Could negative integer factorials be defined in some way?

Question

I know that, calling $F$ such an extension, if we wanted to keep having $$ F(z+1)=(z+1)F(z),$$ letting $ z=-1$ would lead to the absurdity $ 1=0$. Also, $ \Gamma(z)$ has poles at $ z \in \mathbb{Z^-}$. But couldn't we just say this recurrence relation is replaced by some other in $F$ if $z $ is a negative integer? Obviously $ F$ would still derive from some property of factorials.

Answer 1

$\qquad\quad$ I believe you might be interested in Hadamard's $\Gamma$ function, or Luschny's factorial:

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Answer 2

It's called gamma function and expands to rational, real, even complex numbers. (Of course including the negative ones to answer your question correctly)

Answer 3

It is possible to ascribe the values using the following formulas:

$$\Gamma(-n)=\lim_{h\to0} \Re (\Gamma(-n+ih))$$

or, alternatively,

$$\Gamma(-n)=\lim_{h\to0} \frac{\Gamma(-n+h)+\Gamma(-n-h)}2$$

This way we can ascribe values to Gamma function at negative integers: $$\Gamma(0)=-\gamma$$ $$\Gamma(-1)=\gamma-1$$ $$\Gamma(-2)=\frac{3}{4}-\frac{\gamma }{2}$$ $$\Gamma(-3)=\frac{\gamma }{6}-\frac{11}{36}$$ $$\Gamma(-4)=\frac{25}{288}-\frac{\gamma }{24}$$

etc.

This does not preserve the main functional equation however. Although the values are consistent with some other places where Gamma function is used, for example in determining the integration constants in context of differintegral:

$$(1/t)^{(s)}|_{t=1}=\frac{1}{2\pi}\int_{-\infty}^{+\infty} \frac{e^{- i \omega }}{(-i\omega)^s} \int_{-\infty}^{+\infty}(1/t) e^{i\omega t}dt \, d\omega=\Gamma(s+1)\cos(\pi s) $$