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.

There is a long (but elementary) proof.

Claim.
$\forall s \in \mathbb{R}$ such that $\quad 0<\Re(s)<1$ then, $$\quad \Gamma(s)\times \Gamma(1-s)=\frac{\pi}{\sin (s\pi)}$$

Proof. 
$$\Gamma (s)\Gamma (1-s)=\int\limits_{0}^{+\infty }\int\limits_{0}^{+\infty
}u^{s-1}v^{-s}e^{-u}e^{-v}\text{ }dv\text{ }du=\int\limits_{0}^{+\infty }dt
\text{ }v^{-s}e^{-v}(\int\limits_{0}^{+\infty }u^{s-1}e^{-u}\text{ }du)$$
To see this use Fubini to switch the integral and using the substitution $u=\frac{u}{v}$:
\begin{eqnarray*}
\Gamma (s)\Gamma (1-s) &=&\int\limits_{0}^{+\infty }dv\text{ }
e^{-v}(\int\limits_{0}^{+\infty }u^{s-1}e^{-uv}\text{ }du)=\int
\limits_{0}^{+\infty }du\text{ }u^{s-1}e^{-uv}\int\limits_{0}^{+\infty }dv
\text{ }e^{-v}e^{-uv}) \\
&=&\int\limits_{0}^{+\infty }du\text{ }u^{s-1}(\int\limits_{0}^{+\infty }dv
\text{ }e^{-v(1+u)})=\int\limits_{0}^{+\infty }\dfrac{u^{s-1}}{1+u}\text{ }du
\\
&=&\int\limits_{0}^{1}\dfrac{u^{s-1}}{1+u}\text{ }du+\int\limits_{1}^{+
\infty }\dfrac{u^{s-1}}{1+u}\text{ }du
\end{eqnarray*}
To see this use Fubini to switch the integral and the substitution $u=\frac{u}{v}$.
Plus, using $u'=\frac{1}{u}$,
\begin{equation*}
\Gamma (s)\Gamma (1-s)=\int\limits_{0}^{1}\dfrac{u^{s-1}}{1+u}\text{ }
du+\int\limits_{0}^{1}\dfrac{u^{-s}}{1+u}\text{ }du
\end{equation*}
Denote $\quad f(s)=\int\limits_{0}^{1}\dfrac{u^{s-1}}{1+u}$ $du,$
Then,
$$\Gamma (s)\Gamma (1-s)=f(s)+f(1-s). $$
For all  $n,$ we have $\dfrac{1}{1+u}=\sum
\limits_{k=0}^{n-1}(-1)^{k}u^{k}+\dfrac{(-1)^{n}u^{n}}{1+u}$ thus
\begin{eqnarray*}
f(s) &=&\sum\limits_{k=0}^{n-1}(-1)^{k}\int\limits_{0}^{1}u^{k+s-1}\text{ }
du+(-1)^{n}\int\limits_{0}^{1}\dfrac{u^{n+s-1}}{1+u}\text{ }du \\
&=&\sum\limits_{k=0}^{n-1}\dfrac{(-1)^{k}}{k+s}+(-1)^{n}\int\limits_{0}^{1}
\dfrac{u^{n+s-1}}{1+u}\text{ }du
\end{eqnarray*}
The integral tends to $0$ when $n\rightarrow +\infty $ because
\begin{equation*}
0\leqslant \left\vert \int\limits_{0}^{1}\dfrac{u^{n+s-1}}{1+u}du\right\vert
\leqslant \int\limits_{0}^{1}u^{n+\Re(s)s-1}\text{ }du=\dfrac{1}{n+
\Re(s)}\leqslant \dfrac{1}{n}\underset{n\rightarrow +\infty }{
\rightarrow }0.
\end{equation*}
Thus,
\begin{equation*}
\forall s\in \mathbb{C}\text{ such that }0<\Re(s)<1\quad
f(s)=\sum\limits_{n=0}^{+\infty }\dfrac{(-1)^{n}}{n+s}.
\end{equation*}
$$\begin{eqnarray*}
\Gamma (s)\Gamma (1-s) &=&\sum\limits_{n=0}^{+\infty }\dfrac{(-1)^{n}}{n+s}
+\sum\limits_{n=0}^{+\infty }\dfrac{(-1)^{n}}{n+1-s}=\sum\limits_{n=0}^{+
\infty }\dfrac{(-1)^{n}}{n+s}+\sum\limits_{n=1}^{+\infty }\dfrac{(-1)^{n+1}}{
n-s}\text{ (substitution)} \\
&=&\sum\limits_{n=0}^{+\infty }\dfrac{(-1)^{n}}{n+s}+\sum\limits_{n=1}^{+
\infty }\dfrac{(-1)^{n}}{s-n}=\dfrac{1}{s}+\sum\limits_{n=1}^{+\infty }
\dfrac{2(-1)^{n}s}{s^{2}-n^{2}}\text{ (adding the two sums).}
\end{eqnarray*}$$
Now, I am using the development as Fourier serie of $cos(st)$:
$$\forall t\in ]0,1[\quad \cos (st)=\dfrac{a_{0}}{2}+\sum\limits_{n=1}^{+
\infty }a_{n}\cos nt=\dfrac{\sin \pi s}{\pi s}+\sum\limits_{n=1}^{+\infty }
\dfrac{2\sin \pi s}{\pi }\dfrac{(-1)^{n}s}{s^{2}-n^{2}}\cos nt.$$  
By choosing $t=0,$ we get $1=\dfrac{\sin \pi s}{\pi }[\dfrac{1}{s}
+\sum\limits_{n=1}^{+\infty }\dfrac{2(-1)^{n}s}{s^{2}-n^{2}}]$ Therefore,
\begin{equation*}
\forall s\in \mathbb{C}\text{ tel que }0<\Re(s)<1\quad \Gamma
(s)\Gamma (1-s)=\dfrac{1}{s}+\sum\limits_{n=1}^{+\infty }\dfrac{2(-1)^{n}s}{
s^{2}-n^{2}}=\dfrac{\pi }{\sin \pi s}.
\end{equation*}
QED
