47103 Wallis積分與無窮乘積
Wallis積分與無窮乘積

### 1. 約翰--沃利斯 --- 微積分之先驅與皇家學會的創立者

『Foreshadowings of the principles and even of the language of [the infinitesimal] calculus can be found in the writings of Napier, Kepler, Cavalieri, Pascal, Fermat, Wallis, and Barrow. It was Newton's good luck to come at a time when everything was ripe for the discovery, and his ability enabled him to construct almost at once a complete calculus.』

--- W.W.R. Ball In History of Mathematics (3rd Ed., 1901), 366. ---

$${\pi\over 2} = \prod_{k=1}^\infty \bigg({2k\over 2k-1}\cdot {2k\over 2k+1}\bigg) ={2\over 1}\cdot {2\over 3}\cdot {4\over 3}\cdot {4\over 5}\cdot {6\over 5}\cdot {6\over 7}\cdots .$$

### 2. 球體積 --- Wallis 積分之起源

Beta 函數最開始是稱為 Euler 第一類積分! 事實上歐拉 (L. Euler; 1707$\sim$1783) 是研究這個積分的第一人。 這個瑕積分(improper integral)沒有太胖的問題但卻有太高的問題, $t=0, 1$ 這兩點需要討論, 藉由量綱分析(dimensional analysis)可以直觀地看出來 $x, y\gt0$。 首先將積分拆成兩部分

$$\int_0^{1 }t^{x-1}(1-t)^{y-1}dt = \int_0^{1\over 2}t^{x-1}(1-t)^{y-1}dt + \int_{1\over 2}^1 t^{x-1}(1-t)^{y-1}dt.$$

\begin{aligned} \int_0^{1\over 2}t^{x-1}(1-t)^{y-1}dt\approx [t]^{x-1}[dt]= [t]^x, \quad ([t]\to 0) &\quad\Longrightarrow\quad x\gt0, \\[2mm] \int_{1\over 2}^1 t^{x-1}(1-t)^{y-1}dt \approx [1-t]^{y-1}[dt]= [1-t]^y, \quad ([1-t]\to 0) &\quad\Longrightarrow\quad y\gt0. \end{aligned}

\begin{align} \begin{aligned} \Omega_n=V_n(1) &=\mathop{\int\cdots\int}_{0\le x_1^2+\cdots+x_n^2 \le 1}dx_1dx_2\cdots dx_n \\[2mm] &=\int_{-1}^1\left(\mathop{ \int\cdots\int}_{0\le x_1^2+\cdots+x_{n-1}^2 \le 1-x_n^2} dx_1dx_2\cdots dx_{n-1}\right)dx_n \\[2mm] &=\int_{-1}^1V_{n-1}\big(\sqrt{1-x_n^2}\,\big)dx_n \\[2mm] &=\int_{-1}^1(1-x_n^2)^{n-1 \over 2}V_{n-1}(1) dx_n\qquad \hbox{(均勻性)} \\[2mm] &=V_{n-1}(1)\int_0^1(1-t)^{{n+1 \over 2}-1}t^{{1 \over 2}-1}dt,\qquad (t=x_n^2). \end{aligned} \label{2.4} \end{align}

\eqref{2.4} 最後出現的瑕積分就是著名的 Beta 函數

\begin{align} \begin{aligned} B\bigg({n+1 \over 2}, {1\over 2}\bigg) &=\int_0^1(1-t)^{{n+1 \over 2}-1}t^{{1 \over 2}-1}dt \\[2mm] &=2\int_0^{\pi\over 2} \sin^n \theta d\theta,\qquad (t=\cos^2\theta). \end{aligned} \label{2.5} \end{align}

\begin{align} \Omega_n= V_n(1)=V_{n-1}(1)B\bigg({n+1\over 2},{1\over 2}\bigg)= \Omega_{n-1}\cdot 2\int_0^{\pi\over 2} \sin^n \theta d\theta. \label{2.6} \end{align}

\begin{align} \begin{aligned} V_n(1)&= V_{n-1}(1)B\bigg({n+1\over 2},{1\over 2}\bigg) \\[2mm] &= V_{n-2}(1)B\bigg({n\over 2},{1\over 2}\bigg) B\bigg({n+1\over 2},{1\over 2}\bigg) \\[2mm] &= V_{1}(1)B\bigg({3\over 2},{1\over 2}\bigg)\cdots B\bigg({n\over 2},{1\over 2}\bigg) B\bigg({n+1\over 2},{1\over 2}\bigg), \end{aligned} \label{2.7} \end{align}

