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(一)欧拉积分

时间:2014-05-04 11:54:07      阅读:500      评论:0      收藏:0      [点我收藏+]

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欧拉是数学家心目中的英雄,欧拉积分具有重要的应用。先给出欧拉积分的性质以便为进入分数阶微积分打下基础。

1.1 $\beta$函数定义

B(α,β)=bubuko.com,布布扣1bubuko.com,布布扣0bubuko.com,布布扣xbubuko.com,布布扣α?1bubuko.com,布布扣(1?x)bubuko.com,布布扣β?1bubuko.com,布布扣dxbubuko.com,布布扣

易看出$0$和$1$为奇点,积分在$\alpha>0,\beta>0$时收敛.
a.对称性

B(α,β)=B(β,α)bubuko.com,布布扣

只需作积分变量代换$x=1-t$即可.
B(α,β)bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣=bubuko.com,布布扣=bubuko.com,布布扣=bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣1bubuko.com,布布扣0bubuko.com,布布扣xbubuko.com,布布扣α?1bubuko.com,布布扣(1?x)bubuko.com,布布扣β?1bubuko.com,布布扣dxbubuko.com,布布扣bubuko.com,布布扣1bubuko.com,布布扣0bubuko.com,布布扣(1?t)bubuko.com,布布扣α?1bubuko.com,布布扣tbubuko.com,布布扣β?1bubuko.com,布布扣dtbubuko.com,布布扣B(β,α)bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣

b.递推公式
如果$\alpha>1$,那么成立等式
B(α,β)=α?1bubuko.com,布布扣α+β?1bubuko.com,布布扣bubuko.com,布布扣B(α?1,β)bubuko.com,布布扣

证明:利用分部积分法
B(α,β)bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣=bubuko.com,布布扣=bubuko.com,布布扣=bubuko.com,布布扣=bubuko.com,布布扣bubuko.com,布布扣?1bubuko.com,布布扣βbubuko.com,布布扣bubuko.com,布布扣xbubuko.com,布布扣α?1bubuko.com,布布扣(1?x)bubuko.com,布布扣βbubuko.com,布布扣|bubuko.com,布布扣1bubuko.com,布布扣0bubuko.com,布布扣+α?1bubuko.com,布布扣βbubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣1bubuko.com,布布扣0bubuko.com,布布扣(1?x)bubuko.com,布布扣βbubuko.com,布布扣xbubuko.com,布布扣α?2bubuko.com,布布扣dxbubuko.com,布布扣bubuko.com,布布扣1bubuko.com,布布扣0bubuko.com,布布扣(1?x)bubuko.com,布布扣β?1bubuko.com,布布扣(1?x)xbubuko.com,布布扣α?2bubuko.com,布布扣dxbubuko.com,布布扣α?1bubuko.com,布布扣βbubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣1bubuko.com,布布扣0bubuko.com,布布扣(1?x)bubuko.com,布布扣β?1bubuko.com,布布扣xbubuko.com,布布扣α?2bubuko.com,布布扣dx?α?1bubuko.com,布布扣βbubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣1bubuko.com,布布扣0bubuko.com,布布扣(1?x)bubuko.com,布布扣β?1bubuko.com,布布扣xbubuko.com,布布扣α?1bubuko.com,布布扣dxbubuko.com,布布扣α?1bubuko.com,布布扣βbubuko.com,布布扣bubuko.com,布布扣B(α?1,β)?α?1bubuko.com,布布扣βbubuko.com,布布扣bubuko.com,布布扣B(α,β)bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣

从而有
B(α,β)=α?1bubuko.com,布布扣α+β?1bubuko.com,布布扣bubuko.com,布布扣B(α?1,β)bubuko.com,布布扣

一个特例$m,n\in N_{+}$
B(m,n)=(m?1)!(n?1)!bubuko.com,布布扣(m+n?1)!bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣

c.其他变化形式
令$x=\sin^{2}t$,则有
B(α,β)=bubuko.com,布布扣π/2bubuko.com,布布扣0bubuko.com,布布扣sinbubuko.com,布布扣2α?1bubuko.com,布布扣tcosbubuko.com,布布扣2β?1bubuko.com,布布扣tdtbubuko.com,布布扣

令$x=\frac{y}{1+y}$,则有
B(α,β)=bubuko.com,布布扣bubuko.com,布布扣0bubuko.com,布布扣ybubuko.com,布布扣α?1bubuko.com,布布扣bubuko.com,布布扣(1+y)bubuko.com,布布扣α+βbubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣dybubuko.com,布布扣

特别地,
B(α,1?α)=bubuko.com,布布扣bubuko.com,布布扣0bubuko.com,布布扣ybubuko.com,布布扣α?1bubuko.com,布布扣bubuko.com,布布扣1+ybubuko.com,布布扣bubuko.com,布布扣dybubuko.com,布布扣

