分部积分的应用

2020-02-02 本文已影响0人洛玖言

$\displaystyle\int_0^1(1-x)^mx^n\text{d}x$

设
$\text{B}(m+1,n+1)=\displaystyle\int_0^1(1-x)^mx^n\text{d}x$

$\begin{aligned} &\int_0^1(1-x)^mx^n\text{d}x=\text{B}(m+1,n+1)\\ =&-\dfrac{1}{m+1}\int_0^1x^n\text{d}(1-x)^{m+1}\\ =&-\dfrac{1}{m+1}x^n(1-x)^{m+1}\bigg|_0^1+\dfrac{n}{m+1}\int_0^1x^{n-1}(1-x)^{m+1}\text{d}x\\ =&\dfrac{n}{m+1}\int_0^1x^{n-1}(1-x)(1-x)^m\text{d}x\\ =&\dfrac{n}{m+1}\int_0^1x^{n-1}(1-x)^m\text{d}x-\dfrac{n}{m+1}\int_0^1(1-x)^mx^n\text{d}x\\ =&\dfrac{n}{m+1}\left[\text{B}(m+1,n)-\text{B}(m+1,n+1)\right] \end{aligned}$

移项得

$\begin{aligned} \dfrac{n+m+1}{m+1}\text{B}(m+1,n+1)=\dfrac{n}{m+1}\text{B}(m+1,n) \end{aligned}$

因此有
$\begin{aligned} \int_0^1(1-x)^mx^n\text{d}x=\dfrac{n}{n+m+1}\int_0^1(1-x)^mx^{n-1}\text{d}x \end{aligned}$

继续用分部积分有

$\begin{aligned} &\int_0^1(1-x)^mx^{n-1}\text{d}x=\dfrac{n-1}{n+m}\int_0^1(1-x)^mx^{n-2}\text{d}x\\ &\int_0^1(1-x)^mx^{n-2}\text{d}x=\dfrac{n-2}{n+m-1}\int_0^1(1-x)^mx^{n-3}\text{d}x\\ &\cdots \end{aligned}$

得到
$\begin{aligned} \int_0^1(1-x)^mx^n\text{d}x=\dfrac{n(n-1)(n-2)\cdots3\cdot2\cdot1}{(m+n+1)(m+n)(m+n-1)\cdots(m+2)}\int_0^1(1-x)^m\text{d}x \end{aligned}$

又
$\begin{aligned} \int_0^1(1-x)^m\text{d}x=-\dfrac{1}{m+1}(1-x)^{m+1}\bigg|_0^1=\dfrac{1}{m+1} \end{aligned}$

最后得到
$\begin{aligned} &\int_0^1(1-x)^mx^n\text{d}x\\ =&\dfrac{n(n-1)(n-2)\cdots3\cdot2\cdot1}{(m+n+1)(m+n)\cdots(m+2)(m+1)}\\ =&\dfrac{n(n-1)(n-2)\cdots3\cdot2\cdot1\cdot m(m-1)\cdots2\cdot1}{(m+n+1)(m+n)\cdots(m+2)(m+1)\cdot m\cdots2\cdot1}\\ =&\dfrac{m!n!}{(m+n+1)!}\\ =&\dfrac{\Gamma(m+1)\Gamma(n+1)}{\Gamma(m+n+2)} \end{aligned}$

更多有关 $\Gamma$ 函数和 $\Beta$ 函数可参见: Euler积分

分部积分的应用

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