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Martin Sleziak
Martin Sleziak
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The gamma function is not the only meromorphic function satisfying $$f(z+1)=z f(z),\qquad f(1)=1,$$ with no zeroes and no poles other than the points $z=0,-1,-2\dots$.

In fact, there is a whole bunch of such functions, which, in general, have the form $$f(z)=\exp{(-g(z))}\frac{1}{z\prod\limits_{n=1}^{\infty} \left(1+\frac{z}{m}\right)e^{-z/m}},$$ where $g(z)$ is an entire function such that $$g(z+1)-g(z)=\gamma+2k\pi i,\qquad g(1)=\gamma+2l\pi i,\qquad k,l\in\mathbb Z, $$ ($\gamma$ is Euler's constantEuler's constant). The gamma function corresponds to the simplest choice $g(z)=\gamma z$.

The gamma function is not the only meromorphic function satisfying $$f(z+1)=z f(z),\qquad f(1)=1,$$ with no zeroes and no poles other than the points $z=0,-1,-2\dots$.

In fact, there is a whole bunch of such functions, which, in general, have the form $$f(z)=\exp{(-g(z))}\frac{1}{z\prod\limits_{n=1}^{\infty} \left(1+\frac{z}{m}\right)e^{-z/m}},$$ where $g(z)$ is an entire function such that $$g(z+1)-g(z)=\gamma+2k\pi i,\qquad g(1)=\gamma+2l\pi i,\qquad k,l\in\mathbb Z, $$ ($\gamma$ is Euler's constant). The gamma function corresponds to the simplest choice $g(z)=\gamma z$.

The gamma function is not the only meromorphic function satisfying $$f(z+1)=z f(z),\qquad f(1)=1,$$ with no zeroes and no poles other than the points $z=0,-1,-2\dots$.

In fact, there is a whole bunch of such functions, which, in general, have the form $$f(z)=\exp{(-g(z))}\frac{1}{z\prod\limits_{n=1}^{\infty} \left(1+\frac{z}{m}\right)e^{-z/m}},$$ where $g(z)$ is an entire function such that $$g(z+1)-g(z)=\gamma+2k\pi i,\qquad g(1)=\gamma+2l\pi i,\qquad k,l\in\mathbb Z, $$ ($\gamma$ is Euler's constant). The gamma function corresponds to the simplest choice $g(z)=\gamma z$.

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Andrey Rekalo
Andrey Rekalo
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The gamma function is not the only meromorphic function satisfying $$f(z+1)=z f(z),\qquad f(1)=1,$$ with no zeroes and no poles other than the points $z=0,-1,-2\dots$.

In fact, there is a whole bunch of such functions, which, in general, have the form $$f(z)=\exp{(-g(z))}\frac{1}{z\prod\limits_{n=1}^{\infty} \left(1+\frac{z}{m}\right)e^{-z/m}},$$ where $g(z)$ is an entire function such that $$g(z+1)-g(z)=\gamma+2k\pi i,\qquad g(1)=\gamma+2l\pi i,\qquad k,l\in\mathbb Z, $$ ($\gamma$ is Euler's constant). The gamma function corresponds to the simplest choice $g(z)=\gamma z$.