Euler’s Limit Formula for the Gamma Function

This note is based on the Appendix A.7 of (Grafakos, 2008). The gamma function $\Gamma(x)$ is defined by the integral $\int_0^\infty t^{x-1}e^{-t}$. For integer $x$ this defines the factorial up to $x-1$. In general this definition holds for all complex numbers except for non-positive integers, where it has simple poles. This can be seen from the Taylor expansion of $e^{-t}$.

Euler’s limit formula for the gamma function considers the following functions

$$\begin{equation} \Gamma_n(z) = \int_0^n \left(1-\frac{t}{n}\right)^nt^{z-1}dt \end{equation}$$

for integer $n$s and $Re(z)>0$. Because only for $Re(z)>0$, the improper integral is well defined. Obviously the integrand replace $e^{-t}$ by its well known approximation. And Euler’s theorem says that $\lim_{n\to\infty}\Gamma_n(z)=\Gamma(z)$.

The proof considers the difference of two definitions and split the integration interval into 3 parts.

$$\begin{equation} \Gamma(z)-\Gamma_n(z) = I_1(z) + I_2(z) + I_3(z) \end{equation}$$

where

$$\begin{equation} I_1(z) = \int_n^\infty t^{x-1}e^{-t}dt \\ I_2(z) = \int_{n/2}^n \left(e^{-t}-\left(1-\frac{t}{n}\right)^n\right)t^{x-1}dt \\ I_3(z) = \int_0^{n/2} \left(e^{-t}-\left(1-\frac{t}{n}\right)^n\right)t^{x-1}dt\,. \end{equation}$$

$I_1$ goes to $0$ for any $z$ when $n\to\infty$. To study the difference in $I_2$ and $I_3$, we rewrite

$$\begin{equation} \left(1-\frac{t}{n}\right)^n = \exp(n\ln(1-t/n)) = \exp\left(-n\sum_{k=1}^\infty \frac{1}{k}\left(\frac{t}{n}\right)^k\right)\,. \end{equation}$$

Then

$$\begin{align} e^{-t}-\left(1-\frac{t}{n}\right)^n & = e^{-t}\left(1-\exp\left(-\sum_{k=2}^\infty\frac{t^k}{kn^{k-1}}\right)\right) \\ & \le e^{-t}\,. \end{align}$$

By dominant convergence theorem we know $I_2$ also goes to $0$. For $I_3$, we know that

$$\begin{equation} \sum_{k=2}^\infty \frac{t^k}{kn^{k-1}} \le \frac{t^2}{n}\sum_{k=2}^\infty\frac{(1/2)^{k-1}}{k} = c\left(\frac{t^2}{n}\right)\,. \end{equation}$$

So

$$\begin{align} e^{-t}-\left(1-\frac{t}{n}\right)^n & \le e^{-t}(1-\exp(-ct^2/n)) \\ & \le \frac{ct^2}{n}e^{-t} \\ & \to 0\,, \end{align}$$

as $n\to\infty$. Again by dominant convergence theorem, we have $I_3$ goes to $0$.

Reference

1. Grafakos, L. (2008). Classical Fourier Analysis. Springer New York.