【1】引言
前序学习进程中,已经对伽马函数阶乘表达式,积分式和阶乘式等价和阶乘的积分表达式。
今天来一起梳理一下,因为这个学习过程的确翻来覆去。
【2】阶乘式
证明 n ! n! n!可以改写成下式:
n ! = l i m k → + ∞ k n ⋅ k ! ( n + 1 ) ( n + 2 ) . . . ( n + k ) n!=lim_{k\rightarrow+\infty}\frac{k^n\cdot k!}{(n+1)(n+2)...(n+k)} n!=limk→+∞(n+1)(n+2)...(n+k)kn⋅k!这个式子的作用是,用 k k k的幂次抵消乘积的增长,让极限趋向于有限值。
证明这个式子:
第一步:
( n + 1 ) ( n + 2 ) . . . ( n + k ) = ( n + k ) ! n ! (n+1)(n+2)...(n+k)=\frac{(n+k)!}{n!} (n+1)(n+2)...(n+k)=n!(n+k)!
第二步,代入阶乘式有:
n ! = l i m k → + ∞ k n ⋅ k ! ⋅ n ! ( n + k ) ! = n ! l i m k → + ∞ k n ⋅ k ! ( n + k ) ! n!=lim_{k\rightarrow+\infty}\frac{k^n\cdot k!\cdot n!}{(n+k)!}=\\ n!lim_{k\rightarrow+\infty}\frac{k^n\cdot k!}{(n+k)!} n!=limk→+∞(n+k)!kn⋅k!⋅n!=n!limk→+∞(n+k)!kn⋅k!
所以对式子的证明,可以简化为:
l i m k → + ∞ k n ⋅ k ! ( n + k ) ! = 1 lim_{k\rightarrow+\infty}\frac{k^n\cdot k!}{(n+k)!}=1 limk→+∞(n+k)!kn⋅k!=1
第三步:
因为:
( n + k ) ! = [ k ! ] [ ( k + 1 ) ( k + 2 ) ⋅ ⋅ ⋅ ( k + n ) ] (n+k)!=[k!][(k+1)(k+2) \cdot \cdot \cdot(k+n)] (n+k)!=[k!][(k+1)(k+2)⋅⋅⋅(k+n)]
所以:
l i m k → + ∞ k n ⋅ k ! ( n + k ) ! = l i m k → + ∞ k n ⋅ k ! [ k ! ] [ ( k + 1 ) ( k + 2 ) ⋅ ⋅ ⋅ ( k + n ) ] = l i m k → + ∞ k n ⋅ k ! [ k ! ] [ ( k + 1 ) ( k + 2 ) ⋅ ⋅ ⋅ ( k + n ) ] = l i m k → + ∞ k n ( k + 1 ) ( k + 2 ) ⋅ ⋅ ⋅ ( k + n ) lim_{k\rightarrow+\infty}\frac{k^n\cdot k!}{(n+k)!}=\\lim_{k\rightarrow+\infty}\frac{k^n\cdot k!}{[k!][(k+1)(k+2) \cdot \cdot \cdot(k+n)]}=\\lim_{k\rightarrow+\infty}\frac{k^n\cdot k!}{[k!][(k+1)(k+2) \cdot \cdot \cdot(k+n)]}=\\ lim_{k\rightarrow+\infty}\frac{k^n}{(k+1)(k+2)\cdot \cdot \cdot(k+n)} limk→+∞(n+k)!kn⋅k!=limk→+∞[k!][(k+1)(k+2)⋅⋅⋅(k+n)]kn⋅k!=limk→+∞[k!][(k+1)(k+2)⋅⋅⋅(k+n)]kn⋅k!=limk→+∞(k+1)(k+2)⋅⋅⋅(k+n)kn
第四步:分母每个括号中都提取一个 k k k:
l i m k → + ∞ k n ( k + 1 ) ( k + 2 ) ⋅ ⋅ ⋅ ( k + n ) = l i m k → + ∞ k n [ k ( 1 + 1 k ) ] [ k ( 1 + 2 k ) ] ⋅ ⋅ ⋅ [ k ( 1 + n k ) ] = l i m k → + ∞ k n k n ⋅ ( 1 + 1 k ) ( 1 + 2 k ) ⋅ ⋅ ⋅ ( 1 + n k ) = l i m k → + ∞ 1 ( 1 + 1 k ) ( 1 + 2 k ) ⋅ ⋅ ⋅ ( 1 + n k ) lim_{k\rightarrow+\infty}\frac{k^n}{(k+1)(k+2)\cdot \cdot \cdot(k+n)}=\\ lim_{k \rightarrow+\infty}\frac{k^n}{[k(1+\frac{1}{k})][k(1+\frac{2}{k})]\cdot \cdot \cdot [k(1+\frac{n}{k})]}=\\ lim_{k\rightarrow+\infty}\frac{k^n}{k^n\cdot (1+\frac{1}{k})(1+\frac{2}{k})\cdot \cdot \cdot (1+\frac{n}{k})}=\\ lim_{k\rightarrow+\infty}\frac{1}{(1+\frac{1}{k})(1+\frac{2}{k})\cdot \cdot \cdot (1+\frac{n}{k})} limk→+∞(k+1)(k+2)⋅⋅⋅(k+n)kn=limk→+∞[k(1+k1)][k(1+k2)]⋅⋅⋅[k(1+kn)]kn=limk→+∞kn⋅(1+k1)(1+k2)⋅⋅⋅(1+kn)kn=limk→+∞(1+k1)(1+k2)⋅⋅⋅(1+kn)1
