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First let's state a well-known characterization of gamma function.

If f is a positive function on positive real numbers such that: (1).f(x+1)=xf(x); (2).f(1)=1; (3).logf is convex, then f(x) is gamma function.

Now here I'm wondering how many ways can we characterize gamma function like the above? Especially if we consider it as a function on complex plane with poles.

ps: I'm not asking different ways to express gamma function explicitly, but the abstraction of it.

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  • $\begingroup$ I think this question would be more liekly to attract answers if you gave some reason for wanting such a list. The gamma function is interesting, but abstract characterizations of it are mostly interesting if you're using for something. If you are, we'd be curious to know what. $\endgroup$
    – Ben Webster
    Feb 16, 2010 at 17:57
  • $\begingroup$ I'd like to point out this this is a terrible question, because it admits the awful answer: "many". More seriously, see Ben's comment. $\endgroup$ Feb 18, 2010 at 7:17

2 Answers 2

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Have a look here: http://dlmf.nist.gov/5/

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maybe I can give you some help. Gamma function is also called the second Euler integral.

Here comes some characterizations.

a f(s)= $$t(x)=\int_{0}^{+\infty}{t^(s-1)}{exp(-t)}dt$$ s>0

b f(s)=$$\lim n!n^s/[s(s+1)...(s+n)] $$ $$n\rightarrow +\infty$$

c $$B(p,q)=\Gamma(p)\Gamma(q)/\Gamma(pq)$$ p>0 q>0

d $$\Gamma(2s)=2^(2s-1)\Gamma(s)\Gamma(s+1/2)/\sqrt(2\pi) $$ s>0

e $$\Gamma(s)\Gamma(1-s)=\pi/sin(s\pi)$$ 0

May it help!

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  • $\begingroup$ actually ,I don't think there is a particularly difference between definitions and characterizations. That is my opinion! $\endgroup$
    – DarkLight
    Mar 7, 2010 at 16:38
  • $\begingroup$ I disagree. A characterization would be a set of properties that turned out to be uniquely satisfied by the gamma function. It's like the difference between defining the reals as Dedekind cuts and characterizing them as the unique complete ordered field. There is a characterization of the gamma function as the unique function that satisfies f(n)=(n-1)f(n-1) and has some other smoothness property. Unfortunately, I can't remember what that other property is. $\endgroup$
    – gowers
    Mar 7, 2010 at 20:57
  • $\begingroup$ Maybe it is true.But the definition is also unique. $\endgroup$
    – DarkLight
    Mar 8, 2010 at 13:20

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