\begin{align} \begin{aligned} V_n(1)&= V_{n-1}(1)\cdot 2\int_0^{\pi\over 2} \sin^n \theta d\theta \\[2mm] &= V_{n-2}(1)\cdot 2\int_0^{\pi\over 2} \sin^{n-1} \theta d\theta \cdot 2\int_0^{\pi\over 2} \sin^{n} \theta d\theta \\[2mm] &=\cdots \cdots \cdots \\[2mm] &= V_{1}(1)\cdot 2\int_0^{\pi\over 2} \sin^{2} \theta d\theta \cdots \cdot 2\int_0^{\pi\over 2} \sin^{n} \theta d\theta. \end{aligned} \label{2.8} \end{align}

$$\int_0^{\pi\over 2} \sin \theta d\theta =1 \quad\Longrightarrow\quad V_1(1)=2= 2\int_0^{\pi\over 2} \sin \theta d\theta=B\bigg({1+1\over 2},{1\over 2}\bigg),$$

\begin{align} \begin{aligned} V_n(1)&= B\bigg({1+1\over 2},{1\over 2}\bigg)\cdots B\bigg({n\over 2},{1\over 2}\bigg)B\bigg({n+1\over 2},{1\over 2}\bigg) \\[2mm] &=2\int_0^{\pi\over 2} \sin \theta d\theta\cdot 2\int_0^{\pi\over 2} \sin^{2} \theta d\theta \cdots \cdot 2\int_0^{\pi\over 2} \sin^{n} \theta d\theta \\[2mm] &=\int_0^{\pi} \sin \theta d\theta\cdot \int_0^{\pi} \sin^{2} \theta d\theta \cdots \cdot \int_0^{\pi}\sin^n \theta d\theta \\[2mm] &=\prod_{k=1}^n \int_0^{\pi} \sin^{k} \theta d\theta. \end{aligned} \label{2.9} \end{align}

\begin{align} B(x,y) = \frac{\Gamma(x)\Gamma (y)}{\Gamma(x+y)},\qquad x, y\gt0, \label{2.10} \end{align}

\begin{align} V_n(1)&=B\bigg({1+1\over 2},{1\over 2}\bigg)\cdots B\bigg({n\over 2},{1\over 2}\bigg)B\bigg({n+1\over 2},{1\over 2}\bigg) \nonumber\\[2mm] &={\Gamma(1)\Gamma({1\over 2})\over \Gamma({3\over 2})} \cdots {\Gamma({n\over 2})\Gamma({1\over 2})\over \Gamma({n+1\over 2})} {\Gamma({n+1\over 2})\Gamma({1\over 2})\over \Gamma({n+2\over 2})} \label{2.11}\\[2mm] &={\left[\Gamma({1\over 2})\right]^n \over \Gamma({n+2\over 2})} ={(\sqrt{\pi})^n \over \Gamma({n+2 \over 2})} ={\pi^{n\over 2} \over {n\over 2}\Gamma({n \over 2})}.\tag*{$\Box$} \end{align}

$$\int_0^{2\pi}\int_0^1 (1-r^2)^{{n\over 2}-1} r dr d\theta ={2\pi\over n} \quad\Longrightarrow\quad \Omega_n={2\pi\over n} \Omega_{n-2}.$$

$\Box$

\begin{align} \Omega_n =V_n(1) = 2^n W_1 W_2 \cdots W_n. \label{2.14}\end{align} 由 \eqref{2.5} 與 \eqref{2.13} 可得 \begin{align} W_n = {1\over 2} B\bigg({n+1\over 2}, {1\over 2}\bigg)={\Gamma({n+1\over 2})\Gamma({1\over 2})\over {n\over 2}\Gamma({n\over 2})}. \label{2.15}\end{align}

(1) $\{W_n\}$是一非負的遞減數列。

(2) $W_n$滿足遞迴關係(差分方程)

(3) 如果分開奇數與偶數則 ($p\ge 0$)

\begin{align} \left\{ \begin{aligned} W_{2p}&= {2p-1\over 2p}\cdot {2p-3\over 2p-2}\cdots {1\over 2}\cdot W_0 = {(2p)!\over 2^{2p}(p!)^2} \cdot {\pi\over 2}, \\[2mm] W_{2p+1}&= {2p\over 2p+1}\cdot {2p-2\over 2p-1}\cdots {2\over 3}\cdot W_1 = {2^{2p}(p!)^2\over (2p+1)!}. \end{aligned} \right. \label{2.17}\end{align}

(4) 漸近行為: 當 $n\to \infty$ 時 $W_n \sim \sqrt{\dfrac{\pi}{2n}}$

\begin{align} \lim_{n\to \infty} \sqrt{n}W_n = \sqrt{\pi\over 2}. \label{2.18}\end{align}

