# Tips for proving: $\Gamma(n-a)/\Gamma(-a)=(-1)^na(a-1)(a-2)…(a-n+1),~$ for $~n = 1, 2, …$?

I am trying to prove how $$\Gamma(n-a)/\Gamma(-a)=(-1)^na(a-1)(a-2)...(a-n+1),~$$ for $$~n = 1, 2, ...$$? So I believe that the numerator is $$\Gamma(n-a)=(n-a-1)!=((n-a)!)/(a-n)$$, and I proved this by using an analogy to the recurrence relation, and checked it with Wolfram Alpha. However, I do not know what to do with the denominator. The lecture notes state, $$1/\Gamma(-n)=0$$, where n = 0, 1, 2... However, the domain of "a" is not specified in the homework (the problem is given just as I have stated it in the title), so I am assuming that "a" can be any number, complex, negative, noninteger, etc.. So now I don't know where to start.

One route I was thinking of is using integration by parts and using a pattern that may come up to solve the integral (since simply integration by parts will go on endless without recognizing a pattern). In class today, he implied that the answer should be relatively simple to get, so it makes me think that integration by parts is not what he intended. Another route I was wondering about is if there is a relevant identity/formula for the Gamma function that I can use that may help me by analogy? Any starting point would be great, and especially in generic terms/concepts so that I can apply those tips to any similar problem (rather than an answer to the question I have).

This is my first time posting so hopefully I included all relevant information.

• Great first post! Welcome to the site! – Greg Martin Feb 11 at 18:19

## 2 Answers

The right-hand side is $$\prod_{i=0}^{n-1}(i-a)$$. Write $$i-a$$ in the form $$\frac{\Gamma(z+1)}{\Gamma(z)}$$, then telescope.

You're right that there's an identity that will be very helpful here: it's the functional equation for the Gamma function, $$\Gamma(z+1) = z\Gamma(z),$$ which is valid wherever $$\Gamma$$ is defined. (Indeed it can be used to define $$\Gamma(z)$$ wherever $$\Gamma(z+1)$$ is defined, which is even stronger.)

And you're also right about integrating by parts, in that this functional equation is proved precisely by integrating by parts.

• Sorry for the double comment. Now I see how to use it. I think the first idea that I had in the first comment will work while using the identity you suggested. Thank you so much for your help! – Charlotte Feb 11 at 18:38