1.2 $\Gamma$函数
定义
Γ(s)=bubuko.com,布布扣bubuko.com,布布扣0bubuko.com,布布扣xbubuko.com,布布扣s?1bubuko.com,布布扣ebubuko.com,布布扣?xbubuko.com,布布扣dxbubuko.com,布布扣

a.可微性
$\Gamma$函数无限次可微且
Γbubuko.com,布布扣(n)bubuko.com,布布扣(s)=bubuko.com,布布扣+bubuko.com,布布扣0bubuko.com,布布扣xbubuko.com,布布扣s?1bubuko.com,布布扣lnbubuko.com,布布扣nbubuko.com,布布扣xebubuko.com,布布扣?xbubuko.com,布布扣dxbubuko.com,布布扣

b.递推公式
Γ(s+1)=sΓ(s)bubuko.com,布布扣

证明:利用分部积分法
Γ(s+1)bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣=bubuko.com,布布扣=bubuko.com,布布扣=bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣+bubuko.com,布布扣0bubuko.com,布布扣xbubuko.com,布布扣sbubuko.com,布布扣ebubuko.com,布布扣?xbubuko.com,布布扣dxbubuko.com,布布扣?xbubuko.com,布布扣sbubuko.com,布布扣ebubuko.com,布布扣?xbubuko.com,布布扣|bubuko.com,布布扣+bubuko.com,布布扣0bubuko.com,布布扣+sbubuko.com,布布扣+bubuko.com,布布扣0bubuko.com,布布扣xbubuko.com,布布扣s?1bubuko.com,布布扣ebubuko.com,布布扣?xbubuko.com,布布扣dxbubuko.com,布布扣sΓ(s)bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣

一个特例
Γ(n)=(n?1)!bubuko.com,布布扣

c.极限表达式(欧拉公式)
Γ(s)=limbubuko.com,布布扣nbubuko.com,布布扣nbubuko.com,布布扣sbubuko.com,布布扣(n?1)!bubuko.com,布布扣s(s+1)?(s+n?1)bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣

证明:
Γ(s)bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣=bubuko.com,布布扣=bubuko.com,布布扣=bubuko.com,布布扣=bubuko.com,布布扣=bubuko.com,布布扣=bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣0bubuko.com,布布扣ebubuko.com,布布扣?xbubuko.com,布布扣xbubuko.com,布布扣s?1bubuko.com,布布扣dtbubuko.com,布布扣limbubuko.com,布布扣nbubuko.com,布布扣bubuko.com,布布扣nbubuko.com,布布扣0bubuko.com,布布扣(1?xbubuko.com,布布扣nbubuko.com,布布扣bubuko.com,布布扣)bubuko.com,布布扣nbubuko.com,布布扣xbubuko.com,布布扣s?1bubuko.com,布布扣dtbubuko.com,布布扣limbubuko.com,布布扣nbubuko.com,布布扣nbubuko.com,布布扣sbubuko.com,布布扣bubuko.com,布布扣1bubuko.com,布布扣0bubuko.com,布布扣(1?τ)bubuko.com,布布扣nbubuko.com,布布扣τbubuko.com,布布扣s?1bubuko.com,布布扣dτbubuko.com,布布扣limbubuko.com,布布扣nbubuko.com,布布扣nbubuko.com,布布扣sbubuko.com,布布扣B(n+1,s)bubuko.com,布布扣limbubuko.com,布布扣nbubuko.com,布布扣nbubuko.com,布布扣sbubuko.com,布布扣Γ(n+1)Γ(s)bubuko.com,布布扣Γ(n+s+1)bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣limbubuko.com,布布扣nbubuko.com,布布扣nbubuko.com,布布扣sbubuko.com,布布扣(n?1)!bubuko.com,布布扣s(s+1)?(s+n?1)bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣

d.余元公式
Γ(s)Γ(1?s)=πbubuko.com,布布扣sinπsbubuko.com,布布扣bubuko.com,布布扣(0<s<1)bubuko.com,布布扣

证明:
利用上式所得到的极限表达式,则得
Γ(s)Γ(1?s)bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣=bubuko.com,布布扣=bubuko.com,布布扣=bubuko.com,布布扣bubuko.com,布布扣limbubuko.com,布布扣nbubuko.com,布布扣nbubuko.com,布布扣sbubuko.com,布布扣(n?1)!bubuko.com,布布扣s(s+1)?(s+n?1)bubuko.com,布布扣bubuko.com,布布扣nbubuko.com,布布扣1?sbubuko.com,布布扣(n?1)!bubuko.com,布布扣(1?s)(2?s)?(n?s)bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣1bubuko.com,布布扣sbubuko.com,布布扣bubuko.com,布布扣limbubuko.com,布布扣nbubuko.com,布布扣n1bubuko.com,布布扣(1+s)(1+2bubuko.com,布布扣sbubuko.com,布布扣bubuko.com,布布扣)?(1+sbubuko.com,布布扣n?1bubuko.com,布布扣bubuko.com,布布扣)bubuko.com,布布扣bubuko.com,布布扣1bubuko.com,布布扣(1?s)(1?sbubuko.com,布布扣2bubuko.com,布布扣bubuko.com,布布扣?(1?sbubuko.com,布布扣n?1bubuko.com,布布扣bubuko.com,布布扣)(n?s)bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣1bubuko.com,布布扣sbubuko.com,布布扣bubuko.com,布布扣1bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣n=1bubuko.com,布布扣(1?sbubuko.com,布布扣2bubuko.com,布布扣bubuko.com,布布扣nbubuko.com,布布扣2bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣)bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣

