Gamma Euler Reflection formula proof

$$\Gamma \left({z}\right) \Gamma \left({1 – z}\right) = \frac \pi {\sin \left({\pi z}\right)}\,\,\, \, , \forall z \notin \mathbb{Z}$$

$$\textit{proof}$$

We can use the sine product formula

$$\frac{\pi}{\sin(\pi z)} = \frac{1}{z}\prod_{n=1}^\infty\left(1-\frac{z^2}{n^2}\right)^{-1}$$

Now we start by noting that

$$\Gamma(z)\Gamma(1-z) = -z\Gamma(z)\Gamma(-z) $$

Now using the Weierstrass formula we have

$$-z\Gamma(z)\Gamma(-z) = -z\cdot \frac{e^{-\gamma z}}{z} \prod_{n=1}^\infty \left(1 + \frac{z}{n}\right)^{-1} e^{z/n} \cdot \frac{e^{\gamma z}}{-z} \prod_{n=1}^\infty \left(1 – \frac{z}{n}\right)^{-1} e^{-z/n} $$

This simplifies to

$$ \frac{1}{z}\prod_{n=1}^\infty\left(1-\frac{z^2}{n^2}\right)^{-1} = \frac{\pi}{\sin(\pi z)} \,\,$$

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