Let's say you just derived some formula that included integer parameters $m$ and $n$, and you wound up with something that had $m!$ or $n!$ or something similar in it, like the following example: $$ \operatorname{I}(m,\ n\ |\ x) = \int_0^x{t^m(1-t)^n\text{d}t} = m!n!\sum_{k=0}^{n}{(-1)^k\frac{x^{m+k+1}(1-x)^{n-k}}{(n-k)!(m+k+1)!}}, \quad m, n \in \Bbb{N} $$ Is it perfectly reasonable to say $$ \operatorname{I}(m,\ n\ |\ x) = \Gamma(m+1)\Gamma(n+1)\sum_{k=0}^{n}{(-1)^k\frac{x^{m+k+1}(1-x)^{n-k}}{\Gamma(n-k+1)\Gamma(m+k+2)}}, \quad m \in \Bbb{R}? $$ My intuition says yes because there is nothing in the equation which requires that either $m$ to be an integer ($n$ must be an integer for the sum to be finite as Yves Daoust pointed out). If it were something like $$ \text{something} = \sum_{k=0}^{n!}{\text{another thing}} $$ then obviously it would make no sense to say $$ \text{something} = \sum_{k=0}^{\Gamma(n+1)}{\text{another thing}} $$ So in general, if the only thing in an equation that requires a parameter to be an integer is factorial, then is it perfectly fine to replace it with the gamma function? Or are there very rare and tricky scenarios where using the gamma function would be false but using factorial would be true?
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"there is nothing in the equation that requires that either m or n be integers": for the sum to be finite, $n$ must be integer. – Feb 21 '15 at 16:53
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I don't think you get it. When $n$ isn't integer, the sum is infinite. – Feb 21 '15 at 16:56
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Oh my goodness I am such an idiot, I see it now. – user3002473 Feb 21 '15 at 16:58
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You need to use the generalized binomial formula. The identity you wrote is wrong. – Feb 21 '15 at 16:58
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@YvesDaoust how's that, better? – user3002473 Feb 21 '15 at 16:59
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Yep, I think it now agrees with the generalized binomial theorem. – Feb 21 '15 at 17:04
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Once I saw a notation for $a\ge 0$ $$a! = \prod_{n\in \Bbb N,\,n< \lfloor a\rfloor}(a-n).$$ In this case, $a!\ne \Gamma(a+1)$ for non-integer $a$.
If we don't count this extreme case of redefinition of factorial, I don't really see a scenario where we can't replace factorial by gamma-function.
TZakrevskiy
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Here is a contrived example. The formula $n! = \cos(2\pi n) \Gamma(n+1)$ is true for all non-negative integers, but it isn't true for non-integer values of $n$.
Stephen Montgomery-Smith
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