$\Gamma(x)$ function evaluation

Question

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...

Cameron Williams · Accepted Answer · 2015-10-07T16:25:27.040

3

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 $\infty$.

edited Oct 07 '15 at 16:25

answered Oct 07 '15 at 16:19

Cameron Williams

29,432

In a word, no. :/ Where do I go from here? – Michael Roberts Oct 07 '15 at 16:21
Actually, in a word, YES! Thank you. Took a while before the penny dropped. – Michael Roberts Oct 07 '15 at 16:22
Glad you got it :) – Cameron Williams Oct 07 '15 at 16:23
No wait, I was wrong. So for $\alpha=-1$?? How is that then evaluated as? – Michael Roberts Oct 07 '15 at 16:24
When $\alpha = -1$, your expression isn't well-defined since $\Gamma(0)$ is not defined, but people often just call it (complex) infinity. See my edit. – Cameron Williams Oct 07 '15 at 16:24

$\Gamma(x)$ function evaluation

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