Prove that $\Gamma(p) \cdot \Gamma(1-p)=\frac{\pi}{\sin (p\pi)}$ for $p \in (0,\: 1)$ Prove that $$\Gamma(p)\cdot \Gamma(1-p)=\frac{\pi}{\sin (p\pi)},\: \forall  p \in (0,\: 1),$$
where $$\Gamma (p)=\int_{0}^{\infty} x^{p-1} e^{-x}dx.$$
Here's what I tried:
We have
$$B(p, q)=\int_{0}^{1} x^{p-1} (1-x)^{q-1}dx=\frac{\Gamma(p)\cdot \Gamma(q)}{\Gamma(p+q)}$$
Hence
$$B(p, 1-p)=\frac{\Gamma(p)\cdot \Gamma(1-p)}{\Gamma(1)}=\Gamma(p)\cdot \Gamma(1-p)=\int_{0}^{1} x^{p-1} (1-x)^{-p}dx$$
But from here I don't know how to proceed.

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The usual argument is as follows:
$\ds{\Gamma\pars{z}\Gamma\pars{1 - z}\sin\pars{\pi z}}$ is analytical in
${\mathbb C}$ and is bounded. Then, it is a constant. Setting $\ds{z =\half}$ we can discover that constant value:
$$
\Gamma\pars{z}\Gamma\pars{1 - z}\sin\pars{\pi z}=
\Gamma\pars{\half}\Gamma\pars{1 - \half}\sin\pars{\pi\,\half}=
\Gamma^{2}\pars{\half} = \pi 
$$
