Product Form of Euler’s Limit Formula for the Gamma Function

This note is a continuation of last one. By the definition of $\Gamma_n(z)$, for $Re(z)>0$, we can rewrite it in the following form,

$$\begin{align} \Gamma_n(z) & = \int_0^n\left(1-\frac{t}{n}\right)^n t^{z-1}dt \\ & = \frac{n!n^z}{z\cdots(z+n)} \\ 1 & = \Gamma_n(z)z\exp(z\gamma_n)\prod_{k=1}^n\left(1+\frac{z}{k}\right)e^{-z/k}\,, \tag{*} \end{align}$$

where $\gamma_n = \sum_{k=1}^n 1/k - \ln n$ and $\lim_{n\to\infty}\gamma_n$ is the Euler’s constant. So by sending $n$ to infinity, we get

$$\begin{equation} 1 = \Gamma(z)z\exp(z\gamma)\prod_{k=1}^\infty\left(1+\frac{z}{k}\right)e^{-z/k}\,. \end{equation}$$

Assume that $\vert z\vert < N$. Then the infinite product

$$\begin{equation} \prod_{k=N}^\infty\left(1+\frac{z}{k}\right)e^{-z/k} = \exp\left(\sum_{k=N}^\infty -z/k+\ln(1+z/k)\right)\,. \end{equation}$$

And for any $k\ge N+1$,

$$\begin{align} \left\vert -\frac{z}{k}+\ln(1+\frac{z}{k})\right\vert & = \left\vert\sum_{i=2}^\infty \frac{1}{i}\left(-\frac{z}{k}\right)^i\right\vert \\ & \le \frac{\vert z\vert^2}{k^2}\sum_{i=2}^\infty\frac{1}{i}\frac{\vert z\vert^i}{k^i} \\ & \le \frac{c_N\vert z\vert^2}{k^2}\,. \end{align}$$

Therefore the infinite product from $N$ to $\infty$ converges uniformly for all $\vert z\vert<N$. And the limit must be also analytic. The product of first $N-1$ terms is also analytic. So we prove that the infinite product defines an analytic function over the whole complex plane. It has roots at negative integers.

However, the equality (*) only holds for $Re(z)>0$ since we start from the definition of $\Gamma_n$. But we know that $\Gamma(z)$ has simple poles at non-positive integers and is analytic elsewhere (this can be seen from the analytic continuation from the half plane $$ Re(z)>0 $$ and the iteration relation $$ \Gamma(z)=\Gamma(z+1)/z $$) . So by analytic continuation we know that (*) holds for the whole complex plane. And we also get the fact that $\Gamma(z)\ne 0$ and that $\Gamma^{-1}(z)$ is entire.

An immediate consequence of the product form of Euler’s limit formula is the identity

$$\begin{equation} \frac{1}{\vert\Gamma(x+iy)\vert^2}=\frac{1}{\vert\Gamma(x)\vert^2} \prod_{k=0}^\infty\left(1+\frac{y^2}{(k+x)^2}\right)\,, \end{equation}$$

and further

$$\begin{align} \frac{1}{\vert\Gamma(x+iy)\vert^2} & \le \frac{1}{\vert\Gamma(x)\vert^2} \prod_{k=0}^\infty\exp\left(\frac{y^2}{(k+x)^2}\right) \\ & = \frac{1}{\vert\Gamma(x)\vert^2} \exp\left(y^2\sum_{k=0}^\infty\frac{1}{(k+x)^2}\right) \\ & \le \frac{1}{\vert\Gamma(x)\vert^2} \exp\left(y^2\left(\frac{1}{x^2}+\int_0^\infty\frac{1}{(k+x)^2}dk\right)\right) \\ & = \frac{1}{\vert\Gamma(x)\vert^2} \exp\left(y^2\left(\frac{1}{x^2}+\frac{1}{x}\right)\right) \\ & \le \frac{1}{\vert\Gamma(x)\vert^2} \exp\left(y^2\max\left(\frac{2}{x^2},\frac{2}{x}\right)\right)\,. \end{align}$$

The inequality will be used in asymptotic expansion of Bessel functions.

Reference