Derivation of Euler's Limit Formula for the Gamma Function
Formula 1
For the gamma function $\Gamma : (0, \infty) \to \mathbb{R}$, the following holds. $$ \Gamma (x) = \lim_{n \to \infty} {{n^x n!} \over {x(x+1)(x+2) \cdots (x+n) }} $$
Explanation
The gamma function as we already know it has the integral form $$ \Gamma (x) = \int_{0}^{\infty} t^{x-1} e^{-t} dt $$ and does not resemble the above at all, but in 1729 Euler proved that the two expressions are exactly the same. The derivation introduced in this article is somewhat more informal than the original, but there should be no essential problem in understanding it.
Derivation
Let $\displaystyle \Gamma_{n}(x) := \int_{0}^{n} t^{x-1} \left( 1 - { t \over n } \right) ^{n} dt$. Since $\displaystyle e^{-t} = \lim_{n \to \infty } \left( 1 - { t \over n } \right) ^{-n}$, we have $$ \begin{align*} \lim_{n \to \infty} \Gamma_{n}(x) =& \lim _{n \to \infty} \int_{0}^{n} t^{x-1} \left( 1 - { t \over n } \right) ^{n} dt \\ =& \int_{0}^{\infty} t^{x-1} e^{-t} dt \\ =& \Gamma (x) \end{align*} $$ Meanwhile, substituting $u = {t \over n}$ in $\displaystyle \Gamma_{n}(x) = \int_{0}^{n} t^{x-1} \left( 1 - { t \over n } \right) ^{n} dt$ gives $$ \begin{align*} \Gamma_{n}(x) =& \int_{0}^{n} t^{x-1} \left( 1 - { t \over n } \right) ^{n} dt \\ =& \int_{0}^{1} (nu) ^{x-1} ( 1 - u ) ^{n} n du \\ =& n^{x} \int_{0}^{1} u^{x-1} ( 1 - u ) ^{n} du \end{align*} $$ By integration by parts, $$ \begin{align*} & \int_{0}^{1} u^{x-1} ( 1 - u ) ^{n} du \\ =& \left[ \left( { 1 \over x} \right) u^x (1-u)^n \right] _{0}^{1} - \int_{0}^{1} - \left( { 1 \over x} \right) u^x n ( 1 - u ) ^{n-1} du \\ =& \left( { n \over x} \right) \int_{0}^{1} u^x ( 1 - u ) ^{n-1} du \\ =& \left( { n \over x} \right) \left( { {n-1} \over {x+1}} \right) \int_{0}^{1} u^{x+1} ( 1 - u ) ^{n-2} du \\ \vdots& \\ =& \left( { n \over x} \right) \cdots \left( { 2 \over {x+n-2}} \right) \int_{0}^{1} u^{x+n-2} ( 1 - u ) ^{1} du \\ =& \left( { n \over x} \right) \cdots \left( { 2 \over {x+n-2}} \right) \left( { 1 \over {x+n-1}} \right) \int_{0}^{1} u^{x+n-1} ( 1 - u ) ^{0} du \\ =& \left( { n \over x} \right) \cdots \left( { 2 \over {x+n-2}} \right) \left( { 1 \over {x+n-1}} \right) \int_{0}^{1} u^{x+n-1} du \\ =& \left( { n \over x} \right) \cdots \left( { 2 \over {x+n-2}} \right) \left( { 1 \over {x+n-1}} \right) \left[ {{ 1 } \over { x+n }} u^{x+n} \right]_{0}^{1} \\ =& \left( { n \over x} \right) \cdots \left( { 2 \over {x+n-2}} \right) \left( { 1 \over {x+n-1}} \right) \left( { 1 \over {x+n}} \right) \\ =& { {n(n-1)(n-2) \cdots 2 \cdot 1 \cdot 1 } \over {x(x+1)(x+2) \cdots (x + n ) } } \\ =& { { n! } \over {x(x+1)(x+2) \cdots (x + n ) } } \end{align*} $$ and therefore we obtain $$ \Gamma_{n}(x) = n^{x} { { n! } \over {x(x+1)(x+2) \cdots (x + n ) } } $$ Since we showed earlier that $\displaystyle \Gamma (x) = \lim_{n \to \infty} \Gamma_{n}(x)$, the following holds. $$ \begin{align*} \Gamma (x) =& \lim_{n \to \infty} \Gamma_{n}(x) \\ =& \lim_{n \to \infty} { { n^{x} n! } \over {x(x+1)(x+2) \cdots (x + n ) } } \end{align*} $$
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