I have the following likelihood calculation:
\begin{align}\mathcal{L}(s|\alpha) = \sum_{i=1}^O\Biggl\{\ln\frac{ \Gamma(\alpha_0 )}{ \Gamma( B )}- \sum_{k=1}^K \ln \Gamma(\hat{\mathcal{S}}_k^i+1) + \ln \biggl[ \sum_{k=1}^K \Bigl( \ln \Gamma(\hat{\mathcal{S}}_k^i + \alpha_k) - \ln \Gamma(\alpha_k) \Bigr) \biggr] \Biggr\}\end{align}
There are several $\Gamma$ functions in there. $\Gamma$ is not defined for negative integers. When one of the $x$ values in $\Gamma(x)$ is a negative integer, what should I do? Can I add a small number to $x$ (e.g. $x = x+1e^{-131} \quad \text{if}\ x\ \text{is a negative integer}$?
Edit: https://math.stackexchange.com/a/263755/96592 could be the answer? So if $x$ is negative $\Gamma(x) = \frac{\Gamma(x+\epsilon)}{\Gamma(\epsilon)}$ with $\epsilon$ close to $0$? Not sure if I understand that correctly.
$\Gamma(x)$ has poles at non-positive integers, so you won't be able to solve the problem by adding a small value to $x$, since the smaller the value you use, the larger (or more negative depending on if x is odd or even) $\Gamma(x)$ will be.
In other words, for even $x \in \mathbb{Z}_{\leq 0}$
$\lim_{\epsilon \rightarrow 0^+}\Gamma(x+\epsilon)=\infty$ and $\lim_{\epsilon \rightarrow 0^-}\Gamma(x+\epsilon)=-\infty$
and vice versa for odd $x \in \mathbb{Z}_{\leq 0}$