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Euler's Limit Formula Derivation for the Gamma Function 📂Functions

Euler's Limit Formula Derivation for the Gamma Function

Formula 1

The following holds for the Gamma function Γ:(0,∞)→R\Gamma : (0, \infty) \to \mathbb{R}: Γ(x)=lim⁡n→∞nxn!x(x+1)(x+2)⋯(x+n) \Gamma (x) = \lim_{n \to \infty} {{n^x n!} \over {x(x+1)(x+2) \cdots (x+n) }}

Explanation

The Gamma function that we previously knew in the form of integral Γ(x)=∫0∞tx−1e−tdt \Gamma (x) = \int_{0}^{\infty} t^{x-1} e^{-t} dt looks entirely different, yet in 1729 Euler proved that both expressions are exactly the same. The derivation introduced in this article is somewhat simplified compared to the original, but there will be no essential problem in understanding.

Derivation

If we set Γn(x):=∫0ntx−1(1−tn)ndt\displaystyle \Gamma_{n}(x) := \int_{0}^{n} t^{x-1} \left( 1 - { t \over n } \right) ^{n} dt as e−t=lim⁡n→∞(1−tn)−n\displaystyle e^{-t} = \lim_{n \to \infty } \left( 1 - { t \over n } \right) ^{-n}, then lim⁡n→∞Γn(x)=lim⁡n→∞∫0ntx−1(1−tn)ndt=∫0∞tx−1e−tdt=Γ(x) \begin{align*} \lim_{n \to \infty} \Gamma_{n}(x) =& \lim _{n \to \infty} \int_{0}^{n} t^{x-1} \left( 1 - { t \over n } \right) ^{n} dt \\ =& \int_{0}^{\infty} t^{x-1} e^{-t} dt \\ =& \Gamma (x) \end{align*} follows. Meanwhile, substituting from Γn(x)=∫0ntx−1(1−tn)ndt\displaystyle \Gamma_{n}(x) = \int_{0}^{n} t^{x-1} \left( 1 - { t \over n } \right) ^{n} dt to u=tnu = {t \over n} yields Γn(x)=∫0ntx−1(1−tn)ndt=∫01(nu)x−1(1−u)nndu=nx∫01ux−1(1−u)ndu \begin{align*} \Gamma_{n}(x) =& \int_{0}^{n} t^{x-1} \left( 1 - { t \over n } \right) ^{n} dt \\ =& \int_{0}^{1} (nu) ^{x-1} ( 1 - u ) ^{n} n du \\ =& n^{x} \int_{0}^{1} u^{x-1} ( 1 - u ) ^{n} du \end{align*} By the method of integration by parts, ∫01ux−1(1−u)ndu=[(1x)ux(1−u)n]01−∫01−(1x)uxn(1−u)n−1du=(nx)∫01ux(1−u)n−1du=(nx)(n−1x+1)∫01ux+1(1−u)n−2du⋮=(nx)⋯(2x+n−2)∫01ux+n−2(1−u)1du=(nx)⋯(2x+n−2)(1x+n−1)∫01ux+n−1(1−u)0du=(nx)⋯(2x+n−2)(1x+n−1)∫01ux+n−1du=(nx)⋯(2x+n−2)(1x+n−1)[1x+nux+n]01=(nx)⋯(2x+n−2)(1x+n−1)(1x+n)=n(n−1)(n−2)⋯2⋅1⋅1x(x+1)(x+2)⋯(x+n)=n!x(x+1)(x+2)⋯(x+n) \begin{align*} & \int_{0}^{1} u^{x-1} ( 1 - u ) ^{n} du \\ =& \left[ \left( { 1 \over x} \right) u^x (1-u)^n \right] _{0}^{1} - \int_{0}^{1} - \left( { 1 \over x} \right) u^x n ( 1 - u ) ^{n-1} du \\ =& \left( { n \over x} \right) \int_{0}^{1} u^x ( 1 - u ) ^{n-1} du \\ =& \left( { n \over x} \right) \left( { {n-1} \over {x+1}} \right) \int_{0}^{1} u^{x+1} ( 1 - u ) ^{n-2} du \\ \vdots& \\ =& \left( { n \over x} \right) \cdots \left( { 2 \over {x+n-2}} \right) \int_{0}^{1} u^{x+n-2} ( 1 - u ) ^{1} du \\ =& \left( { n \over x} \right) \cdots \left( { 2 \over {x+n-2}} \right) \left( { 1 \over {x+n-1}} \right) \int_{0}^{1} u^{x+n-1} ( 1 - u ) ^{0} du \\ =& \left( { n \over x} \right) \cdots \left( { 2 \over {x+n-2}} \right) \left( { 1 \over {x+n-1}} \right) \int_{0}^{1} u^{x+n-1} du \\ =& \left( { n \over x} \right) \cdots \left( { 2 \over {x+n-2}} \right) \left( { 1 \over {x+n-1}} \right) \left[ {{ 1 } \over { x+n }} u^{x+n} \right]_{0}^{1} \\ =& \left( { n \over x} \right) \cdots \left( { 2 \over {x+n-2}} \right) \left( { 1 \over {x+n-1}} \right) \left( { 1 \over {x+n}} \right) \\ =& { {n(n-1)(n-2) \cdots 2 \cdot 1 \cdot 1 } \over {x(x+1)(x+2) \cdots (x + n ) } } \\ =& { { n! } \over {x(x+1)(x+2) \cdots (x + n ) } } \end{align*} is obtained and, therefore, Γn(x)=nxn!x(x+1)(x+2)⋯(x+n) \Gamma_{n}(x) = n^{x} { { n! } \over {x(x+1)(x+2) \cdots (x + n ) } } is acquired. Since we have previously shown that Γ(x)=lim⁡n→∞Γn(x)\displaystyle \Gamma (x) = \lim_{n \to \infty} \Gamma_{n}(x), the following is established: Γ(x)=lim⁡n→∞Γn(x)=lim⁡n→∞nxn!x(x+1)(x+2)⋯(x+n) \begin{align*} \Gamma (x) =& \lim_{n \to \infty} \Gamma_{n}(x) \\ =& \lim_{n \to \infty} { { n^{x} n! } \over {x(x+1)(x+2) \cdots (x + n ) } } \end{align*}

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See Also

  1. https://gubner.ece.wisc.edu/notes/GammaFunctionStirling.pdf ↩︎