−sinθ2⋮00sinθ2cosθ3cosθ2cosθ3⋮00⋯⋯⋱⋯⋯i=2∏n−2sinθicosθn−1cosθ2i=3∏n−2sinθicosθn−1⋮cosθn−2i=2∏n−3sinθicosθn−1−i=2∏n−1sinθii=2∏n−1sinθicosθ2i=3∏n−1sinθi⋮cosθn−2i=n−2i=2∏n−1sinθicosθn−1i=2∏n−2sinθi∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣。(20)我们发现公式 ( 20 ) (20) (20)中的行列式与公式 ( 17 ) (17) (17)中的行列式有着相似的形状，因此可以递推得到最终结果
J = r n − 1 ⋅ D 1 = r n − 1 sin n − 2 θ 1 ⋅ D 2 = r n − 1 sin n − 2 θ 1 sin n − 3 θ 2 ⋅ D 3 ⋯ = r n − 1 sin n − 2 θ 1 sin n − 3 θ 2 ⋯ sin θ n − 2 。 (21) \begin{aligned} \mathcal{J} &= r^{n-1}\cdot D_1 \\ &= r^{n-1}\sin^{n-2}\theta_1\cdot D_2 \\ &= r^{n-1}\sin^{n-2}\theta_1\sin^{n-3}\theta_2\cdot D_3 \\ &\cdots \\ &= r^{n-1}\sin^{n-2}\theta_1 \sin^{n-3}\theta_2\cdots \sin\theta_{n-2} \thinspace。 \end{aligned} \tag{21} J=rn−1⋅D1=rn−1sinn−2θ1⋅D2=rn−1sinn−2θ1sinn−3θ2⋅D3⋯=rn−1sinn−2θ1sinn−3θ2⋯sinθn−2。(21)
如何暴力求解 c n c_n cn
公式 ( 9 ) (9) (9)中对于 c n c_n cn的计算巧妙地运用了 lim x → + ∞ P ( X < x ) = 1 \lim\limits_{x\rightarrow+\infty}P(X<x) = 1 x→+∞limP(X<x)=1这一性质。实际上，我们也可以直接计算 c n c_n cn。
引理1
∫ 0 π 2 sin n x d x = ∫ 0 π 2 cos n x d x = { ( n − 1 ) ! ! ( n ) ! ! ⋅ π 2 , n = 2 k ( n − 1 ) ! ! ( n ) ! ! , n = 2 k + 1 ( k = 0 , 1 , 2 , ⋯ ) 。 (22) \int_0^{\frac{\pi}{2}}\sin^nx \text{d} x = \int_0^{\frac{\pi}{2}}\cos^nx \text{d} x = \left\{\begin{array}{l} \frac{(n-1)!!}{(n)!!}\cdot \frac{\pi}{2}, & n=2k \\ \frac{(n-1)!!}{(n)!!}, & n=2k+1 \end{array}\right.(k=0,1,2,\cdots) \thinspace。\tag{22} ∫02πsinnxdx=∫02πcosnxdx={
(n)!!(n−1)!!⋅2π,(n)!!(n−1)!!,n=2kn=2k+1(k=0,1,2,⋯)。(22)
证明
引理1及其证明摘自费定晖、周学圣《吉米多维奇数学分析习题集题解（第四版）》第2281题和第2282题，进一步的探究可以参考第2011题。
先证公式 ( 22 ) (22) (22)的第一个等号。令 x = π 2 − t x=\frac{\pi}{2}-t x=2π−t，则 d x = − d t \text{d} x=-\text{d} t dx=−dt，且 cos x = cos ( π 2 − t ) = sin t \cos x=\cos(\frac{\pi}{2}-t)=\sin t cosx=cos(2π−t)=sint，于是证得
