階乘的進(jìn)步一延拓——Gamma函數(shù)幾個公式

2021-12-25 13:17 作者:子瞻Louis 0人讀過 | 我要投稿

上一期：實數(shù)的階乘——歐拉積分中介紹了歐拉兩類積分，

本期主要討論歐拉第二類積分，即Gamma函數(shù)

$%5CGamma(s)%3D%5Cint_0%5E%5Cinfty%20x%5E%7Bs-1%7De%5E%7B-x%7D%5Cmathrm%20dx$

歐拉-馬歇羅尼常數(shù)

這個十分重要的常數(shù)是在調(diào)和級數(shù)中產(chǎn)生的，我們都知道

$%5Cint_1%5Em%5Cfrac1x%5Cmathrm%20dx%3D%5Cln%20m$

將積分區(qū)間切開，寫成以下形式：

$%5Csum_%7Bn%3D1%7D%5E%7Bm-1%7D%5Cint_%7Bn%7D%5E%7Bn%2B1%7D%5Cfrac1x%5Cmathrm%20dx%3D%5Cln%20m$

因為 $%5Cfrac1x$ 在正半軸上單調(diào)遞減，根據(jù)微積分中值定理

$%5Cfrac1%7Bn%2B1%7D%EF%BC%9C%5Cint_%7Bn%7D%5E%7Bn%2B1%7D%5Cfrac1x%5Cmathrm%20dx%EF%BC%9C%5Cfrac1n$

$%5Csum_%7Bn%3D2%7D%5Em%5Cfrac1%7Bn%7D%EF%BC%9C%5Csum_%7Bn%3D1%7D%5E%7Bm-1%7D%5Cint_%7Bn%7D%5E%7Bn%2B1%7D%5Cfrac1x%5Cmathrm%20dx%3D%5Cln%20m%EF%BC%9C%5Csum_%7Bn%3D1%7D%5E%7Bm-1%7D%5Cfrac1%7Bn%7D$

令 $m%5Crightarrow%20%5Cinfty$ ，得到

$%5Clim_%7Bm%5Cto%5Cinfty%7D%5Ccolor%7Bblue%7D%7B%5Csum_%7Bn%3D1%7D%5Em%5Cfrac1%7Bn%7D%7D-1%EF%BC%9C%5Csum_%7Bn%3D1%7D%5E%7Bm-1%7D%5Cint_%7Bn%7D%5E%7Bn%2B1%7D%5Cfrac1x%5Cmathrm%20dx%3D%5Ccolor%7Bblue%7D%7B%5Cln%20m%7D%EF%BC%9C%5Ccolor%7Bblue%7D%7B%5Csum_%7Bn%3D1%7D%5E%7Bm%7D%5Cfrac1%7Bn%7D%7D-%5Cfrac1m$

$%5CRightarrow%20%5Clim_%7Bm%5Cto%5Cinfty%7D%5Cfrac1m%EF%BC%9C%5Ccolor%7Bpurple%7D%7B%5Csum_%7Bn%3D1%7D%5Em%5Cfrac1n-%5Cln%20m%7D%EF%BC%9C1$

即它倆極限的差收斂到一個0到1之間的常數(shù)，此外還可以由此輕松得到調(diào)和級數(shù)成對數(shù)狀發(fā)散，而這個常數(shù)就是歐拉-馬歇羅尼常數(shù)（Euler—Mascheroni constant），有時也叫歐拉常數(shù)，記為

$%5Cgamma%3D%5Clim_%7BN%5Cto%5Cinfty%7D%5Csum_%7Bn%3D1%7D%5EN%5Cfrac1n-%5Cln%20N$

由于它是在極限中產(chǎn)生的常數(shù)，所以它的超越性是尚不好確定的，并且事實上這個常數(shù)的無理性的證明都是十分棘手的，但我們可以通過一下方法計算它的近似值，令

$%5Cgamma_n%3D1%2B%5Cfrac12%2B%E2%80%A6%2B%5Cfrac1n-%5Cln%20n$

則它的極限就是Euler常數(shù)

$%5Clim_%7Bn%5Cto%5Cinfty%7D%5Cgamma_n%3D%5Cgamma$

由此可以計算該常數(shù)約為

$%CE%B3%E2%89%880.577215%E2%80%A6$

復(fù)平面上的解析延拓

我們先前規(guī)定了在Gamma函數(shù)的積分中s為一大于零的實數(shù)，因為這樣右邊的積分才收斂

如果我們把s換成一個復(fù)數(shù)，那么該積分在 $%5CRe(s)%EF%BC%9E0$ 的平面上是收斂的，但是這并不能讓我們滿足，不妨試著將Gamma函數(shù)延拓到整個復(fù)平面，

