阶乘的进步一延拓——Gamma函数几个公式

2021-12-25 13:17 作者:子瞻Louis 0人读过 | 我要投稿

上一期：实数的阶乘——欧拉积分中介绍了欧拉两类积分，

本期主要讨论欧拉第二类积分，即Gamma函数

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

欧拉-马歇罗尼常数

这个十分重要的常数是在调和级数中产生的，我们都知道

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

将积分区间切开，写成以下形式：

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

因为 $%5Cfrac1x$ 在正半轴上单调递减，根据微积分中值定理

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

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

令 $m%5Crightarrow%20%5Cinfty$ ，得到

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

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

即它俩极限的差收敛到一个0到1之间的常数，此外还可以由此轻松得到调和级数成对数状发散，而这个常数就是欧拉-马歇罗尼常数（Euler—Mascheroni constant），有时也叫欧拉常数，记为

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

由于它是在极限中产生的常数，所以它的超越性是尚不好确定的，并且事实上这个常数的无理性的证明都是十分棘手的，但我们可以通过一下方法计算它的近似值，令

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

则它的极限就是Euler常数

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

由此可以计算该常数约为

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

复平面上的解析延拓

我们先前规定了在Gamma函数的积分中s为一大于零的实数，因为这样右边的积分才收敛

如果我们把s换成一个复数，那么该积分在 $%5CRe(s)%EF%BC%9E0$ 的平面上是收敛的，但是这并不能让我们满足，不妨试着将Gamma函数延拓到整个复平面，

对此，我们希望它在复平面上满足以下条件：

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

对于第二个递推条件，取s=0，得到

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

发现它在s=0处出现了极点，再根据该递推公式，可知 $0%2C-1%2C-2%2C%E2%80%A6$ 都是它的极点，这可不是好事，但是又注意到满足上述两个条件的 $%5CGamma(s)$ 在复平面上是没有零点的，那我们就可以取它的倒数了，这样一来 $%5Cfrac1%7B%5CGamma(s)%7D$ 就是复平面上的全纯函数了，并且 $0%2C-1%2C-2%2C%E2%80%A6$ 是它的零点，而这个序列是趋向无穷的，又有

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

刚好这些条件满足Weierstrass分解定理，因此它可以展开为以下乘积：

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

对它取对数，得到

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

又根据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函数的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函数在复平面上的解析延拓

对上面的公式作变换

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

取倒数就又可以得到Gauss公式了，并且它还了告诉我们该乘积公式在复平面上除了 $0%2C-1%2C-2%2C%E2%80%A6$ 外都收敛到解析函数

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

根据上面第二个公式，有

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

观察发现蓝色部分就是我们前几期推导过的 $%5Cfrac%7B%5Csin%20%5Cpi%20s%7D%7B%5Cpi%20s%7D$ 的无穷乘积展开，因此

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

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

最后根据递推公式，可得

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

这就是Gamma函数的余元公式了，代入 $s%3D%5Cfrac12$ ，可得

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

又根据它的积分表式，可得Euler-Poisson积分：

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

倍元公式

（温馨提示：下面的推理可能会"吵"到您的眼睛）

根据Gauss公式，有

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

拎出红色部分

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

再把绿色部分拎出来，

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

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

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

一个积分

最后本期专栏就以一个积分来收尾吧

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

其中的根号看起来十分不友好，那我们不妨设

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

看到这个东西，就不难联想到beta函数了，作代换 $u%3Dx%5E%5Cbeta$ ，

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

再根据余元公式，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积分了：

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

此外还可以得到

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

根据它们就来可以整一些奇怪的积分了，比如

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

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