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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.

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    $\begingroup$ Great first post! Welcome to the site! $\endgroup$ – Greg Martin Feb 11 at 18:19
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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.

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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.

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  • $\begingroup$ 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! $\endgroup$ – Charlotte Feb 11 at 18:38

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