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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!

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    $\begingroup$ From your recursive definition what is $\Gamma(1/2)$? $\endgroup$ Commented Feb 5, 2014 at 13:45
  • $\begingroup$ Not possible to do that. Can we add some "minimum" condition to make it an equivalent definition? $\endgroup$
    – Tim
    Commented Feb 5, 2014 at 13:46
  • $\begingroup$ Yes. The definition above is equivalent to itself ;-) $\endgroup$ Commented Feb 5, 2014 at 13:48
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    $\begingroup$ 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. $\endgroup$ Commented Feb 5, 2014 at 13:53
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    $\begingroup$ It is enough to add the condition of log convexity: en.wikipedia.org/wiki/Bohr%E2%80%93Mollerup_theorem $\endgroup$ Commented Feb 5, 2014 at 13:56

3 Answers 3

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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 $(0,+\infty)$.

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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.

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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).

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    $\begingroup$ This doesn't answer OPs question. $\endgroup$
    – Wojowu
    Commented Jan 13, 2016 at 21:21

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