I'm new to using the Gamma function, $\Gamma(n)$, and I was wondering how I would go about evaluating this, if it is at all possible:
$$ \frac{\Gamma(\alpha+1)}{\Gamma(\alpha+2)} $$ This is relating to a specific case I want to find regarding the average luminosity of all galaxies by using the Schechter function. For this case, let's state $\alpha$=-1. For the top case I get infinity, that is to say -1!. I understand that:
$$ \Gamma(n)=(n-1)! $$
But the top expression is not giving me the answer I expected...
The gamma function satisfies a recursive property:
$$\Gamma(x+1) = x\Gamma(x).$$
In your case, $\Gamma(\alpha+2) = \Gamma((\alpha+1)+1)$ so you are left with $\frac{1}{\alpha+1}$ as the final answer. This assumes that $\alpha+1$ is not a negative integer. In the case that $\alpha$ is a negative integer, your expression is not really well-defined but in the case that $\alpha=-1$, only the numerator is ill-defined (the denominator is perfectly well-defined since $\Gamma(-1+2) = \Gamma(1) = 1$) so it can be interpreted as $\infty$.