TITLE: Residue of $\Gamma^{2}$ and $\Gamma^{3}$
QUESTION [5 upvotes]: Based on wiki, the residues of $\Gamma$ at non positive integers are given by:
$$\text{Res}\left ( \Gamma(z),z=-n \right )=\frac{(-1)^{n}}{n!}.$$
I have been trying to find residue for $\Gamma^{2}$ and $\Gamma^{3}$ and could not find any. I tried deriving but not successful. How to find their residue? Or are they given?

REPLY [2 votes]: For the case of $\Gamma^{4}(x)$ the residue is
\begin{align}
Res_{x \rightarrow -n}( \Gamma^{4}(x)) = \frac{1}{3!} \left( \frac{(-1)^{n}}{n!}\right)^{4} \left[ 8 H_{n,3} + 12 (H_{n} - \gamma) H_{n,2} + 36 (H_{n} - \gamma) \zeta(2) + 46 (H_{n} - \gamma)^{3} - 4 \zeta(3) \right]
\end{align}
where $\gamma$ is the Euler-Mascheroni constant, $H_{n}$ are the Harmonic numbers, and 
\begin{align}
H_{n,r} = \sum_{k=1}^{n} \frac{1}{k^{r}}.
\end{align}