實數(shù)的階乘——歐拉積分：Beta、Gamma函數(shù)

2021-12-22 00:13 作者:子瞻Louis 0人讀過 | 我要投稿

歐拉積分是跟一個序列插值問題密切相關(guān)的：即階乘序列

所謂序列插值，就是將通項公式的定義從整數(shù)集延拓到實數(shù)集

1728年，哥德巴赫在考慮序列插值的問題，當(dāng)他開始處理階乘時，被這玩意給難住了，

可以發(fā)現(xiàn)似乎確實存在一條光滑的曲線能將階乘對應(yīng)的點連接起來，但是哥德巴赫無法解決階乘這個問題，于是寫信給了尼古拉一世·伯努利（NikolausI?Bernoulli）和他的弟弟丹尼爾（Daniel?Bernoulli），而當(dāng)時歐拉（Leonhard?Euler）他倆在一塊，因此他也得知了這個問題，最后他在1729年完美地解決了這個問題

這便是今天要說到的歐拉積分了：（勒讓德的提法）

$B(p%2Cq)%3D%5Cint_%7B0%7D%5E1x%5E%7Bp-1%7D(1-x)%5E%7Bq-1%7D%5Cmathrm%20dx$

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

第一個積分也稱為beta函數(shù)，第二個特別特別常用的積分也稱為Gamma函數(shù)

（ps：本文中的beta，Gamma函數(shù)的變量都是實數(shù)，但是需要注意的是它們更廣泛的應(yīng)用都一定會涉及到變量為復(fù)數(shù)的情況）

歐拉第一類積分

這個積分下限是0，因此收斂的充要條件是p＞0，類似地，上限是1收斂的充要條件是q＞0，

所以beta函數(shù)是在p，q都＞0時才有定義，

在beta函數(shù)的積分中作變量代換 $x%3D1-t$ ，可得其對成性：

$%5Cbegin%7Baligned%7DB(p%2Cq)%26%3D%5Cint_%7B0%7D%5E1x%5E%7Bp-1%7D(1-x)%5E%7Bq-1%7D%5Cmathrm%20dx%20%5C%5C%20%26%3D%5Cint_%7B0%7D%5E1(1-t)%5E%7Bp-1%7Dt%5E%7Bq-1%7D%5Cmathrm%20dx%3DB(q%2Cp)%5Cend%7Baligned%7D$
又作代換 $x%3D%5Cfrac%20y%7B1%2By%7D$ ，可得它另一種積分表達(dá)式：

$%5Cbegin%7Baligned%7DB(p%2Cq)%26%3D-%5Cint_%5Cinfty%5E0%20%5Cfrac%7By%5E%7Bp-1%7D%7D%7B(1%2By)%5E%7Bp-1%7D%7D%5Ccdot%5Cfrac%7B1%7D%7B(1%2By)%5E%7Bq-1%7D%7D%5Ccdot%5Cfrac%7B1%7D%7B(1%2By)%5E2%7D%5Cmathrm%20dy%20%5C%5C%20%26%20%3D%5Cint_0%5E%5Cinfty%20%5Cfrac%7By%5E%7Bp-1%7D%7D%7B(1%2By)%5E%7Bp%2Bq%7D%7D%5Cmathrm%20dy%5Cend%7Baligned%7D$
又又作代換 $x%3D%5Csin%5E2%5Calpha$ ，又得到了：

$%5Cbegin%7Baligned%7DB(p%2Cq)%26%3D%5Cint_0%5E%7B%5Cpi%2F2%7D%5Csin%5E%7B2p-2%7D%5Calpha(1-%5Csin%5E2%5Calpha)%5E%7Bq-1%7D%5Ccdot2%5Csin%5Calpha%5Ccos%5Calpha%5Cmathrm%20d%5Calpha%20%5C%5C%26%3D2%5Cint_0%5E%7B%5Cpi%2F2%7D%5Csin%5E%7B2p-1%7D%5Calpha%5Ccos%5E%7B2q-1%7D%5Calpha%5Cmathrm%20d%5Calpha%5Cend%7Baligned%7D$

根據(jù)第二個積分表達(dá)式，利用分部積分法可得一下遞推公式

$%5Cbegin%7Baligned%7DB(p%2B1%2Cq)%26%3D%5Cint_0%5E%5Cinfty%20%5Cfrac%7By%5E%7Bp%7D%7D%7B(1-y)%5E%7Bp%2Bq%2B1%7D%7D%5Cmathrm%20dy%20%5C%5C%26%3D-%5Cfrac1%7Bp%2Bq%7D%5Cint_0%5E%5Cinfty%20y%5E%7Bp%7D%5Cmathrm%20d%5Cfrac%7B1%7D%7B(1-y)%5E%7Bp%2Bq%7D%7D%20%5C%5C%26%3D%5Cfrac1%7Bp%2Bq%7D%5Cint_0%5E%5Cinfty%20%5Cfrac%7B1%7D%7B(1-y)%5E%7Bp%2Bq%7D%7D%5Cmathrm%20dy%5Ep%20%5C%5C%26%3D%5Cfrac%20p%7Bp%2Bq%7D%5Cint_0%5E%5Cinfty%20%5Cfrac%7By%5E%7Bp-1%7D%7D%7B(1-y)%5E%7Bp%2Bq%7D%7D%5Cmathrm%20dy%3D%5Cfrac%20p%7Bp%2Bq%7DB(p%2Cq)%5Cend%7Baligned%7D$

