# The inverse of the Gamma function at $-\infty$

Let $\Gamma$ be the analytic continuation of the Gamma function $$\Gamma:z\mapsto \int_0^{+\infty} x^{z-1}e^{-x}dx$$ on the complex plane except non-positive integers.

We know that $\Gamma$ has no zeros but simple poles on $\{0,-1,-2,-3,\ldots\}$. So if we define $G:z\mapsto \dfrac{1}{\Gamma(z)}$, this is an entire function with only simple zeros at $\{0,-1,-2,-3,\ldots\}$.

My questions are the followings:

What is the behavior between two zeros of $G$ ? What is the growth of $G$ in $[-(n+1),-n]$? Do we know if, for example, $\displaystyle \dfrac{e^{- t^2}}{\Gamma(t)} \underset{t\to -\infty}{\longrightarrow} 0$ ?

• From the reflection formula $-z\Gamma(z)\Gamma(-z) = \frac{\pi}{\sin(\pi z)}$ you get $$\frac{1}{\Gamma(z)}= -\frac{\sin(\pi z)}{z\pi\Gamma(-z)}$$ Now plugin your favorite formula for $z$ with $\Re z > 0$. – gammatester Apr 13 '15 at 10:28
• Great, thank you. So the limit I was asking is true with Stirling's formula. – Bebop Apr 13 '15 at 10:42