對此，我們希望它在復(fù)平面上滿足以下條件：

$%5CGamma(1)$
$%5CGamma(s%2B1)%3Ds%5CGamma(s)$

對于第二個遞推條件，取s=0，得到

$%5CGamma(1)%3D0%5Ccdot%5CGamma(0)$

發(fā)現(xiàn)它在s=0處出現(xiàn)了極點，再根據(jù)該遞推公式，可知 $0%2C-1%2C-2%2C%E2%80%A6$ 都是它的極點，這可不是好事，但是又注意到滿足上述兩個條件的 $%5CGamma(s)$ 在復(fù)平面上是沒有零點的，那我們就可以取它的倒數(shù)了，這樣一來 $%5Cfrac1%7B%5CGamma(s)%7D$ 就是復(fù)平面上的全純函數(shù)了，并且 $0%2C-1%2C-2%2C%E2%80%A6$ 是它的零點，而這個序列是趨向無窮的，又有

$%5Cvert%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac1%7B-n%7D%5Cvert%3D%5Cinfty%2C%E8%80%8C%5Cvert%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac1%7B(-n)%5E2%7D%5Cvert%EF%BC%9C%5Cinfty$

剛好這些條件滿足Weierstrass分解定理，因此它可以展開為以下乘積：

$%5Cfrac1%7B%5CGamma(s)%7D%3De%5E%7BH(s)%7Ds%5Cprod_%7Bn%3D1%7D%5E%5Cinfty%5Cleft(1%2B%5Cfrac%20sn%5Cright)e%5E%7B-s%2Fn%7D$

對它取對數(shù)，得到

$%5Cbegin%7Baligned%7D%5Cln%5CGamma(s)%26%3D%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac%20sn-%7BH(s)%7D-%5Cln%20s-%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cln%5Cleft(%5Cfrac%20%7Bs%2Bn%7Dn%5Cright)%5C%5C%26%3D%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac%20sn-%7BH(s)%7D-%5Cln%20s-%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cln%5Cleft(s%2Bn%5Cright)%2B%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cln%20n%5Cend%7Baligned%7D$

又根據(jù)Gauss公式，可得：

$%5CGamma(s)%3D%5Clim_%7BN%5Cto%5Cinfty%7D%5Cfrac%7BN%5Es%5Ccdot%20N!%7D%7Bs(s%2B1)%E2%80%A6(s%2BN)%7D$

$%5CRightarrow%20%5Cln%5CGamma(s)%3D%5Clim_%7BN%5Cto%5Cinfty%7Ds%5Cln%20N%2B%5Csum_%7Bn%3D1%7D%5E%7BN%7D%5Cln%20n-%5Csum_%7Bn%3D0%7D%5E%7BN%7D%5Cln(s%2Bn)$

代入到上式中，可得

$%5Clim_%7BN%5Cto%5Cinfty%7Ds%5Cln%20N%2B%5Csum_%7Bn%3D1%7D%5E%7BN%7D%5Cln%20n-%5Csum_%7Bn%3D0%7D%5E%7BN%7D%5Cln(s%2Bn)%3D%5Csum_%7Bn%3D1%7D%5E%7BN%7D%5Cfrac%20sn-%7BH(s)%7D-%5Cln%20s-%5Csum_%7Bn%3D1%7D%5E%7BN%7D%5Cln%5Cleft(s%2Bn%5Cright)%2B%5Csum_%7Bn%3D1%7D%5E%7BN%7D%5Cln%20n$

$%5CRightarrow%20H(s)%3D%5Clim_%7BN%5Cto%5Cinfty%7D%5Csum_%7Bn%3D1%7D%5EN%5Cfrac%20sn-s%5Cln%20N%3D%5Cgamma%20s$

于是就得到了Gamma函數(shù)的Weierstrass公式：

$%5Cfrac1%7B%5CGamma(s)%7D%3De%5E%7B%5Cgamma%20s%7Ds%5Cprod_%7Bn%3D1%7D%5E%5Cinfty%5Cleft(1%2B%5Cfrac%20sn%5Cright)e%5E%7B-s%2Fn%7D$