(1) 因為$\sin x$在$[0,\pi/2]$是非負的, 所以$W_n$也是一非負的實數。另外

$$W_n-W_{n+1}= \int_0^{\pi\over 2} \sin^n x (1-\sin x) dx \ge 0.$$

$$0\le \sin x\le 1 \ \Longrightarrow\ \hbox{\{\sin^n x\}是一遞減數列} \ \Longrightarrow\ \hbox{\{W_n\}是一遞減數列。}$$

(2) 這直接是分部積分的應用

\begin{aligned} W_n&=\int_0^{\pi\over 2} \sin^n x dx=\int_0^{\pi\over 2} \sin^{n-1} x\sin x dx =-\int_0^{\pi\over 2} \sin^{n-1} x d\cos x \\[2mm] &= -\sin^{n-1} x \cos x\bigg|_0^{\pi\over 2} + \int_0^{\pi\over 2} \cos x d \sin^{n-1} x \\[2mm] &= (n-1)\int_0^{\pi\over 2} \cos^2 x \sin^{n-2} x dx = (n-1)\int_0^{\pi\over 2} (1-\sin^2 x) \sin^{n-2} x dx \\[2mm] &= (n-1)W_{n-2} - (n-1) W_n, \end{aligned}

$$W_n= {n-1\over n} W_{n-2}.$$

(3) \eqref{2.17} 直接是 \eqref{2.16} 之推論, 我們留給讀者練習。

(4) 由 \eqref{2.17} 可以觀察到Wallis積分的特殊結構, 假設$n=2p$則

$$W_n W_{n+1} = W_{2p} W_{2p+1} = {2^{2p}(p!)^2\over (2p+1)!}\cdot{(2p)!\over 2^{2p}(p!)^2} \cdot {\pi\over 2}={1\over n+1}{\pi\over 2}.$$

$$\lim_{n\to \infty} n W_n^2 = \lim_{n\to \infty} n W_n W_{n+1}= {\pi\over 2} \quad\Longrightarrow\quad \lim_{n\to \infty} \sqrt{n}W_n = \sqrt{\pi\over 2}.$$

$\Box$

(a) Wallis 積分 $W_n$ 滿足二階差分方程 \eqref{2.16} 這相當於二階微分方程, 因此需要兩個(初始)值 $W_0, W_1$ 以決定所有的 $W_n$。

(b) 由 \eqref{2.15} 可得特殊的 Wallis 積分

\begin{align} {1\over \pi}\int_0^\pi \sin^{2n} x dx = {2\over \pi}\int_0^{\pi\over 2} \sin^{2n} x dx ={1\cdot 3\cdots (2n-1)\over 2\cdot 4\cdots 2n}. \label{2.19}\end{align}

(c) \eqref{2.16} 告訴我們 $W_n$ 是以 $\dfrac 1{\sqrt{n}}$ 的速率衰減。 這個事實由 Wallis 積分的定義 \eqref{2.13} 並不容易看出來(如果利用量綱分析只能猜到 $\dfrac 1{n}$), 而是透過奇、 偶相乘彼此互相抵消 (cancellation) 才得到這個量。

### 3. 高斯 --- Poisson 積分與球表面積

\begin{align} A\triangleq \int_{-\infty}^\infty e^{-x^2} dx = 2\int_0^\infty e^{-x^2} dx. \label{3.1}\end{align}

\begin{align} \int_0^\infty e^{-x^2} dx ={1\over 2} \int_0^\infty e^{-t} {1\over{\sqrt t}} dt =-{1\over 2} \int_0^\infty {1\over{\sqrt t}} de^{-t} =\cdots =? \label{3.2}\end{align}

\begin{aligned} A^2&= \int_{-\infty}^\infty e^{-x^2} dx \cdot \int_{-\infty}^\infty e^{-y^2} dy \\[2mm] &=\int_{-\infty}^\infty \int_{-\infty}^\infty e^{-(x^2+y^2)} dxdy \\[2mm] &=\int_0^{2\pi}\int_0^\infty e^{-r^2} r dr d\theta =\int_0^\infty e^{-r^2} 2\pi r dr \\[2mm] &= \pi \int_0^\infty e^{-r^2} dr^2 = \pi, \end{aligned}

$$A=\int_{-\infty}^\infty e^{-x^2} dx = {\sqrt\pi}.$$

$\Box$

\begin{aligned} D_1&=\{(x,y)\in \Bbb{R}^2| 0\le x^2+y^2\le R^2\}, \\[2mm] D_2&=\{(x,y)\in \Bbb{R}^2| 0\le x^2+y^2\le 2R^2\}, \\[2mm] S&=\{(x,y)\in \Bbb{R}^2| -R\le x\le R, -R\le y\le R\}. \end{aligned}

