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I am just a bit curious about the Weierstrass Factorization theorem on the Gamma function.

The Weierstrass Factorization theorem says this:

Let $f(z)$ be an entire function. Suppose that $f$ vanishes to order $m, m \geq 0$. Let ${a_n}$ be the other zeros of $f$, listed with multiplicities. Then there is an entire function, $g(z)$ such that $$f(z) = z^m e^{g(z)} \prod_{n = 1}^{\infty} E_{n-1}\left(\frac{z}{a_n}\right).$$

From the reciprocal Gamma function, by the Weierstrass factorization theorem, they came up with this:

$$ \frac{1}{\Gamma(z)} = ze^{\gamma*z}\prod_{n=1}^{\infty}\left(1 + \frac{z}{n}\right)e^{-z/n}. $$

Now how did they compute the Weierstrass Elementary Factors? I know the definition of the factors. I have looked this function and it says it has simple zeros at $0$ and at all negative integers. I am just confused how they computed the elementary factors on this function. Thanks for all the help!

Arnaud D.
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Jim Baker
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    It is known that all elementary factors can be taken to be $e^{z/r}$ if the function grows at most like $e^{|z|^\alpha}$ for $\alpha<2$. Stirling's approximation to $\Gamma(z)$ establishes such a bound for $\alpha=1+\varepsilon$. – Greg Martin May 13 '19 at 06:14
  • Im not sure what you mean. – Jim Baker May 13 '19 at 14:18
  • Seems like you need Hadamard’s factorization theorem. – Ovi May 13 '19 at 14:53
  • Are you assuming $1/\Gamma(z)$ is entire ? Its zeros are known to be at negative integers. Estimating its growth isn't difficult thus obtaining the condition for applying Weierstrass factorization. – reuns May 15 '19 at 04:42

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The convexity of $\log(\Gamma(x))$ says that for $x\in[0,1]$, $$ \Gamma(n)\,(n-1+x)^x\le\Gamma(n+x)\le\Gamma(n)\,n^x $$ and the inequalities are reversed for $x\not\in[0,1]$. Note that this is exact for $x\in\{0,1\}$. In any case, $$ \begin{align} 1 &=\lim_{n\to\infty}\frac{(n-1)!\,n^x}{\Gamma(n+x)}\\[6pt] &=\lim_{n\to\infty}\frac{(n-1)!\,n^x}{\Gamma(x)\prod\limits_{k=0}^{n-1}(k+x)} \end{align} $$ Therefore, since $\sum\limits_{k=1}^{n-1}\frac1k\sim\gamma+\log(n)$, we get $n^x=e^{x\log(n)}\sim e^{-\gamma x}\prod\limits_{k=1}^{n-1}e^{x/k}$ $$ \begin{align} \Gamma(x) &=\lim_{n\to\infty}\frac{(n-1)!\,n^x}{\prod\limits_{k=0}^{n-1}(k+x)}\\ &=\lim_{n\to\infty}\frac{n^x}x\prod_{k=1}^{n-1}\frac{k}{k+x}\\[6pt] &=\lim_{n\to\infty}\frac{e^{-\gamma x}}x\prod_{k=1}^{n-1}\frac{e^{x/k}}{1+x/k}\\[6pt] &=\frac{e^{-\gamma x}}x\prod_{k=1}^\infty\frac{e^{x/k}}{1+x/k} \end{align} $$

robjohn
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  • @reuns: Other than log-convexity (which is mentioned), I am only using the usual $\Gamma(x+1)=x\Gamma(x)$ and $\Gamma(1)=1$. – robjohn May 15 '19 at 05:08
  • So $\Gamma(x+1) = x \Gamma(x)$ plus $\log$-convexity on $x > 0$ implies the Weierstrass product for $x > 0$ and by analytic continuation it stays true for $\Gamma(z), z\in \Bbb{C}$. The $\log$-convexity follows from Cauchy-Schwartz. – reuns May 15 '19 at 06:28
  • If we define $\Gamma(x)=\int_0^\infty t^{x-1}e^{-t},\mathrm{d}t$, then the log-convexity follows from Cauchy-Schwarz or Jensen's Inequality, as shown at the end of this answer. However, the Bohr-Mollerup Theorem guarantees that we get the same $\Gamma(x)$ assuming the standard recurrence and log-convexity. – robjohn May 15 '19 at 06:49