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Is the gamma function a unique $f(x)$ such that $$f(x)f(1-x)=\dfrac{\pi}{\sin \pi x}?$$ I looked up the Bohr–Mollerup theorem, though I wonder if it is possible to uniquely characterize the gamma function using less than 3 conditions.

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    $\begingroup$ if we set $f:=-\Gamma$ then the statement holds, but $-\Gamma\neq\Gamma$ $\endgroup$ – Masacroso Jul 7 at 15:49
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Well.. Gamma is not a unique function for which $$f(x)f(1-x)=\frac{\pi}{\sin(\pi x)}$$holds. Instead there exists infinity many functions $f$ for which the above equality holds. A trivial example is take$$f(n) = \begin{cases} \pi k &\mbox{if } n\geq 1 \\ \frac{1}{k\sin(\pi x)} & \mbox{if } n <1\end{cases} \mbox{ for all non-zero real numbers $k$}$$

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