Question

$\Gamma(x) \Gamma(1 - x) = \frac{\pi}{ \sin(\pi x) }$​

Answer 1

$\rm\pink{ \Gamma(x) \Gamma(1 - x) = \frac{\pi}{ \sin(\pi x) }}$

$\displaystyle \rm \Beta(m,n) = \int_{0}^{ \infty } \frac{ {x}^{n - 1} }{(1 + x) {}^{m + n} } dx$

$\displaystyle \rm \Beta(m,n) = \frac{ \Gamma(m)\Gamma(n) }{ \Gamma(m + n)}$

$\displaystyle \rm \displaystyle \rm \frac{ \Gamma(m)\Gamma(n) } {\Gamma (m + n)}= \int_{0}^{ \infty } \frac{ {x}^{n - 1} }{(1 + x) {}^{m + n} } dx$

m=1-n , m>0, 1-n>0

m>0, 0<n<1

$\displaystyle \rm \frac{\Gamma(1 - n)\Gamma(n)}{\Gamma(1 - n + n)} = \int_{0}^{ \infty } \frac{ {x}^{n - 1} }{(1 + x) {}^{ - 1 - n + n} } \: dx$

$\displaystyle \rm \Gamma(1 - x) \Gamma(x) = \int_{0}^{ \infty } \frac{ {x}^{n - 1} }{(1 + x) } \: dx$

$\rm\Gamma(1 - n)\Gamma(n) = \frac{\pi}{ \sin(\pi x) }$

$\rm{ \Gamma(x) \Gamma(1 - x) = \frac{\pi}{ \sin(\pi x) }}$