Editing Gamma function (section)

=== Alternative definitions ===
There are many equivalent definitions.

==== Euler's definition as an infinite product ====
<!-- Linked to from [[Binomial coefficient]] -->For a fixed integer <math>m</math>, as the integer <math>n</math> increases, we have that<ref>{{Cite journal |last=Davis |first=Philip |title=Leonhard Euler's Integral: A Historical Profile of the Gamma Function |url=https://ia800108.us.archive.org/view_archive.php?archive=/24/items/wikipedia-scholarly-sources-corpus/10.2307%252F2287541.zip&file=10.2307%252F2309786.pdf |website=maa.org}}</ref>
<math display="block">\lim_{n \to \infty} \frac{n! \, \left(n+1\right)^m}{(n+m)!} = 1\,.</math>

If <math>m</math> is not an integer, then this equation is meaningless, since in this section the factorial of a non-integer has not been defined yet. However, let us assume that this equation continues to hold when <math>m</math> is replaced by an arbitrary complex number <math>z</math>, in order to define the Gamma function for non-integers:

<math display="block">\lim_{n \to \infty} \frac{n! \, \left(n+1\right)^z}{(n+z)!} = 1\,.</math>
Multiplying both sides by <math>(z-1)!</math> gives
<math display="block">\begin{align}
(z-1)!
      &= \frac{1}{z} \lim_{n \to \infty} n!\frac{z!}{(n+z)!} (n+1)^z \\[8pt]
      &= \frac{1}{z} \lim_{n \to \infty} (1 \cdot2\cdots n)\frac{1}{(1+z) \cdots (n+z)} \left(\frac{2}{1} \cdot \frac{3}{2} \cdots \frac{n+1}{n}\right)^z \\[8pt]
      &= \frac{1}{z} \prod_{n=1}^\infty \left[ \frac{1}{1+\frac{z}{n}} \left(1 + \frac{1}{n}\right)^z \right].
\end{align}</math>This [[infinite product]], which is due to Euler,<ref>{{Cite journal |last=Bonvini |first=Marco |date=October 9, 2010 |title=The Gamma function |url=https://www.roma1.infn.it/~bonvini/math/Marco_Bonvini__Gamma_function.pdf |journal=Roma1.infn.it}}</ref> converges for all complex numbers <math>z</math> except the non-positive integers, which fail because of a division by zero. In fact, the above assumption produces a unique definition of <math>\Gamma(z)</math> as {{tmath|(z-1)!}}.

Intuitively, this formula indicates that <math>\Gamma(z)</math> is approximately the result of computing <math>\Gamma(n+1)=n!</math> for some large integer <math>n</math>, multiplying by <math>(n+1)^z</math> to approximate <math>\Gamma(n+z+1)</math>, and then using the relationship <math>\Gamma(x+1) = x \Gamma(x)</math> backwards <math>n+1</math> times to get an approximation for <math>\Gamma(z)</math>; and furthermore that this approximation becomes exact as <math>n</math> increases to infinity.

The infinite product for the [[reciprocal gamma function|reciprocal]]
<math display="block">\frac{1}{\Gamma(z)} = z \prod_{n=1}^\infty \left[ \left(1+\frac{z}{n}\right) / {\left(1 + \frac{1}{n}\right)^z} \right]</math>
is an [[entire function]], converging for every complex number {{mvar|z}}.

==== Weierstrass's definition ====
The definition for the gamma function due to [[Karl Weierstrass|Weierstrass]] is also valid for all complex numbers&nbsp;<math>z</math> except non-positive integers:
<math display="block">\Gamma(z) = \frac{e^{-\gamma z}} z \prod_{n=1}^\infty \left(1 + \frac z n \right)^{-1} e^{z/n},</math>
where <math>\gamma \approx 0.577216</math> is the [[Euler–Mascheroni constant]].<ref name="Davis" /> This is the [[Entire function#Genus|Hadamard product]] of <math>1/\Gamma(z)</math> in a rewritten form.

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'''Equivalence of the integral definition and Weierstrass definition'''

By the integral definition, the relation <math>\Gamma (z+1)=z\Gamma (z)</math> and [[Hadamard factorization theorem]],
<math display="block>\frac{1}{\Gamma (z)}=ze^{c_1 z+c_2}\prod_{n=1}^\infty e^{-\frac{z}{n}}\left(1+\frac{z}{n}\right)</math>
for some constants <math>c_1,c_2</math> since <math>1/\Gamma</math> is an entire function of order <math>1</math>. Since <math>z\Gamma (z)\to 1</math> as <math>z\to 0</math>, <math>c_2=0</math> (or an integer multiple of <math>2\pi i</math>) and since <math>\Gamma (1)=1</math>,
<math display="block">\begin{align}e^{-c_1}
&=\prod_{n=1}^\infty e^{-\frac{1}{n}}\left(1+\frac{1}{n}\right)\\
&=\exp\left(\lim_{N\to\infty}\sum_{n=1}^N \left(\log\left(1+\frac{1}{n}\right)-\frac{1}{n}\right)\right)\\
&=\exp\left(\lim_{N\to\infty}\left(\log (N+1)-\sum_{n=1}^N \frac{1}{n}\right)\right).\end{align}</math>

where <math>c_1=\gamma+2\pi i k</math> for some integer <math>k</math>. Since <math>\Gamma (z)\in\mathbb{R}</math> for <math>z\in\mathbb{R}\setminus\mathbb{Z}_0^-</math>, we have <math>k=0</math> and
<math display="block>\frac{1}{\Gamma (z)}=ze^{\gamma z}\prod_{n=1}^\infty e^{-\frac{z}{n}}\left(1+\frac{z}{n}\right)</math>

'''Equivalence of the Weierstrass definition and Euler definition'''

<math display="block">\begin{align}\Gamma (z)&=\frac{e^{-\gamma z}}{z}\prod_{n=1}^{\infty}\left(1+\frac{z}{n}\right)^{-1}e^{z/n}\\
&=\frac1z\lim_{n\to\infty}e^{z\left(\log (n+1)-1-\frac{1}{2}-\frac{1}{3}-\cdots-\frac{1}{n}\right)}\frac{e^{z\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)}}{\left(1+z\right)\left(1+\frac{z}{2}\right)\cdots\left(1+\frac{z}{n}\right)}\\
&=\frac1z\lim_{n\to\infty}\frac{1}{\left(1+z\right)\left(1+\frac{z}{2}\right)\cdots\left(1+\frac{z}{n}\right)}e^{z\log\left(n+1\right)}\\
&=\lim_{n\to\infty}\frac{n!(n+1)^z}{z(z+1)\cdots (z+n)},\quad z\in\mathbb{C}\setminus\mathbb{Z}_0^-\end{align}</math>
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