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Question Number 1643 by 112358 last updated on 28/Aug/15

C a l c u l a t e

I (a, b) = \int_{0}^{1} t^{a} {(1 - t)}^{b} d t

g i v e n t h a t I (a, b) = \frac{b}{a + 1} I (a + 1, b - 1)

(a > 0, b > 0) . U s e t h e f a c t t h a t

I (a, b) = I (a + 1, b) + I (a, b + 1)

a n d I (a, b) = I (b, a)

t o h e l p e v a l u a t e I (a, b) .

Answered by 123456 last updated on 29/Aug/15

I (a, b) = \underset{0}{\int^{1}} t^{a} {(1 - t)}^{b} d t = \underset{0}{\int^{1}} t^{a + 1 - 1} {(1 - t)}^{b + 1 - 1} d t

I (a, b) = B (a + 1, b + 1) = \frac{Γ (a + 1) Γ (b + 1)}{Γ (a + b + 2)} = \frac{a! b!}{(a + b + 1)!}

- - - - - - - - - - - - -

I (a + 1, b) + I (a, b + 1) = \frac{Γ (a + 2) Γ (b + 1)}{Γ (a + b + 3)} + \frac{Γ (a + 1) Γ (b + 2)}{Γ (a + b + 3)}

= \frac{Γ (a + 2) Γ (b + 1) + Γ (a + 1) Γ (b + 2)}{Γ (a + b + 3)}

= \frac{Γ (a + 1 + 1) Γ (b + 1) + Γ (a + 1) Γ (b + 1 + 1)}{Γ (a + b + 2 + 1)}

= \frac{(a + 1) Γ (a + 1) Γ (b + 1) + (b + 1) Γ (a + 1) Γ (b + 1)}{(a + b + 2) Γ (a + b + 2)}

= \frac{(a + b + 2) Γ (a + 1) Γ (b + 1)}{(a + b + 2) Γ (a + b + 2)}

= \frac{Γ (a + 1) Γ (b + 1)}{Γ (a + b + 2)}

= I (a, b)

\frac{b}{a + 1} I (a + 1, b - 1) = \frac{b}{a + 1} \cdot \frac{Γ (a + 2) Γ (b)}{Γ (a + b + 2)}

= \frac{Γ (a + 1 + 1) b Γ (b)}{(a + 1) Γ (a + b + 2)}

= \frac{(a + 1) Γ (a + 1) Γ (b + 1)}{(a + 1) Γ (a + b + 2)}

= \frac{Γ (a + 1) Γ (b + 1)}{Γ (a + b + 2)}

= I (a, b)

Commented by 123456 last updated on 29/Aug/15

B (x, y) = \underset{0}{\int^{1}} t^{x - 1} {(1 - t)}^{y - 1} d t

this is the beta function

Γ (x) = \underset{0}{\int^{+ \infty}} t^{x - 1} e^{- t} d t

this is the gamma function

there a relation betwen these two function

B (x, y) = \frac{Γ (x) Γ (y)}{Γ (x + y)}

Γ (x + 1) = x Γ (x)

also there a relation betwen gamma function and factorial

Γ (x + 1) = x!

Commented by 112358 last updated on 29/Aug/15

{T h a t}^{'} s c o o l s t u f f . T h a n k s f o r t h e

i n f o r m a t i o n .

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