Gamma reflection formula for large z

Question

In an expression involving many gamma functions I have a $\frac{1}{\Gamma(-z)}$ where $z\gg 1$. I need to use Stirling formula to approximate this expression. Can I use the following formula? $$\Gamma(z) \, \Gamma(1 − z) = \pi \, \sin (\pi z) $$

Answer 1

Why not? The reciprocal gamma function is entire and thus "well behaved".
The Euler reflection formula on the other side should be : $$\Gamma(z) \, \Gamma(1 − z) = \frac {\pi}{\sin (\pi z)}$$ so that using $\;\Gamma(1-z)=-z\,\Gamma(-z)\,$ you get : $$\frac{1}{\Gamma(-z)}=-\frac{z}{\pi}\,\Gamma(z)\sin (\pi z)$$ and may apply Stirling...

Answer 2

The reflection formula is valid for all $z$ and may be rewritten as $$ \frac{1}{\Gamma(-z)} = -\frac{\sin(\pi z)}{\pi}\Gamma(1+z). $$ However, note that in your case, if you use Strirling approximation, you might be considering integer $z \in \mathbb{Z}$ and then $$\frac{1}{\Gamma(-z)} \equiv 0.$$