至此就完成了Gamma函數(shù)在復(fù)平面上的解析延拓

對上面的公式作變換

$%5Cbegin%7Baligned%7D%5Cfrac1%7B%5CGamma(s)%7D%26%3D%5Clim_%7BN%5Cto%5Cinfty%7De%5E%7B%5Csum_%7Bn%3D1%7D%5EN%5Cfrac%20sn-s%5Cln%20N%7Ds%5Cprod_%7Bn%3D1%7D%5EN%5Cleft(1%2B%5Cfrac%20sn%5Cright)e%5E%7B-s%2Fn%7D%5C%5C%26%3D%5Clim_%7BN%5Cto%5Cinfty%7D%5Cfrac%20s%7BN%5Es%7D%5Cprod_%7Bn%3D1%7D%5EN%5Cleft(1%2B%5Cfrac%20sn%5Cright)e%5E%7B-s%2Fn%2Bs%2Fn%7D%5C%5C%26%3D%5Clim_%7BN%5Cto%5Cinfty%7D%5Cfrac%20s%7BN%5Es%7D%5Cprod_%7Bn%3D1%7D%5EN%5Cleft(%5Cfrac%20%7Bs%2Bn%7Dn%5Cright)%5Cend%7Baligned%7D$

取倒數(shù)就又可以得到Gauss公式了，并且它還了告訴我們該乘積公式在復(fù)平面上除了 $0%2C-1%2C-2%2C%E2%80%A6$ 外都收斂到解析函數(shù)

又有：

$%5Cprod_%7Bn%3D1%7D%5E%7BN-1%7D%5Cleft(1%2B%5Cfrac1n%5Cright)%3D%5Cprod_%7Bn%3D1%7D%5E%7BN-1%7D%5Cleft(%5Cfrac%7Bn%2B1%7Dn%5Cright)%3D%5Cfrac21%5Ccdot%5Cfrac32%5Ccdot%5Cfrac43%E2%80%A6%5Cfrac%7BN-1%7D%7BN-2%7D%5Ccdot%5Cfrac%20N%7Bn-1%7D%3DN$

$%5CRightarrow%5Cfrac1%7BN%5Es%7D%3D%5Cprod_%7Bn%3D1%7D%5E%7BN-1%7D%5Cleft(1%2B%5Cfrac1n%5Cright)%5E%7B-s%7D%3D%5Cprod_%7Bn%3D1%7D%5E%7BN%7D%5Cleft(1%2B%5Cfrac1n%5Cright)%5E%7B-s%7D%5Ccdot%5Ccolor%7Bgray%7D%7B%5Cfrac1%7B%5Cleft(1%2B%5Cfrac1N%5Cright)%5Es%7D%7D$

由此便可得

$%5Cfrac1%7B%5CGamma(s)%7D%3Ds%5Cprod_%7Bn%3D1%7D%5E%5Cinfty%5Cleft(1%2B%5Cfrac%20%7Bs%7Dn%5Cright)%5Cleft(1%2B%5Cfrac1n%5Cright)%5E%7B-s%7D$

余元和倍元公式

余元公式

根據(jù)上面第二個公式，有

$%5Cbegin%7Baligned%7D%5Cfrac1%7B%5CGamma(s)%5CGamma(-s)%7D%26%3D-s%5Ccdot%20s%5Cprod_%7Bn%3D1%7D%5E%5Cinfty%5Cleft(1%2B%5Cfrac%20%7Bs%7Dn%5Cright)%5Cleft(1-%5Cfrac%20%7Bs%7Dn%5Cright)%5Cleft(1%2B%5Cfrac1n%5Cright)%5E%7B-s%2Bs%7D%5C%5C%26%3D-s%5Ccdot%20s%5Ccolor%7Bblue%7D%7B%5Cprod_%7Bn%3D1%7D%5E%5Cinfty%5Cleft(1-%5Cfrac%7Bs%5E2%7D%7Bn%5E2%7D%5Cright)%7D%5Cend%7Baligned%7D$ ????

觀察發(fā)現(xiàn)藍(lán)色部分就是我們前幾期推導(dǎo)過的 $%5Cfrac%7B%5Csin%20%5Cpi%20s%7D%7B%5Cpi%20s%7D$ 的無窮乘積展開，因此

$%5Cfrac1%7B%5CGamma(s)%5CGamma(-s)%7D%3D-s%5Ccdot%5Cfrac%7B%5Csin%20%5Cpi%20s%7D%7B%5Cpi%20%7D$

$%5CRightarrow%5Cfrac1%7B%5CGamma(s)%5Ccdot%5Ccolor%7Bpurple%7D%7B-s%5CGamma(-s)%7D%7D%3D%5Cfrac%7B%5Csin%20%5Cpi%20s%7D%5Cpi$