$$\int\!\!\!\int_{D_1} e^{-(x^2+y^2)}dxdy \le \int\!\!\!\int_{S} e^{-(x^2+y^2)}dxdy \le \int\!\!\!\int_{D_2} e^{-(x^2+y^2)}dxdy.$$

$$\int_0^R 2\pi r e^{-r^2} dr \le \bigg(\int_{-R}^R e^{-x^2} dx\bigg)^2 \le \int_0^{\sqrt{2}R} 2\pi r e^{-r^2} dr.$$ $$\pi\Big(1-e^{-R^2}\big) \le \bigg(\int_{-R}^R e^{-x^2} dx\bigg)^2 \le \pi\Big(1-e^{-2R^2}\Big).$$

$$\bigg(\int_{-\infty}^\infty e^{-x^2} dx\bigg)^2 = \pi \quad\Longrightarrow\quad \int_{-\infty}^\infty e^{-x^2} dx = \sqrt{\pi}.$$

$\Box$

\begin{align} \Gamma(x)=\int_0^{\infty}t^{x-1}e^{-t} dt, \qquad x\gt0. \label{3.4}\end{align}

$\Gamma$-函數最開始是稱為Euler第二類積分。這個瑕積分(improper integral)有《太胖》與《太高》的問題, $t=0, \infty$ 這兩點需要討論, 藉由量綱分析(dimensional analysis)可以直觀地看出來 $x\gt0$。 首先將積分拆成兩部分(這是處理瑕積分的典型手法)

$$\int_0^{\infty}t^{x-1}e^{-t} dt=\int_0^{1}t^{x-1}e^{-t} dt+\int_1^{\infty}t^{x-1}e^{-t} dt,$$

$$\lim_{t\to \infty}{t^{x-1}e^{-t}\over t^2} = 0,$$ $$\int_1^\infty {1\over t^2} dt\lt\infty \quad\Longrightarrow\quad \int_1^{\infty}t^{x-1}e^{-t} dt\lt\infty.$$

$$\int_0^{1}t^{x-1}e^{-t} dt\le \int_0^{1}t^{x-1} dt \approx [t]^{x-1}[dt]= [t]^x, \quad ([t]\to 0) \quad\Longrightarrow\quad x\gt0.$$

\begin{aligned} n&=2\quad \hbox{時} \quad V_2(r)=\pi r^2, \qquad \Bbb{S}^{1}(r)=2\pi r, \\ n&=3\quad \hbox{時} \quad V_3(r)={4 \over 3}\pi r^3,\,\quad \Bbb{S}^{2}(r)=4\pi r^2. \end{aligned}

\begin{align} {d\over dr} V_n(r)=\Bbb{S}^{n-1}(r). \label{3.6}\end{align}

$$\int_{-\infty}^\infty \int_{-\infty}^\infty e^{-(x^2+y^2)} dxdy =\int_0^\infty e^{-r^2} 2\pi r dr,$$

\begin{align} V_2(a)=\int_0^a 2\pi rdr=\pi a^2. \label{3.7}\end{align}

\begin{align} V_n(a)=\int_0^a \Bbb{S}^{n-1}(r)dr=\Bbb{S}^{n-1}(1)\int_0^a r^{n-1}dr ={\Bbb{S}^{n-1}(1)\over n} a^n. \label{3.8}\end{align}

$$f(\pmb{x})=f(|\pmb{x}|)=f\bigg(\sqrt{x_1^2+\cdots+x_n^2}\,\bigg)=f(r),$$

\begin{aligned} \mathop{\int\cdots\int}_{0\le x_1^2+\cdots+x_n^2 \le a^2} &f\bigg(\sqrt{x_1^2+\cdots+x_n^2}\,\bigg) dx_1dx_2\cdots dx_n \\[2mm] &=\int_0^a f(r) \Bbb{S}^{n-1}(r)dr =\Bbb{S}^{n-1}(1) \int_0^a f(r)r^{n-1}dr. \end{aligned}

\begin{align} \omega_n \triangleq \Bbb{S}^{n-1}(1)=n \cdot{(\sqrt{\pi})^n \over \Gamma({n+2\over 2})} ={2 \pi^{n\over 2} \over \Gamma({n \over 2})}. \label{3.10}\end{align}

\begin{align} \Bbb{S}^{n-1}(a)=\Bbb{S}^{n-1}(1)a^{n-1}=\omega_n a^{n-1} ={2 \pi^{n\over 2}\over \Gamma({n \over 2})}a^{n-1}. \label{3.11}\end{align}

$$\int_{-\infty}^{\infty}\cdots \int_{-\infty}^{\infty}e^{-(x_1^2+\cdots+x_n^2)} dx_1\cdots dx_n =\prod_{k=1}^n \int_{-\infty}^{\infty}e^{-x_k^2}dx_k = \pi^{n\over 2}.$$

