Prove that $\Gamma(p)\times \Gamma(1-p)=\frac{\pi}{\sin (p\pi)},\: \forall p \in (0,\: 1)$

Solution 1:

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

$$ \begin{align} \int_{0}^{\infty} \frac{t^{p-1}}{1+t} \ dt &= \frac{- 2 \pi i e^{\pi i p} }{1-e^{2 \pi i p}} \\ &= \pi \frac{2i}{e^{\pi i p} - e^{- \pi i p}} \\ &= \frac{\pi}{\sin \pi p} . \end{align}$$

Solution 2:

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 $\pi$.

$$\prod_{n=1}^N\bigg(1-\frac{x^2}{n^2}\bigg)=\prod_{n=1}^N\frac{n^2-x^2}{n^2}=\prod_{n=1}^N\bigg(\frac{n-x}n\cdot\frac{n+x}n\bigg)=\frac{\displaystyle\prod_{n=1}^N(n-x)}{\displaystyle\prod_{n=1}^Nn}\cdot\frac{\displaystyle\prod_{n=1}^N(n+x)}{\displaystyle\prod_{n=1}^Nn}=$$

$$=\frac{\displaystyle\prod_{n=1}^N(n-x)}{N!}\cdot\frac{\displaystyle\prod_{n=1}^N(n+x)}{N!}=\frac{(-x)!\cdot\displaystyle\prod_{n=1}^N(n-x)}{(-x)!\cdot N!}\cdot\frac{x!\cdot\displaystyle\prod_{n=1}^N(n+x)}{x!\cdot N!}=$$

$$=\frac{(N-x)!}{(-x)!\cdot N!}\cdot\frac{(N+x)!}{x!\cdot N!}=\frac1{(-x)!\cdot x!}\cdot\underbrace{\frac{(N-x)!\cdot(N+x)!}{(N!)^2}}_{\to1\text{ as }N\to\infty}$$

Solution 3:

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle #1 \right\rangle} \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace} \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left( #1 \right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ 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: $$ \Gamma\pars{z}\Gamma\pars{1 - z}\sin\pars{\pi z}= \Gamma\pars{\half}\Gamma\pars{1 - \half}\sin\pars{\pi\,\half}= \Gamma^{2}\pars{\half} = \pi $$