最后根據(jù)遞推公式，可得

$%5CGamma(s)%5CGamma(1-s)%3D%5Cfrac%5Cpi%7B%5Csin%20%5Cpi%20s%7D$

這就是Gamma函數(shù)的余元公式了，代入 $s%3D%5Cfrac12$ ，可得

$%5CGamma%5Cleft(%5Cfrac12%5Cright)%3D%5Csqrt%5Cpi$

又根據(jù)它的積分表式，可得Euler-Poisson積分：

$%5Cint_%5Cmathbb%20Re%5E%7B-x%5E2%7D%5Cmathrm%20%20dx%3D%5Csqrt%5Cpi$

倍元公式

（溫馨提示：下面的推理可能會"吵"到您的眼睛）

根據(jù)Gauss公式，有

$%5Cbegin%7Baligned%7D%5CGamma(s)%5CGamma%5Cleft(s%2B%5Cfrac12%5Cright)%26%3D%5Clim_%7BN%5Cto%5Cinfty%7D%5Cfrac%7BN%5E%7B2s%2B%5Cfrac12%7D%5Ccdot%20N!%5E2%7D%7Bs(s%2B%5Cfrac12)(s%2B1)%E2%80%A6(s%2BN)(s%2BN%2B%5Cfrac12)%7D%5C%5C%26%3D%5Clim_%7BN%5Cto%5Cinfty%7D%5Cfrac%7BN%5E%7B2s%2B%5Cfrac12%7D%5Ccdot2%5E%7B2N%2B2%7D%5Ccdot%20N!%5E2%7D%7B2s(2s%2B1)%E2%80%A6(2s%2B2N)(2s%2B2N%2B1)%7D%5C%5C%26%3D%5Clim_%7BN%5Cto%5Cinfty%7D%5Ccolor%7Bred%7D%7B%5Cfrac%7BN%5E%5Cfrac12%5Ccdot%202%5E%7B2N%2B1%7D%5Ccdot%20N!%5E2%7D%7B(2N%2B1)!%7D%7D%5Ccolor%7Bgreen%7D%7B%5Cfrac%7B2%5E%7B-2s%2B1%7D(2N)%5E%7B2s%7D%5Ccdot%20(2N)!(2N%2B1)%7D%7B2s(2s%2B1)%E2%80%A6(2s%2B2N)(2s%2B2N%2B1)%7D%7D%5Cend%7Baligned%7D$

拎出紅色部分

$%5Cbegin%7Baligned%7D%5Clim_%7BN%5Cto%5Cinfty%7D%5Cfrac%7BN%5E%5Cfrac12%5Ccdot2%5E%7B2N%2B1%7D%5Ccdot%20N!%5E2%7D%7B(2N%2B1)!%7D%26%3D%5Clim_%7BN%5Cto%5Cinfty%7D%5Cfrac%7BN%5E%5Cfrac12%5Ccdot2%5E%7B2N%2B1%7D%5Ccdot%20N!%5Ccdot1%5Ccdot2%E2%80%A6N%7D%7B1%5Ccdot3%E2%80%A6(2N%2B1)%5Ccdot2%5Ccdot4%E2%80%A6(2N)%7D%5C%5C%26%3D%5Clim_%7BN%5Cto%5Cinfty%7D%5Cfrac%7BN%5E%5Cfrac12%5Ccdot%20N!%5Ccdot1%5Ccdot2%E2%80%A6N%7D%7B%5Cfrac12%5Ccdot(%5Cfrac12%2B1)%E2%80%A6(%5Cfrac12%2BN)%5Ccdot1%5Ccdot2%E2%80%A6N%7D%5C%5C%26%3D%5CGamma%5Cleft(%5Cfrac12%5Cright)%3D%5Csqrt%5Cpi%5Cend%7Baligned%7D$

再把綠色部分拎出來，

$%5Cbegin%7Baligned%7D%5Clim_%7BN%5Cto%5Cinfty%7D%5Cfrac%7B2%5E%7B-2s%2B1%7D(2N)%5E%7B2s%7D%5Ccdot%20(2N)!(2N%2B1)%7D%7B2s(2s%2B1)%E2%80%A6(2s%2B2N)(2s%2B2N%2B1)%7D%26%3D%5Cfrac1%7B2%5E%7B2s-1%7D%7D%5CGamma(2s)%5Clim_%7BN%5Cto%5Cinfty%7D%5Cfrac%7B(2N%2B1)%7D%7B(2s%2B2N%2B1)%7D%5C%5C%26%3D%5Cfrac1%7B2%5E%7B2s-1%7D%7D%5CGamma(2s)%5Cend%7Baligned%7D$

代回上式中，即可得

$%5CGamma(s)%5CGamma%5Cleft(s%2B%5Cfrac12%5Cright)%3D%5Cfrac1%7B2%5E%7B2s-1%7D%7D%5CGamma%5Cleft(%5Cfrac12%5Cright)%5CGamma(2s)$

