# Relation Gamma function and products

I have been thinking about a specific problem for quite some time. Imagine we have the following product, where $$x \in \mathbb{N}$$ and $$a \in \mathbb{Z^+}$$ then we now that the following holds: \begin{align} \prod_{j=k}^{n-1} (x +a +j) = \frac{(x+a+n-1)!}{(x+a+k-1)!} = \frac{\Gamma(x+a+n)}{\Gamma(x+a+k)}. \end{align} What if this $$a$$ were not to be in $$\mathbb{Z}^+$$, but rather in $$\mathbb{R}^+$$, could we then still have a similar expression in terms of the Gamma function?

• Yes, $\Gamma(z+1)=z\Gamma(z)$ holds even for complex numbers.
– user65203
Jul 10, 2019 at 14:36
• You can group $x+a$ in a single variable, for clarity.
– user65203
Jul 10, 2019 at 14:38

Absolutely. This type of product is common enough in mathematics to have its own notation, the Pochammer symbol, defined as $$(a)_n \equiv \prod_{j=0}^{n-1}(a+j).$$ From the Gamma recursion identity $$\Gamma(z+1) = z\Gamma(z)$$, we can write the symbol in terms of the Gamma function: $$\begin{multline} \frac{\Gamma(a+n)}{\Gamma(a)} = (a+n-1)\frac{\Gamma(a+n-1)}{\Gamma(a)} = (a+n-1)(a+n-2)\frac{\Gamma(a+n-2)}{\Gamma(a)}=\,\,... \\ = (a+n-1)(a+n-2)...(a+1)a\frac{\Gamma(a)}{\Gamma(a)} = \prod_{j=0}^{n-1}(a+j) = (a)_n. \end{multline}$$ Since your product can be written as the Pochhammer symbol $$(x+a+k)_{n-k}$$, we see that indeed, $$\prod_{j=k}^{n-1}(x+a+j) = (x+a+k)_{n-k} = \frac{\Gamma(x+a+n)}{\Gamma(x+a+k)}.$$