对于上述计算式,当 k → + ∞ k \rightarrow+\infty k→+∞时,分母的乘积为1,所以:
l i m k → + ∞ k n ( k + 1 ) ( k + 2 ) ⋅ ⋅ ⋅ ( k + n ) = 1 lim_{k\rightarrow+\infty}\frac{k^n}{(k+1)(k+2)\cdot \cdot \cdot(k+n)}=1 limk→+∞(k+1)(k+2)⋅⋅⋅(k+n)kn=1
第五步,反过来再直接推一遍式子:
因为:
l i m k → + ∞ k n ( k + 1 ) ( k + 2 ) ⋅ ⋅ ⋅ ( k + n ) = 1 = l i m k → + ∞ k n ⋅ k ! k ! ⋅ ( k + 1 ) ( k + 2 ) ⋅ ⋅ ⋅ ( k + n ) = l i m k → + ∞ k n ⋅ k ! ( k + n ) ! = 1 lim_{k\rightarrow+\infty}\frac{k^n}{(k+1)(k+2)\cdot \cdot \cdot(k+n)}=1\\= lim_{k\rightarrow+\infty}\frac{k^n\cdot k!}{k!\cdot (k+1)(k+2)\cdot \cdot \cdot(k+n)}=\\ lim_{k\rightarrow+\infty}\frac{k^n\cdot k!}{(k+n)!}=1 limk→+∞(k+1)(k+2)⋅⋅⋅(k+n)kn=1=limk→+∞k!⋅(k+1)(k+2)⋅⋅⋅(k+n)kn⋅k!=limk→+∞(k+n)!kn⋅k!=1
所以
n ! = n ! ⋅ l i m k → + ∞ k n ⋅ k ! ( n + k ) ! = l i m k → + ∞ k n ⋅ k ! ⋅ n ! ( n + k ) ! = l i m k → + ∞ k n ⋅ k ! ( n + 1 ) ( n + 2 ) . . . ( n + k ) n!=n! \cdot lim_{k\rightarrow+\infty}\frac{k^n\cdot k!}{(n+k)!}=\\lim_{k\rightarrow+\infty}\frac{k^n\cdot k!\cdot n!}{(n+k)!}=\\ lim_{k\rightarrow+\infty}\frac{k^n\cdot k!}{(n+1)(n+2)...(n+k)} n!=n!⋅limk→+∞(n+k)!kn⋅k!=limk→+∞(n+k)!kn⋅k!⋅n!=limk→+∞(n+1)(n+2)...(n+k)kn⋅k!
【3】对数积分表达式的指数形式
证明当 s s s为正整数 n n n时,
∫ 0 1 ( − l n t ) s d t = ∫ 0 + ∞ u s e − u d u \int_{0}^{1}(-lnt)^sdt=\int_{0}^{+\infty}u^se^{-u}du ∫01(−lnt)sdt=∫0+∞use−udu
首先令 u = − l n t u=-ln t u=−lnt,有: d u = − 1 t d t d t = − t d u t = e − u du=-\frac{1}{t}dt\\ dt=-tdu \\t=e^{-u} du=−t1dtdt=−tdut=e−u
此时被积函数变换为:
( − l n t ) s = u s (-lnt)^s=u^s (−lnt)s=us
当 t → 0 + t\rightarrow 0^+ t→0+时, u = − l n t = + ∞ u=-lnt=+\infty u=−lnt=+∞
当 t → 1 t\rightarrow 1 t→1时, u = − l n t = 0 u=-lnt=0 u=−lnt=0
将上述变换代入积分式:
∫ 0 1 ( − l n t ) s d t = ∫ + ∞ 0 u s ( − t ) d u = ∫ + ∞ 0 u s ( − e u ) d u = ∫ 0 + ∞ u s e − u d u \int_{0}^{1}(-lnt)^sdt=\int_{+\infty}^{0}u^s(-t)du=\\ \int_{+\infty}^{0}u^s(-e^u)du=\int_{0}^{+\infty}u^se^{-u}du ∫01(−lnt)sdt=∫+∞0us(−t)du=∫+∞0us(−eu)du=∫0+∞use−udu