\begin{aligned} &\int_{-\infty}^{\infty}\cdots\int_{-\infty}^{\infty} e^{-(x_1^2+\cdots+x_n^2)} dx_1\cdots dx_n =\int_{0}^{\infty}e^{-r^2}\Bbb{S}^{n-1}(r)dr \\[2mm] &=\Bbb{S}^{n-1}(1)\int_0^{\infty}e^{-r^2}r^{n-1}dr =\Bbb{S}^{n-1}(1){1\over 2}\int_0^{\infty}e^{-t}t^{{n\over 2}-1}dt \\[2mm] &=\Bbb{S}^{n-1}(1){1 \over 2} \Gamma\Big({n\over 2}\Big). \end{aligned}

$$\omega_n = \Bbb{S}^{n-1}(1) ={2 \pi^{n\over 2} \over \Gamma({n \over 2})}.$$

$\Box$

(a) 如果與定理 2.2 比較, 感覺上定理 3.5 並不是直接得到單位球體積(或表面積)。 這樣的認知不完全是正確的! 對於圓應該有的常識(common sense)

$$\hbox{『有圓就有 \pi、 有 \pi 就有圓。』}$$

(b) 由單位球表面積 \eqref{3.10} 可得單位球體積

\begin{align} \Omega_n=V_n(1) = {\omega_n\over n}={\pi^{n\over 2} \over {n\over 2}\Gamma({n \over 2})}. \label{3.12}\end{align}

\begin{align} \omega_n = n \Omega_n = n 2^n W_1 W_2 \cdots W_n. \label{3.13}\end{align}

(c) Wallis積分 \eqref{2.13} 可以藉由單位球表面積重新表示為

\begin{align} W_n = \int_0^{\pi\over 2} \sin^n x dx = {\Omega_n\over 2\Omega_{n-1}}= {n-1\over 2n}\cdot {\omega_n\over \omega_{n-1}}. \label{3.14}\end{align}

(d) 由 Wallis 積分可以回頭推得高斯--Poisson 積分。 首先由 Taylor 級數或直接微分有不等式

$$1-t \le e^{-t}, \quad 1+t \le e^t, \qquad 0\le t\lt\infty,$$

\begin{align} 1-t \le e^{-t} \le {1\over 1+t},\qquad 0\le t\lt \infty. \label{3.15}\end{align}

\begin{align} \bigg(1-{x^2\over n}\bigg)^n \le e^{-x^2}\le \bigg(1+{x^2\over n}\bigg)^{-n}, \label{3.16}\end{align}

\begin{align} \int_0^{\sqrt{n} }\bigg(1-{x^2\over n}\bigg)^n dx \le \int_0^{\sqrt{n} }e^{-x^2}dx \le \int_0^{\sqrt{n} }\bigg(1+{x^2\over n}\bigg)^{-n}dx. \label{3.17}\end{align}

\eqref{3.17} 左右兩個積分可以透過三角變數變換轉換為 Wallis 積分

\left\{ \begin{aligned} \int_0^{\sqrt{n} }\bigg(1-{x^2\over n}\bigg)^n dx &=\sqrt{n}\int_0^1 \cos^{2n+1} t dt =\sqrt{n} W_{2n+1},\qquad x =\sqrt{n} \sin t, \\[2mm] \int_0^{\sqrt{n} }\bigg(1+{x^2\over n}\bigg)^{-n} dx &=\sqrt{n}\int_0^1 \cos^{2n-2} t dt =\sqrt{n} W_{2n-2},\qquad x =\sqrt{n} \tan t, \end{aligned} \right.

\begin{align} \sqrt{n} W_{2n+1} \le \int_0^{\sqrt{n} }e^{-x^2}dx \le \sqrt{n} W_{2n-2}. \label{3.18}\end{align}

\begin{align} \lim_{n\to \infty} \sqrt{n} W_{2n+1}= \lim_{n\to \infty} \sqrt{n} W_{2n-2}={1\over \sqrt{2}}\sqrt{\pi\over 2}= {\sqrt{\pi}\over 2}. \label{3.19}\end{align}

$$\int_{-\infty}^\infty e^{-x^2}dx= 2\int_0^\infty e^{-x^2}dx=\sqrt{\pi},$$

\begin{align} \lim_{n\to \infty} \sqrt{n} W_{2n+1}= \lim_{n\to \infty} \sqrt{n} W_{2n-2}=\int_0^\infty e^{-x^2} dx. \label{3.20}\end{align}

