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.
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
$$ \begin{align} \int_{0}^{\infty} \frac{t^{p-1}}{1+t} \ dt &= \frac{- 2 \pi i e^{\pi i p} }{1-e^{2 \pi i p}} \\ &= \pi \frac{2i}{e^{\pi i p} - e^{- \pi i p}} \\ &= \frac{\pi}{\sin \pi p} . \end{align}$$
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 $\pi$.
$$\prod_{n=1}^N\bigg(1-\frac{x^2}{n^2}\bigg)=\prod_{n=1}^N\frac{n^2-x^2}{n^2}=\prod_{n=1}^N\bigg(\frac{n-x}n\cdot\frac{n+x}n\bigg)=\frac{\displaystyle\prod_{n=1}^N(n-x)}{\displaystyle\prod_{n=1}^Nn}\cdot\frac{\displaystyle\prod_{n=1}^N(n+x)}{\displaystyle\prod_{n=1}^Nn}=$$
$$=\frac{\displaystyle\prod_{n=1}^N(n-x)}{N!}\cdot\frac{\displaystyle\prod_{n=1}^N(n+x)}{N!}=\frac{(-x)!\cdot\displaystyle\prod_{n=1}^N(n-x)}{(-x)!\cdot N!}\cdot\frac{x!\cdot\displaystyle\prod_{n=1}^N(n+x)}{x!\cdot N!}=$$
$$=\frac{(N-x)!}{(-x)!\cdot N!}\cdot\frac{(N+x)!}{x!\cdot N!}=\frac1{(-x)!\cdot x!}\cdot\underbrace{\frac{(N-x)!\cdot(N+x)!}{(N!)^2}}_{\to1\text{ as }N\to\infty}$$
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
$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle #1 \right\rangle} \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace} \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left( #1 \right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ 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: $$ \Gamma\pars{z}\Gamma\pars{1 - z}\sin\pars{\pi z}= \Gamma\pars{\half}\Gamma\pars{1 - \half}\sin\pars{\pi\,\half}= \Gamma^{2}\pars{\half} = \pi $$
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.
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 $\displaystyle\int_{0}^{\infty}\frac{x^{p-1}}{1+x}dx$.
