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The Gamma function is defined in terms of an integral as

The notation $Γ(t)$ is due to Legendre. If the real part of the complex number $t$ is positive $(Re(t) > 0)$, then the integral $$ \Gamma(t) = \int_0^\infty x^t e^{-x}\,\frac{{\rm d}x}{x} $$ converges absolutely, and is known as the Euler integral of the second kind (the Euler integral of the first kind defines the Beta function).

Can it be equivalently defined in terms of a recursive relation as $$ \Gamma(t+1)=t \Gamma(t)$$ $$\Gamma(1)=1$$ with some non-redundant conditions?

Thanks!

Tim
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    From your recursive definition what is $\Gamma(1/2)$? – JP McCarthy Feb 05 '14 at 13:45
  • Not possible to do that. Can we add some "minimum" condition to make it an equivalent definition? – Tim Feb 05 '14 at 13:46
  • Yes. The definition above is equivalent to itself ;-) – JP McCarthy Feb 05 '14 at 13:48
  • By "minimum" condition, I mean a condition which is not overlapping the recursive relation and can complement the recursive relation into an equivalent definition. – Tim Feb 05 '14 at 13:50
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    I would look at the Wikipedia entry; there are some possibilities there for defining Gamma on the fractions using the values on the natural numbers. – JP McCarthy Feb 05 '14 at 13:53
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    It is enough to add the condition of log convexity: http://en.wikipedia.org/wiki/Bohr%E2%80%93Mollerup_theorem – Cardboard Box Feb 05 '14 at 13:56

3 Answers3

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Baby Rudin gives the following definition:

$\forall x\in(0,+\infty):\ \Gamma(x+1)=x\Gamma(x);$

$\forall n\in{\Bbb N}:\ \Gamma(n+1)=n!;$

$\log\Gamma$ is convex in $(0,+\infty)$.

Martín-Blas Pérez Pinilla
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The two definitions are equivalent if $t$ is a an integer. If $t$ is an integer less than equal to zero neither object is defined and if $t$ is an integer greater or equal than one then the two definitions are equivalent. In all other cases the standard definition is defined whereas yours is not defined.

JPi
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There are an infinitude of functions interpolating $n!$, a few interesting alternatives are discussed here (Bernoulli's, Hadamard's, and Luschny's).

vonbrand
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