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:
$$
\Gamma\pars{z}\Gamma\pars{1 - z}\sin\pars{\pi z}=
\Gamma\pars{\half}\Gamma\pars{1 - \half}\sin\pars{\pi\,\half}=
\Gamma^{2}\pars{\half} = \pi 
$$
A: 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
A: Another definiton of Gamma function is
$$\Gamma(p)=\lim_{n\rightarrow \infty}\dfrac{n!\ n^p}{p(p+1)\cdots(p+n)}$$
Then, $\Gamma(x)\Gamma(1-x)$ is:
$$\Gamma(x)\Gamma(1-x)=\lim_{n\rightarrow \infty}\dfrac{n!\ n^x}{x(x+1)\cdots(x+n)}\dfrac{n!\ n^{1-x}}{(1-x)(2-x)\cdots(1-x+n)}$$
Simplifying and reorganizing:
$$\Gamma(x)\Gamma(1-x)=\lim_{n\rightarrow \infty}\dfrac{(n!)^2 \ n}{x(x+1)(1-x)(x+2)(2-x)\cdots(n-x)(n+x)(1-x+n)}$$
$$=\lim_{n\rightarrow \infty}\dfrac{(n!)^2 \ n}{x(n-x+1)\prod\limits_{i=1}^n (i^2-x^2)}=\lim_{n\rightarrow \infty}\dfrac{(n!)^2 \ n}{x(n-x+1)\prod\limits_{i=1}^n i^2(1-\dfrac{x^2}{i^2})}$$
$$= \lim_{n\rightarrow \infty}\dfrac{(n!)^2 \ n}{x(n-x+1)(n!)^2\prod\limits_{i=1}^n (1-\dfrac{x^2}{i^2})}=\Bigg(\lim_{n\rightarrow \infty}\dfrac{n}{n-x+1}\Bigg)\Bigg(\lim_{n\rightarrow \infty}\dfrac{1}{x\prod\limits_{i=1}^n (1-\dfrac{x^2}{i^2})} \Bigg)$$
The first limit is one, so,
$$\Gamma(x)\Gamma(1-x)=\lim_{n\rightarrow \infty}\dfrac{1}{x\prod\limits_{i=1}^n (1-\dfrac{x^2}{i^2})}$$
But, remember that we can write $\sin(\pi x)$ as
$$\sin(\pi x)=\pi x \prod\limits_{i=1}^\infty(1-\dfrac{x^2}{i^2})$$
Then,
$$\Gamma(x)\Gamma(1-x)=\dfrac{\pi}{\pi x\prod\limits_{i=1}^\infty (1-\dfrac{x^2}{i^2})}=\dfrac{\pi}{\sin(\pi x)}$$
What we wanted to show.
A: Another way that could be considered, for what it's worth. Kind of a mix between Julien and RV's methods. But, I am using the same idea one commonly sees when evaluating the Gaussian integral. 
Begin with the alternate Gamma integral $\displaystyle \Gamma(x)=2\int_{0}^{\infty}t^{2x-1}e^{-t^{2}}dt$.
this can be shown by simply making the sub $u=t^{2}$ into the 'usual' integral for Gamma,  $\Gamma(x)=\displaystyle\int_{0}^{\infty}u^{x-1}e^{-u}du$
Anyway, continuing..........
Then, $\displaystyle\Gamma(n)\Gamma(1-n)=2\int_{0}^{\infty}x^{2n-1}e^{-x^{2}}dx\cdot 2\int_{0}^{\infty}y^{1-2n}e^{-y^{2}}dy$
$\displaystyle=4\int_{0}^{\infty}\int_{0}^{\infty}x^{2m-1}y^{1-2n}e^{-(x^{2}+y^{2})}dxdy$
Now, use polar coordinates: $x=r\cos\theta, \;\ y=r\sin\theta$
$\displaystyle\Gamma(n)\Gamma(1-n)=4\int_{0}^{\frac{\pi}{2}}\int_{0}^{\infty}\tan^{1-2n}\theta re^{-r^{2}}drd\theta$
integrating w.r.t r gives:
$\displaystyle\Gamma(n)\Gamma(1-n)=2\int_{0}^{\frac{\pi}{2}}\tan^{1-2n}\theta d\theta $
With a sub or two, this can now be shown to be equal to the same integral RV gave $\displaystyle\int_{0}^{\infty}\frac{x^{p-1}}{1+x}dx$.  
A: Another common integral representation of the beta function is $$B(x,y) = \int_{0}^{\infty} \frac{t^{x-1}}{(1+t)^{x+y}} \ dt .$$
So
$$\Gamma(p)\Gamma(1-p) = B(p,1-p) = \int_{0}^{\infty} \frac{t^{p-1}}{1+t} \ dt .$$
That integral can be evaluated by considering $ \displaystyle f(z) = \frac{z^{p-1}}{1+z}$ and integrating around a keyhole contour with the branch cut for $z^{p-1}$ along the positive real axis.  
Then
$$ \begin{align} \int_{0}^{\infty} \frac{t^{p-1}}{1+t} \ dt + \int^{0}_{\infty}\frac{(te^{2 \pi i})^{p-1}}{1+t} \ dt &= 2\pi i \ \text{Res} [f(z) ,e^{\pi i}] \\ &= 2 \pi i  \ (e^{\pi i })^{p-1} \\ &= - 2 \pi i e^{\pi i p} \end{align}$$
which implies
$$ \begin{align} \int_{0}^{\infty} \frac{t^{p-1}}{1+t} \ dt &= \frac{- 2 \pi i e^{\pi i p} }{1-e^{2 \pi i p}} \\ &= \pi \frac{2i}{e^{\pi i p} - e^{- \pi i p}} \\ &= \frac{\pi}{\sin \pi p} . \end{align}$$
A: I suppose you are probably already familiar with Euler's ingenious infinite product formula for the sine function, $\displaystyle\prod_{n=1}^\infty\bigg(1-\frac{x^2}{n^2}\bigg)=\frac{\sin\pi x}{\pi x}$, based on the observation that the sine function vanishes when its argument is an integer multiple of $\pi$.
$$\prod_{n=1}^N\bigg(1-\frac{x^2}{n^2}\bigg)=\prod_{n=1}^N\frac{n^2-x^2}{n^2}=\prod_{n=1}^N\bigg(\frac{n-x}n\cdot\frac{n+x}n\bigg)=\frac{\displaystyle\prod_{n=1}^N(n-x)}{\displaystyle\prod_{n=1}^Nn}\cdot\frac{\displaystyle\prod_{n=1}^N(n+x)}{\displaystyle\prod_{n=1}^Nn}=$$
$$=\frac{\displaystyle\prod_{n=1}^N(n-x)}{N!}\cdot\frac{\displaystyle\prod_{n=1}^N(n+x)}{N!}=\frac{(-x)!\cdot\displaystyle\prod_{n=1}^N(n-x)}{(-x)!\cdot N!}\cdot\frac{x!\cdot\displaystyle\prod_{n=1}^N(n+x)}{x!\cdot N!}=$$
$$=\frac{(N-x)!}{(-x)!\cdot N!}\cdot\frac{(N+x)!}{x!\cdot N!}=\frac1{(-x)!\cdot x!}\cdot\underbrace{\frac{(N-x)!\cdot(N+x)!}{(N!)^2}}_{\to1\text{ as }N\to\infty}$$
