how-to-prove-0-1-x-n-1-x-m-dx-m-n-m-n-via-Gamma-function-

Question Number 80227 by jagoll last updated on 01/Feb/20

h o w t o p r o v e

\underset{0}{\int^{1}} x^{n} {(1 - x)}^{m} d x = \frac{m! \times n!}{(m + n)!}

v i a G a m m a f u n c t i o n

Commented by john santu last updated on 01/Feb/20

B e t a f u n c t i o n

Commented by Tony Lin last updated on 01/Feb/20

\int_{0}^{1} x^{n} {(1 - x)}^{m} d x

= B (n + 1, m + 1)

= \frac{Γ (n + 1) Γ (m + 1)}{Γ (n + m + 2)}

= \frac{m! n!}{(m + n + 1)!}

Commented by mathmax by abdo last updated on 01/Feb/20

l e t A_{n, m} = \int_{0}^{1} x^{n} {(1 - x)}^{m} d x b y p s r t s u^{^{'}} = x^{n} a n d v = {(1 - x)}^{m}

A_{n, m} = {[\frac{1}{n + 1} x^{n + 1} {(1 - x)}^{m}]}_{0}^{1} - \int_{0}^{1} \frac{x^{n + 1}}{n + 1} (- m) {(1 - x)}^{m - 1} d x

= \frac{m}{n + 1} A_{n + 1, m - 1} = \frac{m (m - 1)}{(n + 1) (n + 2)} A_{n + 2, m - 2} =

= \frac{m (m - 1) (m - 2) \dots . (m - k + 1)}{(n + 1) (n + 2) \dots (n + k)} A_{n + k, m - k}

k = m \Rightarrow A_{n, m} = \frac{m!}{(n + 1) (n + 2) \dots (n + m)} A_{n + m, 0}

A_{n + m, 0} = \int_{0}^{1} x^{n + m} d x = \frac{1}{n + m + 1} \Rightarrow A_{n, m} = \frac{m!}{(n + 1) (n + 2) \dots . (n + m + 1)}

= \frac{m! \times n!}{(m + n + 1)!}

Commented by mathmax by abdo last updated on 01/Feb/20

e r r o r i n t h e Q u e s t i o n \dots

Commented by mind is power last updated on 03/Feb/20

n i c e S i r