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Prouver-que-a-b-a-b-a-b-0-1-x-a-1-1-x-b-1-dx-

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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
Prouverqueβ(a,b)=Γ(a)Γ(b)Γ(a+b)=∫01xa−1(1−x)b−1dx
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)))
β(a,b)=∫01xa−1(1−x)b−1dxΓ(a)=∫0∞xa−1e−xdxΓ(b)=∫0∞yb−1e−ydyΓ(a+b)=∫0∞xa+b−1e−xdxΓ(a)Γ(b)=∫0∞∫0∞xa−1.yb−1e−(x+y)dydxx=vty=v(1−t)Γ(a)Γ(b)=∫0∞va+b−1e−v∫01ta−1(1−t)b−1dtΓ(a)Γ(b)=Γ(a+b)β(a,b)β(a,b)=Γ(a)Γ(b)Γ(a+b)

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