或者

$B%5Cleft(s%2Cs%2B%5Cfrac12%5Cright)%3D%5Cfrac1%7B2%5E%7B2s-1%7D%7DB%5Cleft(%5Cfrac12%2C2s%5Cright)$

一個積分

最后本期專欄就以一個積分來收尾吧

$I%3D%5Cint_0%5E%5Cinfty%5Cfrac%7B%5Csqrt%20x%7D%7B1%2Bx%5E2%7D%5Cmathrm%20dx$

其中的根號看起來十分不友好，那我們不妨設(shè)

$%5CPhi%20(%5Calpha%2C%5Cbeta)%3D%5Cint_0%5E%5Cinfty%5Cfrac%7Bx%5E%7B%5Calpha-1%7D%7D%7B1%2Bx%5E%5Cbeta%7D%5Cmathrm%20dx$

看到這個東西，就不難聯(lián)想到beta函數(shù)了，作代換 $u%3Dx%5E%5Cbeta$ ，

$%5Cbegin%7Baligned%7D%5CPhi(%5Calpha%2C%5Cbeta)%26%3D%5Cint_0%5E%5Cinfty%5Cfrac%7Bu%5E%5Cfrac%7B%5Calpha-1%7D%7B%5Cbeta%7D%7D%7B1%2Bu%7D%5Cmathrm%20du%5E%7B1%2F%5Cbeta%7D%5C%5C%26%3D%5Cfrac1%5Cbeta%5Cint_0%5E%5Cinfty%5Cfrac%7Bu%5E%7B%5Calpha%2F%5Cbeta-1%7D%7D%7B1%2Bu%7D%5Cmathrm%20du%5C%5C%26%3D%5Cfrac1%5Cbeta%20B%5Cleft(%5Cfrac%7B%5Calpha%7D%7B%5Cbeta%7D%2C1-%5Cfrac%7B%5Calpha%7D%7B%5Cbeta%7D%5Cright)%5C%5C%26%3D%5Cfrac1%5Cbeta%5CGamma%5Cleft(%5Cfrac%7B%5Calpha%7D%7B%5Cbeta%7D%5Cright)%5CGamma%5Cleft(1-%5Cfrac%7B%5Calpha%7D%5Cbeta%5Cright)%5Cend%7Baligned%7D$

再根據(jù)余元公式，可得

$%5CPhi(%5Calpha%2C%5Cbeta)%3D%5Cfrac%7B%5Cpi%7D%7B%5Cbeta%5Csin%5Cfrac%7B%5Calpha%7D%5Cbeta%5Cpi%7D$

所以就能得到上面的積分了：

$%5Cbegin%7Baligned%7DI%3D%5CPhi%5Cleft(%5Cfrac32%2C2%5Cright)%26%3D%5Cfrac%7B%5Cpi%7D%7B2%5Csin%5Cfrac34%5Cpi%7D%5C%5C%26%3D%5Cfrac%7B%5Cpi%7D%7B%5Csqrt2%7D%3D%5Cfrac%7B%5Csqrt2%7D2%5Cpi%5Cend%7Baligned%7D$ ?

此外還可以得到

$%5Cint_0%5E%5Cinfty%5Cfrac%7B%5Cmathrm%20dx%7D%7B1%2Bx%5Ea%7D%3D%5CPhi(1%2Ca)%3D%5Cfrac%7B%5Cpi%7D%7Ba%5Csin%5Cfrac1a%5Cpi%7D$

根據(jù)它們就來可以整一些奇怪的積分了，比如

$%5Cint_0%5E%5Cinfty%5Cfrac%7B%5Cmathrm%20dx%7D%7B1%2Bx%5E%5Cpi%7D%3D%5Cfrac1%7B%5Csin1%7D$

$%5Cint_0%5E%5Cinfty%5Cfrac%7Bx%5E%7Be-1%7D%7D%7B1%2Bx%5E%5Cgamma%7D%5Cmathrm%20dx%3D%5Cfrac%7B%5Cpi%7D%7B%5Cgamma%5Csin%5Cfrac%7Be%5Cpi%7D%7B%5Cgamma%7D%7D$

標(biāo)簽：數(shù)學(xué)階乘歐拉常數(shù)解析延拓歐拉積分 gamma函數(shù)余元公式倍元公式

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階乘的進(jìn)步一延拓——Gamma函數(shù)幾個公式

歐拉-馬歇羅尼常數(shù)

復(fù)平面上的解析延拓

余元和倍元公式

一個積分