calculate Γ(n+(1/2)) with n ∈N.


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Question Number 34719 by abdo mathsup 649 cc last updated on 10/May/18

$c a l c u l a t e Γ (n + \frac{1}{2}) w i t h n \in N .$
Commented by abdo mathsup 649 cc last updated on 12/May/18

$w e h a v e Γ (x) = \int_{0}^{\infty} t^{x - 1} e^{- t} d t \Rightarrow$ $Γ (n + \frac{1}{2}) = \int_{0}^{\infty} t^{n - \frac{1}{2}} e^{- t} d t$ $=_{t = u^{2}} \int_{0}^{\infty} {(u^{2})}^{n - \frac{1}{2}} e^{- u^{2}} 2 u d u$ $= 2 \int_{0}^{\infty} u^{2 n} e^{- u^{2}} d u l e t p u t$ $A_{n} = \int_{0}^{\infty} u^{2 n} e^{- u^{2}} d u b y p a r t s α^{^{'}} = u^{2 n} a n d β = e^{- u^{2}}$ $A_{n} = {[\frac{1}{2 n + 1} u^{2 n + 1} e^{- u^{2}}]}_{0}^{+ \infty} - \int_{0}^{\infty} \frac{1}{2 n + 1} u^{2 n + 1} (- 2 u) e^{- u^{2}} d u$ $= \frac{2}{2 n + 1} \int_{0}^{\infty} u^{2 n + 2} e^{- u^{2}} d u = \frac{2}{2 n + 1} A_{n + 1}$ $\Rightarrow A_{n + 1} = \frac{2 n + 1}{2} A_{n}$ $\prod_{k = 0}^{n - 1} A_{k + 1} = \frac{\prod_{k = 0}^{n - 1} (2 k + 1)}{2^{n}} \prod_{k = 0}^{n - 1} A_{n} \Rightarrow$ $A_{n} = \frac{\prod_{k = 0}^{n - 1} (2 k + 1)}{2^{n}} A_{0} b u t A_{0} = \int_{0}^{\infty} e^{- u^{2}} d u = \frac{\sqrt{π}}{2}$ $\Rightarrow A_{n} = \frac{\sqrt{π}}{2^{n + 1}} (1.3.5 . . . . (2 n - 1))$ $= \frac{\sqrt{π}}{2^{n + 1}} \frac{1.2.3.4 . . . . . (2 n)}{2^{n} n!} = \frac{\sqrt{π} (2 n)!}{2^{2 n + 1} (n!)}$
Commented by abdo mathsup 649 cc last updated on 12/May/18

$Γ (n + \frac{1}{2}) = 2 A_{n} \Rightarrow Γ (n + \frac{1}{2}) = \frac{\sqrt{π}}{2^{2 n}} \frac{(2 n)!}{n!} .$
	Answered by alex041103 last updated on 10/May/18

	$B y d e f i n i t i o n Γ (k) = (k - 1)! = (k - 1) (k - 2)! .$ $T h e v a l u e Γ (n + \frac{1}{2}) = (n - \frac{1}{2})!$ $\Rightarrow (n - \frac{1}{2})! = (n - \frac{1}{2}) (n - 1 - \frac{1}{2})! = . . . =$ $= (\frac{1}{2})! \underset{k = 2}{\prod^{n}} \frac{2 k - 1}{2} = \frac{\sqrt{π}}{2} \frac{1}{2^{n - 1}} \underset{k = 1}{\prod^{n - 1}} (2 k + 1)$ $\Rightarrow Γ (n + \frac{1}{2}) = \frac{\sqrt{π}}{2^{n}} \underset{k = 1}{\prod^{n - 1}} (2 k + 1)$
	Commented by NECx last updated on 10/May/18

	$w e l c o m e b a c k m r A l e x$
	Commented by alex041103 last updated on 10/May/18

	$I^{'} m s o r r y . I m a d e a m i s t a k e .$ $T h e a n s w e r s h o u l d b e t h i s :$ $\frac{\sqrt{π}}{2^{n}} \underset{k = 1}{\prod^{n - 1}} (2 k + 1)$
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