利用由Euler发现的等式
sinπx=πxbubuko.com,布布扣n=1bubuko.com,布布扣bubuko.com,布布扣(1?xbubuko.com,布布扣2bubuko.com,布布扣bubuko.com,布布扣nbubuko.com,布布扣2bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣)bubuko.com,布布扣

于是成立余元公式
Γ(s)Γ(1?s)=πbubuko.com,布布扣sinπsbubuko.com,布布扣bubuko.com,布布扣(0<s<1)bubuko.com,布布扣

特别地,令$s=\frac{1}{2}$
Γ(1bubuko.com,布布扣2bubuko.com,布布扣bubuko.com,布布扣)=πbubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣

e.$\Gamma$函数与$\beta$函数的关系
B(α,β)=Γ(α)Γ(β)bubuko.com,布布扣Γ(α+β)bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣

证明:
作变换$x=u^{2},y=v^{2}$则
Γ(α)Γ(β)bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣=bubuko.com,布布扣=bubuko.com,布布扣=bubuko.com,布布扣=bubuko.com,布布扣=bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣+bubuko.com,布布扣0bubuko.com,布布扣xbubuko.com,布布扣α?1bubuko.com,布布扣ebubuko.com,布布扣?xbubuko.com,布布扣dxbubuko.com,布布扣+bubuko.com,布布扣0bubuko.com,布布扣ybubuko.com,布布扣β?1bubuko.com,布布扣ebubuko.com,布布扣?ybubuko.com,布布扣dybubuko.com,布布扣4bubuko.com,布布扣+bubuko.com,布布扣0bubuko.com,布布扣ububuko.com,布布扣2α?1bubuko.com,布布扣ebubuko.com,布布扣?ububuko.com,布布扣2bubuko.com,布布扣bubuko.com,布布扣dububuko.com,布布扣+bubuko.com,布布扣0bubuko.com,布布扣vbubuko.com,布布扣2β?1bubuko.com,布布扣ebubuko.com,布布扣?vbubuko.com,布布扣2bubuko.com,布布扣bubuko.com,布布扣dvbubuko.com,布布扣4bubuko.com,布布扣+bubuko.com,布布扣0bubuko.com,布布扣bubuko.com,布布扣+bubuko.com,布布扣0bubuko.com,布布扣ububuko.com,布布扣2α?1bubuko.com,布布扣vbubuko.com,布布扣2β?1bubuko.com,布布扣ebubuko.com,布布扣ububuko.com,布布扣2bubuko.com,布布扣+vbubuko.com,布布扣2bubuko.com,布布扣bubuko.com,布布扣dudvbubuko.com,布布扣4bubuko.com,布布扣π/2bubuko.com,布布扣0bubuko.com,布布扣cosbubuko.com,布布扣2α?1bubuko.com,布布扣θsinbubuko.com,布布扣2β?1bubuko.com,布布扣θdθ(Let  u=rcosθ,v=rsinθ)bubuko.com,布布扣B(α,β)Γ(α+β)bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣

f.$\Gamma$函数的推广
Γ(x)=bubuko.com,布布扣n=0bubuko.com,布布扣bubuko.com,布布扣(?1)bubuko.com,布布扣nbubuko.com,布布扣bubuko.com,布布扣n+xbubuko.com,布布扣bubuko.com,布布扣?1bubuko.com,布布扣n!bubuko.com,布布扣bubuko.com,布布扣+bubuko.com,布布扣bubuko.com,布布扣1bubuko.com,布布扣tbubuko.com,布布扣x?1bubuko.com,布布扣ebubuko.com,布布扣?tbubuko.com,布布扣dtbubuko.com,布布扣

这个等式对除去点$0,-1,-2,\cdots$以外的复数$z$定义$\Gamma(z)$.
g.所谓的倍角公式($Legendre$)
Γ(s)Γ(s+1bubuko.com,布布扣2bubuko.com,布布扣bubuko.com,布布扣)=πbubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣2bubuko.com,布布扣1?2sbubuko.com,布布扣Γ(2s)bubuko.com,布布扣

此式可作进一步的推广
 
Γ(s)Γ(s+1bubuko.com,布布扣mbubuko.com,布布扣bubuko.com,布布扣)Γ(s+2bubuko.com,布布扣mbubuko.com,布布扣bubuko.com,布布扣)?Γ(s+m?1bubuko.com,布布扣mbubuko.com,布布扣bubuko.com,布布扣)=(2π)bubuko.com,布布扣(m?1)/2bubuko.com,布布扣mbubuko.com,布布扣1/2?msbubuko.com,布布扣Γ(ms)bubuko.com,布布扣

(一)欧拉积分,布布扣,bubuko.com

(一)欧拉积分

标签:style   class   ext   color   width   int   

原文地址:http://www.cnblogs.com/zhangwenbiao/p/3705462.html

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