Gamma function $(-1)!$

Question

How would you show that $(-1)!$ is infinite? I think you need to use the gamma function but I'm not entirely sure because I don't think the gamma function works for negative reals

Answer 1

Hint. One may recall that $$ \int_0^\infty u^se^{-u}\:du=\Gamma(s+1),\quad s>-1. \tag1 $$ Then, integrating by parts, one gets $$ \Gamma(s+1)=s\:\Gamma(s),\qquad s>0, \tag2 $$ giving $$ \Gamma(s)=\frac{\Gamma(s+1)}s,\qquad s>0, $$ and, as $s \to 0^+$, $$ \Gamma(s)\sim\frac1s $$ yielding $$ \lim_{s \to 0^+}\Gamma(s)=+\infty $$ Then, if one defines $'(-1)!'$ as $'\Gamma(0)'=\lim_{s \to 0^+}\Gamma(s)$ , one obtains the announced result.

Answer 2

One needs two conditions for the factorial:

1. $n!=n(n-1)!$

2. $1!=1$

So it becomes clear that

$$1!=1\times0!\implies0!=1$$

$$0!=0\times(-1)!\implies(-1)!=1/0\to\pm\infty$$

Anytime a division by $0$ occurs, the factorial (and the gamma function) diverge to $\pm\infty$. This occurs at the negative integers.

The gamma function, on the other hand, does not diverge over the negative reals (except at the negative integers). Examine that $\Gamma(-1/2)=-2\sqrt\pi$, for example.

Answer 3

Define the following

$$\Gamma(x+1) = x\Gamma(x)$$

Then we clearly see that

$$\Gamma(x+1) = x(x-1)\cdots(x-n)\Gamma(x-n)$$

That implies the gamma function diverges for non-positive integers.