(e) 以高斯(Carl Friedrich Gauss, 1777$\sim$1855)為頭像的 10 德國馬克是我最愛的紙鈔之一(另一張是歐拉 (Leonhard Euler, 1707$\sim$1783) 的 10 瑞士法郎, 隨時都放在皮夾裡。) 這張鈔票有兩個看點: 正面除了高斯之外還印有高斯的常(正)態分配, 背面則有測量的工具與三角形形成的丈量區域 (這就是著名的單形法(simplex method)構成了微分幾何之逼近理論)。 天才就是天才連做測量這個無聊的工作也可以創出一門學問(非歐幾何與微分幾何)。

### 4. $n$ 維球座標

\begin{align} \left\{ \begin{aligned} z &=\rho\cos\varphi \\[2mm] x &=\rho\sin\varphi\cos\theta \\[2mm] y &=\rho\sin\varphi\sin\theta \end{aligned} \right. \qquad\Longleftrightarrow\qquad \left\{ \begin{aligned} \rho &=\sqrt{x^2+y^2+z^2} \\[2mm] \varphi&=\cos^{-1}\frac{z}{\sqrt{x^2+y^2+z^2}} \\[2mm] \theta &=\tan^{-1}\frac{y}{x} \end{aligned}, \right. \label{4.1}\end{align}

$$0\le \rho \lt\infty,\qquad 0\le \varphi\le \pi,\qquad 0\le \theta \lt2\pi$$

\begin{align} ds_1 = d\rho,\qquad ds_2 = \rho d\varphi,\qquad ds_3 = \rho \sin \varphi d\theta. \label{4.2a}\end{align} 體積元 (volume element) 與球面上的面積元 (area element) 則是 \begin{align} \begin{aligned} dV&= dxdydz = {\partial(x,y,z)\over \partial (\rho, \varphi, \theta)}d\rho d\varphi d\theta= \rho^2 \sin \varphi d\rho d\varphi d\theta, \\[2mm] dS_\rho &= \rho^2 \sin \varphi d\varphi d\theta,\qquad \rho=\hbox{常數。} \end{aligned} \label{4.2b}\end{align}

\begin{align} d\Omega = {dS_\rho \over \rho^2} = \sin \varphi d\varphi d\theta. \label{4.2c}\end{align}

$$\sqrt{x^2+y^2}\ge 0,\quad \rho\ge 0 \quad\Longrightarrow\quad \sin \varphi\ge 0 \quad\Longrightarrow\quad 0\le \varphi \le \pi$$

\begin{align} \left\{ \begin{aligned} x_1 &= r \cos \theta_1, \\[2mm] x_2 &= r \sin \theta_1\cos \theta_2, \\[2mm] &\cdots\cdots \\[2mm] x_{n-1}&= r\sin\theta_1 \sin\theta_2\cdots \sin\theta_{n-2} \cos\theta_{n-1}, \\[2mm] x_n&=r\sin\theta_1 \sin\theta_2\cdots \sin\theta_{n-2} \sin\theta_{n-1}, \end{aligned} \right.\hskip 3cm~ \label{4.4}\\ 0\le r \lt \infty,\qquad 0\le \theta_k\le \pi, \qquad k=1,2,\ldots, n-2,\qquad 0\le \theta_{n-1}\le 2\pi.\nonumber \end{align}

\begin{align} \left\{ \begin{aligned} ds_1 &= dr, \\[2mm] ds_2 &= r d\theta_1, \\[2mm] ds_3 &= r \sin \theta_1 d\theta_2, \\[2mm] &\cdots \cdots \\[2mm] ds_n & = r \sin \theta_1 \sin \theta_2 \cdots \sin \theta_{n-2} d\theta_1 d\theta_2\cdots d\theta_{n-1}. \end{aligned} \right. \label{4.5}\end{align}

\begin{align} \begin{aligned} dV_n&= dx_1dx_2\cdots dx_n = ds_1 ds_2 \cdots ds_n \\[2mm] &= (r^{n-1}dr)(\sin^{n-2}\theta_1 d \theta_1)(\sin^{n-3}\theta_2 d \theta_2)\cdots (\sin\theta_{n-2} d \theta_{n-2})(d\theta _{n-1}) \\[2mm] &= {\partial (x_1, x_2,\ldots, x_n)\over \partial (r,\theta_1,\ldots, \theta_{n-1})} dr d \theta_1d \theta_2\cdots d\theta _{n-1} \\[2mm] &= r^{n-1}\sin^{n-2}\theta_1 \sin^{n-3}\theta_2 \cdots \sin\theta_{n-2}\, dr d \theta_1d \theta_2\cdots d\theta _{n-1}. \end{aligned} \label{4.6}\end{align}

