Editing Gamma function (section)

==== 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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