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Question Number 113600 by eric last updated on 14/Sep/20
Prouver que β(a,b)=((Γ(a)Γ(b))/(Γ(a+b)))=∫_0 ^1 x^(a−1) (1−x)^(b−1) dx
$${Prouver}\:{que} \\ $$$$\beta\left({a},{b}\right)=\frac{\Gamma\left({a}\right)\Gamma\left({b}\right)}{\Gamma\left({a}+{b}\right)}=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{a}−\mathrm{1}} \left(\mathrm{1}−{x}\right)^{{b}−\mathrm{1}} {dx} \\ $$
Answered by Dwaipayan Shikari last updated on 14/Sep/20
β(a,b)=∫_0 ^1 x^(a−1) (1−x)^(b−1) dx Γ(a)=∫_0 ^∞ x^(a−1) e^(−x) dx Γ(b)=∫_0 ^∞ y^(b−1) e^(−y) dy Γ(a+b)=∫_0 ^∞ x^(a+b−1) e^(−x) dx Γ(a)Γ(b)=∫_0 ^∞ ∫_0 ^∞ x^(a−1) .y^(b−1) e^(−(x+y)) dydx x=vt y=v(1−t) Γ(a)Γ(b)=∫_0 ^∞ v^(a+b−1) e^(−v) ∫_0 ^1 t^(a−1) (1−t)^(b−1) dt Γ(a)Γ(b)=Γ(a+b)β(a,b) β(a,b)=((Γ(a)Γ(b))/(Γ(a+b)))
$$\beta\left({a},{b}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{a}−\mathrm{1}} \left(\mathrm{1}−{x}\right)^{{b}−\mathrm{1}} {dx} \\ $$$$\Gamma\left({a}\right)=\int_{\mathrm{0}} ^{\infty} {x}^{{a}−\mathrm{1}} {e}^{−{x}} {dx} \\ $$$$\Gamma\left({b}\right)=\int_{\mathrm{0}} ^{\infty} {y}^{{b}−\mathrm{1}} {e}^{−{y}} {dy} \\ $$$$\Gamma\left({a}+{b}\right)=\int_{\mathrm{0}} ^{\infty} {x}^{{a}+{b}−\mathrm{1}} {e}^{−{x}} {dx} \\ $$$$\Gamma\left({a}\right)\Gamma\left({b}\right)=\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} {x}^{{a}−\mathrm{1}} .{y}^{{b}−\mathrm{1}} {e}^{−\left({x}+{y}\right)} {dydx} \\ $$$${x}={vt} \\ $$$${y}={v}\left(\mathrm{1}−{t}\right) \\ $$$$\Gamma\left({a}\right)\Gamma\left({b}\right)=\int_{\mathrm{0}} ^{\infty} {v}^{{a}+{b}−\mathrm{1}} {e}^{−{v}} \int_{\mathrm{0}} ^{\mathrm{1}} {t}^{{a}−\mathrm{1}} \left(\mathrm{1}−{t}\right)^{{b}−\mathrm{1}} {dt} \\ $$$$\Gamma\left({a}\right)\Gamma\left({b}\right)=\Gamma\left({a}+{b}\right)\beta\left({a},{b}\right) \\ $$$$\beta\left({a},{b}\right)=\frac{\Gamma\left({a}\right)\Gamma\left({b}\right)}{\Gamma\left({a}+{b}\right)} \\ $$

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