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Question Number 33915 by prof Abdo imad last updated on 27/Apr/18

l e t Γ (x) = \int_{0}^{\infty} t^{x - 1} e^{- t} d t w i t h x > 0

1) f i n d Γ^{(n)} (x) w i t h n \in N^{★}

2) c a l c u l a t e Γ (n + \frac{3}{2}) f o r n i n t e g r .

Commented by prof Abdo imad last updated on 29/Apr/18

1) w e h a v e Γ (x) = \int_{0}^{\infty} e^{(x - 1) l n t} e^{- t} d t \Rightarrow

Γ^{^{'}} (x) = \int_{0}^{\infty} l n (t) e^{(x - 1) l n (t)} e^{- t} d t

Γ^{^{″}} (x) = \int_{0}^{\infty} {(l n t)}^{2} e^{(x - 1) l n (t)} e^{- t} d t a n d i t s e a s y

t o p r o v e b y r e c u r r e n c e t h a t

Γ^{(n)} (x) = \int_{0}^{\infty} {(l n t)}^{n} t^{x - 1} e^{- t} d t \forall n \in N (Γ^{(0)} = Γ)

Commented by prof Abdo imad last updated on 29/Apr/18

w e k n e w t h a t Γ (x + 1) = x Γ (x) \Rightarrow

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

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

= (n + \frac{1}{2}) (n - \frac{1}{2}) (n - \frac{3}{2}) Γ (n - \frac{3}{2})

= (n + \frac{1}{2}) (n - \frac{1}{2}) (n - \frac{3}{2}) \dots . (n - \frac{2 n - 1}{2}) Γ (n - \frac{2 n - 1}{2})

= (n + \frac{1}{2}) (n - \frac{1}{2}) (n - \frac{3}{2}) \dots \frac{1}{2} Γ (\frac{1}{2})

b u t

Γ (\frac{1}{2}) = \int_{0}^{\infty} t^{- \frac{1}{2}} e^{- t} d t = \int_{0}^{\infty} \frac{e^{- t}}{\sqrt{t}} d t

=_{\sqrt{t} = x} \int_{0}^{\infty} \frac{e^{- x^{2}}}{x} 2 x d c = 2 \int_{0}^{\infty} e^{- x^{2}} d x = 2 \frac{\sqrt{π}}{2} = \sqrt{π}

Γ (n + \frac{3}{2}) = \sqrt{π} (n + \frac{1}{2}) (n - \frac{1}{2}) (n - \frac{3}{2}) \dots . \frac{3}{2} \frac{1}{2}

a n d t h i s q u a n t i t y c a n b e g i v e n b y f a c t o r i e l s .

Commented by prof Abdo imad last updated on 29/Apr/18

Γ (n + \frac{3}{2}) = \frac{\sqrt{π}}{2} \prod_{k = 1}^{n} \frac{2 k + 1}{2}

= \frac{\sqrt{π}}{2} \frac{1}{2^{n}} (3.5.7 \dots . . (2 n + 1))

= \frac{\sqrt{π}}{2^{n + 1}} (2.3.4.5.6 \dots . . (2 n) (2 n + 1)) {(2.4.6 \dots . (2 n))}^{- 1}

= \frac{\sqrt{π}}{2^{n + 1}} \frac{(2 n + 1)!}{2^{n} n!} = \frac{\sqrt{π}}{2^{2 n + 1}} \frac{(2 n + 1)!}{n!} .

Answered by tanmay.chaudhury50@gmail.com last updated on 27/Apr/18

2 . ⌈ (n + 3 / 2)

= ⌈ (n + 1 + 1 / 2)

= (n + 1 / 2) ⌈ (n + 1 / 2)

= (n + 1 / 2) ⌈ (n - 1 / 2 + 1)

= (n + 1 / 2) (n - 1 / 2) ⌈ (n - 1 / 2)

= (n + 1 / 2) (n - 1 / 2) (n - 3 / 2) ⌈ (n - 3 / 2)

= (n + 1 / 2) (n - 1 / 2) (n - 3 / 2) (n - 5 / 2) ⌈ (n - 5 / 2

t h e e x p r e s s i o n c a n n o t b e i n f a c t o r i a l f o r m

s i n c e e x t r e m e r i g h t f a c t o r c a n n o t b e 1

Answered by MJS last updated on 27/Apr/18

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

2 . \frac{\sqrt{π}}{2} \times \underset{i = 1}{\prod^{n}} \frac{2 i + 1}{2}

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