$B(p%2B1%2Cq)%3D%5Cfrac%20p%7Bp%2Bq%7DB(p%2Cq)$

當(dāng)n為整數(shù)時，由遞推公式，有

$%5Cbegin%7Baligned%7Dn%5E%5Calpha%20B(n%2C%5Calpha)%26%3Dn%5E%5Calpha%5Cfrac%7Bn-1%7D%7B%5Calpha%2Bn-1%7DB(n-1%2C%5Calpha)%5C%5C%26%3Dn%5E%5Calpha%5Cfrac%7B(n-2)(n-1)%7D%7B(%5Calpha%2Bn-2)(%5Calpha%2Bn-1)%7DB(n-2%2C%5Calpha)%20%5C%5C%26%3D%E2%80%A6%3Dn%5E%5Calpha%5Cfrac%7B1%5Ccdot2%E2%80%A6(n-1)%7D%7B(%5Calpha%2B1)%E2%80%A6(%5Calpha%2Bn-1)%7DB(1%2C%5Calpha)%5C%5C%26%3D%5Cfrac%7Bn%5E%5Calpha%5Ccdot1%5Ccdot2%E2%80%A6(n-1)%7D%7B%5Calpha(%5Calpha%2B1)%E2%80%A6(%5Calpha%2Bn-1)%7D%5Cend%7Baligned%7D$

令 $n%5Crightarrow%20%5Cinfty$ ，可得高斯公式（Gauss Formula）：

$%5CPi(%5Calpha)%3D%5Clim_%7Bn%5Cto%5Cinfty%7D%5Cfrac%7Bn%5E%5Calpha%5Ccdot1%5Ccdot2%E2%80%A6(n-1)%7D%7B%5Calpha(%5Calpha%2B1)%E2%80%A6(%5Calpha%2Bn-1)%7D$

我們來驗證當(dāng) $%5Calpha$ 是非負(fù)整數(shù)時它其實就是階乘：

首先有 $%5CPi(1)%3D0!%3D1$

又有（α＞0）：

$%5Cbegin%7Baligned%7D%5CPi(%5Calpha%2B1)%26%3D%5Clim_%7Bn%5Cto%5Cinfty%7D%5Cfrac%7Bn%5E%7B%5Calpha%2B1%7D%5Ccdot1%5Ccdot2%E2%80%A6(n-1)%7D%7B(%5Calpha%2B1)(%5Calpha%2B2)%E2%80%A6(%5Calpha%2Bn)%7D%5C%5C%20%26%3D%5Calpha%5Clim_%7Bn%5Cto%5Cinfty%7D%5Cfrac%7Bn%5E%7B%5Calpha%2B1%7D%5Ccdot1%5Ccdot2%E2%80%A6(n-1)%7D%7B%5Calpha(%5Calpha%2B1)%E2%80%A6(%5Calpha%2Bn)%7D%5C%5C%26%3D%5Calpha%5CPi(%5Calpha)%5Clim_%7Bn%5Cto%5Cinfty%7D%5Cfrac%7Bn%7D%7B%5Calpha%2Bn%7D%3D%5Calpha%5CPi(%5Calpha)%5Cend%7Baligned%7D$

$%5CPi(%5Calpha%2B1)%3D%5Calpha%5CPi(%5Calpha)$

因此 $%5CPi(n%2B1)%3Dn!%2Cn%5Cin%5Cmathbb%20N$

而當(dāng)把變量換成正實數(shù)時，仍然滿足遞推性質(zhì)，所以它作為階乘的延拓是良好定義的

第二類歐拉積分

在下面的公式中作代換 $x%3Dnu$ ，根據(jù)beta函數(shù)的遞推公式，有

$%5Cbegin%7Baligned%7D%5CPi(n%2Cs)%26%3D%5Cint_0%5Enx%5E%7Bs-1%7D%5Ccolor%7Bblue%7D%7B%5Cleft(1-%5Cfrac%20xn%5Cright)%5E%7Bn%7D%7D%5Cmathrm%20dx%5C%5C%20%26%3Dn%5Es%5Cint_0%5E1u%5E%7Bs-1%7D%5Cleft(1-u%5Cright)%5E%7Bn%7D%5Cmathrm%20du%20%5C%5C%26%3Dn%5EsB(n%2B1%2Cs)%5C%5C%26%3D%5Cfrac%7Bn%5Es%5Ccdot1%5Ccdot2%E2%80%A6n%7D%7Bs(s%2B1)%E2%80%A6(s%2Bn)%7D%5Cend%7Baligned%7D$