\begin{align} V_n(1)&=\mathop{\int\cdots\int}_{0\le x_1^2+\cdots+x_n^2 \le 1}dV_n\nonumber\\[2mm] &= \int_0^1 \!\!r^{n-1}dr\int_0^\pi \!\!\sin^{n-2}\theta_1 d \theta_1 \cdot\int_0^\pi \!\!\sin^{n-3}\theta_2 d \theta_2\cdots \int_0^{\pi}\!\!\sin\theta_{n-2} d \theta_{n-2}\int_0^{2\pi}\!\!d\theta _{n-1} \nonumber\\[2mm] &= {2\pi\over n} 2^{n-2} W_1W_2\cdots W_{n-2} = {2\pi\over n} \Omega_{n-2} = \Omega_n. \label{4.7}\end{align}

\begin{align} \Bbb{S}^{n-1}(1)&= \mathop{\int\cdots\int}_{x_1^2+\cdots+x_n^2 = 1}d \Bbb{S}^{n-1} \nonumber\\[2mm] &= \int_0^\pi \sin^{n-2}\theta_1 d \theta_1 \cdot\int_0^\pi \sin^{n-3}\theta_2 d \theta_2\cdots \int_0^{\pi}\sin\theta_{n-2} d \theta_{n-2}\int_0^{2\pi}d\theta _{n-1} \nonumber\\[2mm] &= 2\pi \cdot 2^{n-2} W_1 W_2\cdots W_{n-2} = n \Omega_n = \omega_n. \label{4.8}\end{align}

\begin{align} dS_r = r^{n-1}(\sin^{n-2}\theta_1 d \theta_1)(\sin^{n-3}\theta_2 d \theta_2)\cdots (\sin\theta_{n-2} d \theta_{n-2})(d\theta _{n-1}), \label{4.9}\end{align}

\begin{align} d\Omega = {dS_r \over r^{n-1}} = (\sin^{n-2}\theta_1 d \theta_1)(\sin^{n-3}\theta_2 d \theta_2)\cdots (\sin\theta_{n-2} d \theta_{n-2})(d\theta _{n-1}). \label{4.10}\end{align}

### 5. Wallis 無窮乘積

\begin{align} {\pi\over 2} = \prod_{k=1}^\infty \bigg({2k\over 2k-1}\cdot {2k\over 2k+1}\bigg) ={2\over 1}\cdot {2\over 3}\cdot {4\over 3}\cdot {4\over 5}\cdot {6\over 5}\cdot {6\over 7}\cdots. \label{5.1}\end{align}

$$\sin^{2n+1} x \le \sin^{2n} x \le \sin^{2n-1} x \quad\Longrightarrow\quad W_{2n+1}\le W_{2n}\le W_{2n-1}.$$

$$1\le {W_{2n}\over W_{2n+1}} \le {W_{2n-1}\over W_{2n+1}}= {2n+1\over 2n},$$

\begin{align} \lim_{n\to \infty} {W_{2n}\over W_{2n+1}}={\pi\over 2}\prod_{k=1}^\infty \bigg({2k-1\over 2k}\cdot {2k+1\over 2k}\bigg)=1. \label{5.2}\end{align}

$\Box$

\begin{align} \sin x =x\prod_{n=1}^\infty\bigg(1-{x^2\over n^2 \pi^2}\bigg) =x\bigg( 1-{x^2 \over \pi^2}\bigg) \bigg( 1-{x^2 \over 4\pi^2}\bigg) \bigg( 1-{x^2 \over 9\pi^2}\bigg) \cdots. \label{5.3}\end{align}

$$\bigg(1-{1\over 2^2}\bigg)\bigg(1-{1\over 4^2}\bigg)\bigg(1-{1\over 6^2}\bigg) \cdots = {2\over \pi}.$$

$$\bigg(1-{1\over 2}\bigg)\bigg(1+{1\over 2}\bigg)\bigg(1-{1\over 4}\bigg)\bigg(1-{1\over 4}\bigg)\bigg(1-{1\over 6}\bigg)\bigg(1+{1\over 6}\bigg) \cdots = {2\over \pi},$$

$${1\over 2}\cdot{3\over 2}\cdot{3\over 4}\cdot{5\over 4}\cdot{5\over 6}\cdot {7\over 6}\cdots = {2\over \pi}.$$

(a) 藉由分部積分推得 Wallis 積分滿足遞迴公式, 我們觀察到不同次數的 Wallis 積分相差一個常數, 所以分部積分這個動作相當於乘積(product)。 所以不難想像 Wallis 無窮乘積是 Wallis 積分經過無窮多次的分部積分而來的結果, 同樣的論證也出現在將瑕積分形式的 $\Gamma$-函數轉化為高斯形式的無窮乘積。

