find-I-n-m-0-1-x-n-1-x-m-dx-with-n-m-N-2-and-calculate-n-0-I-n-m-

Question Number 27999 by abdo imad last updated on 18/Jan/18

f i n d I_{n, m} = \int_{0}^{1} x^{n} {(1 - x)}^{m} d x w i t h

(n, m) \in N^{★^{2}} a n d c a l c u l a t e \sum_{n = 0}^{\propto} I_{n, m} .

Commented by abdo imad last updated on 22/Jan/18

d u e t o u n i f o r m c o n v e r g e n c e w e h a v e

\sum_{n = 0}^{+ \infty} I_{n, m} = \int_{0}^{1} {(1 - x)}^{m} (\sum_{n = 0}^{\propto} x^{n})

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

= {[\frac{- 1}{m} {(1 - x)}^{m}]}_{0}^{1} = \frac{1}{m} f o r m ⩾ 1 . l e t c a l c u l a t e I_{n, m}

b y p a r t s w e h a v e

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

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

I_{n, m} = \frac{n}{m + 1} I_{n - 1, m + 1} = \frac{n}{m + 1} \frac{n - 1}{m + 2} I_{n - 2, m + 2}

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

= \frac{n!}{(m + 1) (m + 2) \dots . (m + n)} I_{0, m + n} b u t

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

= \frac{1}{m + n + 1} s o

I_{n, m} = \frac{n!}{(m + 1) (m + 2) \dots (m + n + 1)}

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