The Gamma function has the property
$$\Gamma(z+1) = z\Gamma(z) $$
In case I have a positive real number $x$, then can I also say that:
$$ \Gamma(-x) = \frac{\Gamma(1-x)}{-x}$$
I know that $\Gamma(x)$ is not defined when $x$ is a negative integer. But is the above relation true?
For sure, the integral definition of the Gamma function is not defined for $x<0$, since the integral $$\int_0^\infty t^{x-1}e^{-t}\mathrm dt$$ Won't converge for $x<0$. But we can extend (or analytically continue) this definition to these values. We know that for $x>1$, that $\Gamma(x+1)=x\Gamma(x)$. As long as $x$ is not an integer (why?) we can use the this recurrence as many times as we wish. For example, to compute $\Gamma(-1/2)$ we would write $$\Gamma(1/2)=\Gamma(-1/2+1)=\frac{-1}{2}\Gamma(-1/2)$$ And using the well known $\Gamma(1/2)=\sqrt{\pi}$ we get $$\frac{-1}{2}\Gamma(-1/2)=\sqrt{\pi}$$ Hence $$\Gamma(-1/2)=-2\sqrt{\pi}$$ My question to you - why does this process fail for negative integers?