∫ 0 π 2 cos n x d x = − ∫ π 2 0 sin n t d t = ∫ 0 π 2 sin n x d x 。 (23) \begin{aligned} &\quad \int_0^{\frac{\pi}{2}}\cos^nx \text{d} x \\ &= -\int_{\frac{\pi}{2}}^0\sin^nt \text{d} t \\ &= \int_0^{\frac{\pi}{2}}\sin^nx \text{d} x \thinspace。 \end{aligned} \tag{23} ∫02πcosnxdx=−∫2π0sinntdt=∫02πsinnxdx。(23)再证公式 ( 22 ) (22) (22)的第二个等号。令 I n = ∫ 0 π 2 sin n x d x I_n = \int_0^{\frac{\pi}{2}}\sin^nx \text{d} x In=∫02πsinnxdx，于是有
I n = − ∫ 0 π 2 sin n − 1 x d ( cos x ) = − sin n − 1 x cos x ∣ 0 π 2 + ( n − 1 ) ∫ 0 π 2 sin n − 2 x cos 2 x d x = 0 + ( n − 1 ) ∫ 0 π 2 sin n − 2 x d x − ( n − 1 ) ∫ 0 π 2 sin n x d x = ( n − 1 ) I n − 2 − ( n − 1 ) I n 。 (24) \begin{aligned} I_n &= -\int_0^{\frac{\pi}{2}}\sin^{n-1}x \text{d}(\cos x) \\ &= -\sin^{n-1}x\cos x {\Bigg|}_0^{\frac{\pi}{2}} + (n-1)\int_0^{\frac{\pi}{2}}\sin^{n-2}x\cos^2x \text{d} x \\ &= 0 + (n-1)\int_0^{\frac{\pi}{2}}\sin^{n-2}x\text{d} x - (n-1)\int_0^{\frac{\pi}{2}}\sin^nx\text{d} x \\ &= (n-1)I_{n-2} - (n-1)I_n \thinspace。 \end{aligned} \tag{24} In=−∫02πsinn−1xd(cosx)=−sinn−1xcosx∣∣∣∣∣02π+(n−1)∫02πsinn−2xcos2xdx=0+(n−1)∫02πsinn−2xdx−(n−1)∫02πsinnxdx=(n−1)In−2−(n−1)In。(24)移项合并得
I n = n − 1 n I n − 2 。 (25) I_n = \frac{n-1}{n}I_{n-2} \thinspace。\tag{25} In=nn−1In−2。(25)利用公式 ( 25 ) (25) (25)进行递推即可得证。
引理2
∫ 0 π sin n x d x = 2 ∫ 0 π 2 sin n x d x 。 (26) \int_0^{\pi}\sin^nx \text{d}x = 2\int_0^{\frac{\pi}{2}}\sin^nx \text{d}x \thinspace。\tag{26} ∫0πsinnxdx=2∫02πsinnxdx。(26)
证明
∫ 0 π sin n x d x = ∫ 0 π 2 sin n x d x + ∫ π 2 π sin n x d x = ∫ 0 π 2 sin n x d x + ∫ 0 π 2 sin n ( t + π 2 ) d ( t + π 2 ) = ∫ 0 π 2 sin n x d x + ∫ 0 π 2 cos n t d t = 2 ∫ 0 π 2 sin n x d x 。 (27) \begin{aligned} \int_0^{\pi}\sin^nx \text{d} x &= \int_0^{\frac{\pi}{2}}\sin^nx \text{d} x + \int_{\frac{\pi}{2}}^{\pi}\sin^nx \text{d} x \\ &= \int_0^{\frac{\pi}{2}}\sin^nx \text{d} x + \int_0^{\frac{\pi}{2}}\sin^n(t+\frac{\pi}{2}) \text{d}(t+\frac{\pi}{2}) \\ &= \int_0^{\frac{\pi}{2}}\sin^nx \text{d} x + \int_0^{\frac{\pi}{2}}\cos^nt \text{d} t \\ &= 2\int_0^{\frac{\pi}{2}}\sin^nx \text{d} x \thinspace。 \end{aligned} \tag{27} ∫0πsinnxdx=∫02πsinnxdx+∫2ππsinnxdx=∫02πsinnxdx+∫02πsinn(t+2π)d(t+2π)=∫02πsinnxdx+∫02πcosntdt=2∫02πsinnxdx。(27)
计算 c n c_n cn
n = 1 n=1 n=1和 n = 2 n=2 n=2时的情况由公式 ( 2 ) (2) (2)和公式 ( 4 ) (4) (4)得到结果。下面针对 n ⩾ 3 n \geqslant 3 n⩾3的情况进行计算。
当 n = 2 k , ( k ⩾ 2 ) n = 2k,\space(k \geqslant 2) n=2k, (k⩾2)时，
c n = ( 2 π ) − n 2 ∫ 0 π sin n − 2 θ 1 d θ 1 ⋯ ∫ 0 π sin θ n − 2 d θ n − 2 ∫ 0 2 π d θ n − 1 = ( 2 π ) 1 − k ∫ 0 π sin 2 k − 2 θ 1 d θ 1 ⋯ ∫ 0 π sin θ 2 k − 2 d θ 2 k − 2 = ( 2 π ) 1 − k ⋅ 2 2 k − 2 ⋅ ( ( 2 k − 3 ) ! ! ( 2 k − 2 ) ! ! ⋅ π 2 ) ( ( 2 k − 4 ) ! ! ( 2 k − 3 ) ! ! ) ⋯ ( 1 ! ! 2 ! ! ⋅ π 2 ) ( 0 ! ! 1 ! ! ) = ( 2 π ) 1 − k ⋅ 2 2 k − 2 ⋅ ( π 2 ) k − 1 ⋅ 1 ( 2 k − 2 ) ! ! = 2 1 − k ⋅ 1 ( k − 1 ) ! = 2 1 − n 2 ⋅ 1 Γ ( n / 2 ) 。 (28) \begin{aligned} c_n &= (2\pi)^{-\frac{n}{2}} \int_0^{\pi}\sin^{n-2}\theta_1\text{d}\theta_1\cdots \int_0^{\pi}\sin\theta_{n-2}\text{d}\theta_{n-2} \int_0^{2\pi}\text{d}\theta_{n-1} \\ &= (2\pi)^{1-k} \int_0^{\pi}\sin^{2k-2}\theta_1\text{d}\theta_1\cdots \int_0^{\pi}\sin\theta_{2k-2}\text{d}\theta_{2k-2} \\ &= (2\pi)^{1-k}\cdot 2^{2k-2}\cdot (\frac{(2k-3)!!}{(2k-2)!!}\cdot\frac{\pi}{2})(\frac{(2k-4)!!}{(2k-3)!!})\cdots(\frac{1!!}{2!!}\cdot\frac{\pi}{2})(\frac{0!!}{1!!}) \\ &= (2\pi)^{1-k}\cdot 2^{2k-2}\cdot (\frac{\pi}{2})^{k-1}\cdot \frac{1}{(2k-2)!!} \\ &= 2^{1-k}\cdot \frac{1}{(k-1)!} \\ &= 2^{1-\frac{n}{2}}\cdot \frac{1}{\Gamma(n/2)} \thinspace。 \end{aligned} \tag{28} cn=(2π)−2n∫0πsinn−2θ1dθ1⋯∫0πsinθn−2dθn−2∫02πdθn−1=(2π)1−k∫0πsin2k−2θ1dθ1⋯∫0πsinθ2k−2dθ2k−2=(2π)1−k⋅22k−2⋅((2k−2)!!(2k−3)!!⋅2π)((2k−3)!!(2k−4)!!)⋯(2!!1!!⋅2π)(1!!0!!)=(2π)1−k⋅22k−2⋅(2π)k−1⋅(2k−2)!!1=21−k⋅(k−1)!1=21−2n⋅Γ(n/2)1。(28)