(b) 細心觀察 Wallis 無窮乘積各項為

$${2\over 1}, \quad {2\over 3},\quad {4\over 3},\quad {4\over 5},\quad {6\over 5},\quad {6\over 7},\quad \cdots,$$

(c) Wallis 無窮乘積可以表示為

\begin{align} \begin{aligned} \hskip 2cm~\lim_{n\to \infty}\bigg({2\over 1}\bigg)^2\bigg({4\over 3}\bigg)^2 \cdots \bigg({2n\over 2n-1}\bigg)^2 {1\over 2n+1}={\pi \over 2},\hskip 2cm~\\ {\hbox{或}} \lim_{n\to \infty} {2\over 1}\cdot{4\over 3}\cdot {6\over 5}\cdot \cdots {2n \over 2n-1}{1\over \sqrt{n}} =\sqrt{\pi}.\hskip 2cm~ \end{aligned}\label{5.4} \end{align}

\begin{align} {1\cdot 3\cdot 5\cdot\cdots (2n-1)\over 2\cdot 4\cdot 6\cdot\cdots 2n} ={1\over 2^{2n}} \begin{pmatrix} 2n \\ n\end{pmatrix} = (-1)^n \begin{pmatrix}-1/2 \\ n \end{pmatrix} \approx {1\over \sqrt{n \pi}}. \label{5.5}\end{align}

\eqref{5.5} 有漂亮的機率解釋: 假設丟硬幣正面、反面的機率都是 $1/2$。 當 $n$ 相當大則丟 $2n$ 次硬幣剛好出現 $n$ 次正面 $n$ 次反面的機率趨近於 $\dfrac {1}{\sqrt{n \pi}}$;

$${1\over 2^{n}}\cdot {1\over 2^{n}}\begin{pmatrix} 2n \\ n\end{pmatrix} \approx {1\over \sqrt{n \pi}}.$$

(d) 由 Wallis 乘積可以推得高斯--Poisson 積分。 考慮函數 $e^{-x^2}$ 的 $n$ 次矩 ($n$-th moment)

\begin{align} I_n =\int_0^\infty x^n e^{-x^2} dx, \label{5.6}\end{align}

\begin{align} 2I_n =(n-1) I_{n-2}\,,\qquad n\ge 2. \label{5.7}\end{align}

\begin{align} 2^k I_{2k}=1\cdot 3\cdot 5\cdots (2k-1) I_0\,,\qquad 2I_{2k+1} = k!\,,\qquad k\in \Bbb{N} . \label{5.8}\end{align}

$$I_n^2 \lt I_{n-1} I_{n+1}\qquad\hbox{或}\qquad 2I_n^2 \lt n I_{n-1}^2,$$

$${(k!)^2 \over 4k+2}= {2\over 2k+1} I_{2k+1}^2\lt I_{2k}^2 \lt I_{2k-1} I_{2k+1} = {(k!)^2 \over 4k},$$

$$I_{2k}^2 ={(k!)^2 \over 4k+2}(1+\varepsilon_k)\,,\qquad 1\lt\varepsilon_k \lt{1\over 2k},$$

$$2I_0^2 = {[2\cdot 4\cdot 6\cdots (2k)]^2 \over [1\cdot 3\cdot 5\cdots (2k-1)]^2} (1+\varepsilon_k).$$

$$2I_0^2 ={\pi \over 2} \qquad\hbox{或}\qquad \int_0^\infty e^{-x^2} dx = {\sqrt \pi \over 2}.$$

### 參考文獻

Amir Alexander, Infinitesimal: How a Dangerous Mathematics Theory Shaped the Modern World, Oneworld Publications, 2014.

E. Artin, The Gamma Function, Dover Publications; Reprint edition, Jan. 28, 2015。

《I feel that this monograph will help to show that the gamma function can be thought of as one of the elementary functions, and that all of its basic properties can be established using elementary methods of the calculus.》

Richard Courant and Fritz John, Introduction to Calculus and Analysis, Vol. I, II, Springer-Verlag New-York, Inc, 1989. Peter Duran, Invitation to Classical Analysis, Pure and Applied Undergraduate Texts; Vol. 17, American Mathematical Society, 2012.

Laurent Schwartz, Mathematics for the Physical Sciences, Hermann Paris, 1966.

O. Hijab, Introduction to Calculus and Classical Analysis, 2nd Edition, Springer-Verlag, 2007. Eli Maor, Trigonometric Delights, Princeton University Press, Princeton, New Jersey, 1998。 中譯本: 毛起來說三角。 鄭惟厚譯。 天下文化出版, 2000。 E. T. Whittaker and G. N. Watson, A Course of Modern Analysis, 4th Edition, Cambridge Mathematical Library, Cambridge University Press, 1962.