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https://www.desmos.com/calculator/jpbffl8fhe

Why do we have $\frac{x}{(x!(-x)!)}$ = $\frac{\sin \left(\pi x\right)}{\pi }$ ?

With gamma functions that gives us : $\frac{x}{(\Gamma(x+1)\Gamma(-x+1))}$ = $\frac{\sin \left(\pi x\right)}{\pi }$

Prove that $\Gamma(p)\times \Gamma(1-p)=\frac{\pi}{\sin (p\pi)},\: \forall p \in (0,\: 1)$ tells us that $\left(x-1\right)!\left(-x\right)!$ = $\frac{\pi }{\sin \left(\pi x\right)}$, maybe I can make something out of it ?

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Hippalectryon
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The proof given in math.stackexchange.com/questions/714482 gives a direct answer

It proves that $$\Gamma(p)\times \Gamma(1-p)=\frac{\pi}{\sin (p\pi)},\: \forall p \in (0,\: 1)$$Which is exactly the question here knowing that $\Gamma(x)=\frac{\Gamma(x+1)}{x}$

Hippalectryon
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