Given the function Γ defined by Γ(x)=∫_0 ^(+∞) t^(x−1) e^(−t) dt 1. What is the domain of definition of Γ ? 2. Show that ∀x∈ DΓ, xΓ(x)=Γ(x+1) and deduce the value of Γ(n), n∈N^∗ 3. Assuming ∫_0 ^(+∞) e^(−u^2 ) =((√π)/2), calculate Γ((1/2)) and deduce that Γ(n+(1/2))=(((2n)!(√π))/(2^2^n n!))


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Question Number 108283 by Ar Brandon last updated on 16/Aug/20

$Given the function Γ defined by Γ (x) = \int_{0}^{+ \infty} t^{x - 1} e^{- t} dt$ $1 . What is the domain of definition of Γ ?$ $2 . Show that \forall x \in D Γ, x Γ (x) = Γ (x + 1) and deduce the value of Γ (n), n \in N^{*}$ $3 . Assuming \int_{0}^{+ \infty} e^{- u^{2}} = \frac{\sqrt{π}}{2}, calculate Γ (\frac{1}{2}) and deduce that$ $Γ (n + \frac{1}{2}) = \frac{(2 n)! \sqrt{π}}{2^{2^{n}} n!}$
	Answered by mathmax by abdo last updated on 16/Aug/20

	$1) Γ (x) = \int_{0}^{\infty} t^{x - 1} e^{- t} dt = \int_{0}^{1} t^{x - 1} e^{- t} dt + \int_{1}^{+ \infty} t^{x - 1} e^{- t} dt$ $at v (0) t^{x - 1} e^{- t} \sim t^{x - 1} and \int_{0}^{1} t^{x - 1} dt = \int_{0}^{1} \frac{dt}{t^{1 - x}} conv \Leftrightarrow 1 - x < 1 \Rightarrow x > 0$ $at + \infty we have \lim_{t \to + \infty} t^{2} t^{x - 1} e^{- t} = \lim_{t \to 0} t^{x + 1} e^{- t} = 0 \Rightarrow$ $\int_{1}^{\infty} t^{x - 1} e^{- t} dt converge for x > 0 \Rightarrow D_{Γ} =] 0, + \infty [$ $2) Γ (x + 1) = \int_{0}^{\infty} t^{x} e^{- t} dt =_{bypsrts} {[- t^{x} e^{- t}]}_{t = 0}^{\infty} + \int_{0}^{\infty} {xt}^{x - 1} e^{- t} dt$ $= x \int_{0}^{\infty} t^{x - 1} e^{- t} dt = x Γ (x) \Rightarrow for n natural$ $Γ (n + 1) = n Γ (n - 1) = n (n - 1) Γ (n - 1) = . . . = n! Γ (1) = n!$ $generaly Γ (x + n) = (x + n - 1) (x + n - 2) . . . . (x + 1) x Γ (x)$ $3) Γ (\frac{1}{2}) = \int_{0}^{\infty} t^{- \frac{1}{2}} e^{- t} dt = \int_{0}^{\infty} \frac{e^{- t}}{\sqrt{t}} dt =_{\sqrt{t} = u} \int_{0}^{\infty} \frac{e^{- u^{2}}}{u} (2 u) du$ $= 2 \int_{0}^{\infty} e^{- u^{2}} du = 2 . \frac{\sqrt{π}}{2} = \sqrt{π} \Rightarrow Γ (\frac{1}{2}) = \sqrt{π}$ $4) Γ (\frac{1}{2} + n) = (\frac{1}{2} + n - 1) (\frac{1}{2} + n - 2) . . . . . (\frac{1}{2} + 1) \frac{1}{2} Γ (\frac{1}{2})$ $= \frac{2 n - 1}{2} . \frac{2 n - 3}{2} . . . . . . . \frac{3}{2} . \frac{1}{2} \sqrt{π}$ $= \frac{1.3.5 . . . . . (2 n - 3) (2 n - 1)}{2^{n}} \sqrt{π}$ $= \frac{1.2.3.4.5 . . . . . (2 n - 1) (2 n)}{2^{n} (2.4.6 . . . . . . (2 n))} \sqrt{π} = \frac{(2 n)!}{2^{2 n} n!} \sqrt{π}$
	Commented by Ar Brandon last updated on 16/Aug/20
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