【4】阶乘和指数形式的积分表达式相等
当 s s s为正整数 n n n时,积分先写作:
∫ 0 1 ( − l n t ) s d t = ∫ 0 + ∞ u n e − u d u \int_{0}^{1}(-lnt)^sdt=\int_{0}^{+\infty}u^ne^{-u}du ∫01(−lnt)sdt=∫0+∞une−udu
令 v = u n , d w = e − u d u v=u^n,dw=e^{-u}du v=un,dw=e−udu,有:
d v = n u n − 1 d u , w = − e − u dv=nu^{n-1}du,w=-e^{-u} dv=nun−1du,w=−e−u
此时积分式转化为:
∫ 0 1 ( − l n t ) s d t = ∫ 0 + ∞ u n e − u d u = ∫ 0 + ∞ v d w = v w ∣ 0 + ∞ − ∫ 0 + ∞ w d v = ( u n ( − e − u ) ) ∣ 0 + ∞ + ∫ 0 + ∞ n u n − 1 e − u d u = 0 + ∫ 0 + ∞ n u n − 1 e − u d u = n ∫ 0 + ∞ u n − 1 e − u d u \int_{0}^{1}(-lnt)^sdt=\int_{0}^{+\infty}u^ne^{-u}du=\\ \int_{0}^{+\infty}vdw=vw|{0}^{+\infty}-\int{0}^{+\infty}wdv=\\ (u^n(-e^{-u}))|{0}^{+\infty}+\int{0}^{+\infty}nu^{n-1}e^{-u}du=\\ 0+\int_{0}^{+\infty}nu^{n-1}e^{-u}du=n\int_{0}^{+\infty}u^{n-1}e^{-u}du ∫01(−lnt)sdt=∫0+∞une−udu=∫0+∞vdw=vw∣0+∞−∫0+∞wdv=(un(−e−u))∣0+∞+∫0+∞nun−1e−udu=0+∫0+∞nun−1e−udu=n∫0+∞un−1e−udu
这时候先暂停一下,根据前述推导有:
∫ 0 1 ( − l n t ) s d t = ∫ 0 + ∞ u n e − u d u = n ∫ 0 + ∞ u n − 1 e − u d u \int_{0}^{1}(-lnt)^sdt=\int_{0}^{+\infty}u^ne^{-u}du=n\int_{0}^{+\infty}u^{n-1}e^{-u}du ∫01(−lnt)sdt=∫0+∞une−udu=n∫0+∞un−1e−udu按照这个形式,会有:
∫ 0 1 ( − l n t ) s d t = ∫ 0 + ∞ u n e − u d u = n ∫ 0 + ∞ u n − 1 e − u d u = n ( n − 1 ) ∫ 0 + ∞ u n − 2 e − u d u = . . . = n ( n − 1 ) . . . 2 ∫ 0 + ∞ u 1 e − u d u = n ( n − 1 ) . . . 2 ⋅ 1 ∫ 0 + ∞ u 0 e − u d u = n ! \int_{0}^{1}(-lnt)^sdt=\int_{0}^{+\infty}u^ne^{-u}du=n\int_{0}^{+\infty}u^{n-1}e^{-u}du=\\ n(n-1)\int_{0}^{+\infty}u^{n-2}e^{-u}du=...=\\ n(n-1)...2\int_{0}^{+\infty}u^{1}e^{-u}du=\\ n(n-1)...2\cdot 1\int_{0}^{+\infty}u^{0}e^{-u}du=n! ∫01(−lnt)sdt=∫0+∞une−udu=n∫0+∞un−1e−udu=n(n−1)∫0+∞un−2e−udu=...=n(n−1)...2∫0+∞u1e−udu=n(n−1)...2⋅1∫0+∞u0e−udu=n!至此可知,当 s s s为正整数 n n n时,
∫ 0 1 ( − l n t ) s d t = s ! \int_{0}^{1}(-lnt)^sdt=s! ∫01(−lnt)sdt=s!
【5】说明
实际上这里展示了两种阶乘表达式:
n ! = l i m k → + ∞ k n ⋅ k ! ( n + 1 ) ( n + 2 ) . . . ( n + k ) n!=lim_{k\rightarrow+\infty}\frac{k^n\cdot k!}{(n+1)(n+2)...(n+k)} n!=limk→+∞(n+1)(n+2)...(n+k)kn⋅k!
∫ 0 1 ( − l n t ) s d t = ∫ 0 + ∞ u n e − u d u = s ! ( s 为整数,用 n 代表整数 ) \int_{0}^{1}(-lnt)^sdt=\int_{0}^{+\infty}u^ne^{-u}du=s!(s为整数,用n代表整数) ∫01(−lnt)sdt=∫0+∞une−udu=s!(s为整数,用n代表整数)
【6】总结
学习了伽马函数的推导过程。