以我們熟知的藍(lán)色部分當(dāng) $n%5Crightarrow%20%5Cinfty$ 時一致收斂到 $e%5E%7B-u%7D$ ，而最下面的乘積則收斂到 $%5CPi(s)$ ，又不難驗證該函數(shù)到Gamma函數(shù)，于是

$%5CGamma(s)%3D%5CPi(s)$

因此，有以下性質(zhì)

$%5CGamma(1)%3D0!%3D1$
$%5CGamma(s%2B1)%3Ds%5CGamma(s)$
$%5CGamma(n%2B1)%3Dn!%2Cn%5Cin%5Cmathbb%20N$
$%5CGamma(s)%3D%5Clim_%7Bn%5Cto%5Cinfty%7D%5Cfrac%7Bn%5Es%5Ccdot1%5Ccdot2%E2%80%A6(n-1)%7D%7Bs(s%2B1)%E2%80%A6(s%2Bn-1)%7D$

在積分中作代換 $x%3Du%5E2$ ，有Gamma函數(shù)的另一積分表達(dá)式為：

$%5Cbegin%7Baligned%7D%5CGamma(s)%3D2%5Cint_0%5E%5Cinfty%20x%5E%7B2s-1%7De%5E%7B-x%5E2%7D%5Cmathrm%20dx%5Cend%7Baligned%7D$

又作代換 $x%3Du%5En$ ，又有積分表式為：

$%5Cbegin%7Baligned%7D%5CGamma(s)%3Dn%5Cint_0%5E%5Cinfty%20x%5E%7Bns-1%7De%5E%7B-x%5En%7D%5Cmathrm%20dx%5Cend%7Baligned%7D$

也許有的人看到積分中減了個1會有些疑惑，為什么不直接定義

$%5CGamma(s)%3D%5Cint_0%5E%5Cinfty%20x%5E%7Bs%7De%5E%7B-x%7D%5Cmathrm%20dx$
呢？這樣既好看了些，還能在s是整數(shù)時直接有 $%5CGamma(s)%3Ds!$ 我說如果這樣想的話那你就格局小了（被打），其實不難發(fā)現(xiàn)上面這個積分右邊僅僅在s＞-1時才收斂，數(shù)學(xué)家們不太喜歡這樣的收斂域，而正好-1這種用法十九世紀(jì)末期在法國十分流行，于是當(dāng)時由勒讓德（Legendre）介紹了Gamma（s）=（s-1）！，這樣收斂范圍變成了s＞0，這樣的收斂域既省事又挺美觀

beta函數(shù)與Gamma之間函數(shù)的聯(lián)系

設(shè) $y%EF%BC%9E0$ ，通過變量代換 $u%3Dxy$ ，有以下等式：

$%5Cbegin%7Baligned%7D%5Cint_0%5E%5Cinfty%20x%5E%7B%5Calpha-1%7De%5E%7B-xt%7D%5Cmathrm%20dx%26%3D%5Cfrac1%20%7Bt%5E%7B%5Calpha%7D%7D%5Cint_0%5E%5Cinfty%20u%5E%7B%5Calpha-1%7De%5E%7B-u%7D%5Cmathrm%20du%3D%5Cfrac1%7Bt%5E%5Calpha%7D%5CGamma(%5Calpha)%5Cend%7Baligned%7D$

令 $t%3D1%2By%2C%5Calpha%3Dp%2Bq%2C(p%2Cq%EF%BC%9E0)$ ，則有

$%5Cfrac%7B%5CGamma(p%2Bq)%7D%7B(1%2By)%5E%7Bp%2Bq%7D%7D%3D%5Cint_0%5E%5Cinfty%20x%5E%7Bp%2Bq-1%7De%5E%7B-x(1%2By)%7D%5Cmathrm%20dx$

用上式和beta函數(shù)的第二積分表式，有

$%5Cbegin%7Baligned%7D%5CGamma(p%2Bq)B(p%2Cq)%26%3D%5Cint_0%5E%5Cinfty%20y%5E%7Bp-1%7D%5Ccolor%7Bgreen%7D%7B%5Cfrac%7B%5CGamma(p%2Bq)%7D%7B(1%2By)%5E%7Bp%2Bq%7D%7D%7D%5Cmathrm%20dy%20%5C%5C%26%3D%5Cint_0%5E%5Cinfty%20y%5E%7Bp-1%7D%5Ccolor%7Bgreen%7D%7B%5Cint_0%5E%5Cinfty%20x%5E%7Bp%2Bq-1%7De%5E%7B-x(1%2By)%7D%5Cmathrm%20dx%7D%5Cmathrm%20dy%5C%5C%26%3D%5Cint_0%5E%5Cinfty(xy)%5E%7Bp-1%7De%5E%7B-xy%7D%5Cmathrm%20d(xy)%5Cint_0%5E%5Cinfty%20x%5E%7Bq-1%7De%5E%7B-x%7D%5Cmathrm%20dx%5C%5C%26%3D%5CGamma(p)%5CGamma(q)%5Cend%7Baligned%7D$