当 n = 2 k + 1 , ( k ⩾ 1 ) n = 2k+1,\space(k \geqslant 1) n=2k+1, (k⩾1)时，
c n = ( 2 π ) − n 2 ∫ 0 π sin n − 2 θ 1 d θ 1 ⋯ ∫ 0 π sin θ n − 2 d θ n − 2 ∫ 0 2 π d θ n − 1 = ( 2 π ) 1 2 − k ∫ 0 π sin 2 k − 1 θ 1 d θ 1 ⋯ ∫ 0 π sin θ 2 k − 1 d θ 2 k − 1 = ( 2 π ) 1 2 − k ⋅ 2 2 k − 1 ⋅ ( ( 2 k − 2 ) ! ! ( 2 k − 1 ) ! ! ) ( ( 2 k − 3 ) ! ! ( 2 k − 2 ) ! ! ⋅ π 2 ) ( ( 2 k − 4 ) ! ! ( 2 k − 3 ) ! ! ) ⋯ ( 1 ! ! 2 ! ! ⋅ π 2 ) ( 0 ! ! 1 ! ! ) = ( 2 π ) 1 2 − k ⋅ 2 2 k − 1 ⋅ ( π 2 ) k − 1 ⋅ 1 ( 2 k − 1 ) ! ! = 2 1 2 − k ⋅ 1 2 k − 1 2 2 k − 3 2 ⋯ 1 2 π = 2 1 − n 2 ⋅ 1 Γ ( n / 2 ) 。 (29) \begin{aligned} c_n &= (2\pi)^{-\frac{n}{2}} \int_0^{\pi}\sin^{n-2}\theta_1\text{d}\theta_1\cdots \int_0^{\pi}\sin\theta_{n-2}\text{d}\theta_{n-2} \int_0^{2\pi}\text{d}\theta_{n-1} \\ &= (2\pi)^{\frac{1}{2}-k} \int_0^{\pi}\sin^{2k-1}\theta_1\text{d}\theta_1\cdots \int_0^{\pi}\sin\theta_{2k-1}\text{d}\theta_{2k-1} \\ &= (2\pi)^{\frac{1}{2}-k}\cdot 2^{2k-1}\cdot (\frac{(2k-2)!!}{(2k-1)!!})(\frac{(2k-3)!!}{(2k-2)!!}\cdot\frac{\pi}{2})(\frac{(2k-4)!!}{(2k-3)!!})\cdots(\frac{1!!}{2!!}\cdot\frac{\pi}{2})(\frac{0!!}{1!!}) \\ &= (2\pi)^{\frac{1}{2}-k}\cdot 2^{2k-1}\cdot (\frac{\pi}{2})^{k-1}\cdot \frac{1}{(2k-1)!!} \\ &= 2^{\frac{1}{2}-k}\cdot \frac{1}{\frac{2k-1}{2}\frac{2k-3}{2}\cdots\frac{1}{2}\sqrt{\pi}} \\ &= 2^{1-\frac{n}{2}}\cdot \frac{1}{\Gamma(n/2)} \thinspace。 \end{aligned} \tag{29} cn=(2π)−2n∫0πsinn−2θ1dθ1⋯∫0πsinθn−2dθn−2∫02πdθn−1=(2π)21−k∫0πsin2k−1θ1dθ1⋯∫0πsinθ2k−1dθ2k−1=(2π)21−k⋅22k−1⋅((2k−1)!!(2k−2)!!)((2k−2)!!(2k−3)!!⋅2π)((2k−3)!!(2k−4)!!)⋯(2!!1!!⋅2π)(1!!0!!)=(2π)21−k⋅22k−1⋅(2π)k−1⋅(2k−1)!!1=221−k⋅22k−122k−3⋯21π1=21−2n⋅Γ(n/2)1。(29)
综上，
c n = 2 1 − n 2 ⋅ 1 Γ ( n / 2 ) 。 (30) c_n = 2^{1-\frac{n}{2}}\cdot \frac{1}{\Gamma(n/2)} \thinspace。\tag{30} cn=21−2n⋅Γ(n/2)1。(30)