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B-a-b-0-1-x-a-1-1-x-b-1-dx-s-0-t-s-1-e-t-dt-Why-B-a-b-a-b-a-b-

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Question Number 173976 by savitar last updated on 22/Jul/22
B(a,b)=∫_0 ^1 x^(a−1) (1−x)^(b−1) dx Γ(s)= ∫_0 ^∞ t^(s−1) e^(−t) dt Why B(a,b)= ((Γ(a)Γ(b))/(Γ(a+b))) ?
B(a,b)=∫01xa−1(1−x)b−1dxΓ(s)=∫0∞ts−1e−tdtWhyB(a,b)=Γ(a)Γ(b)Γ(a+b)?
Answered by aleks041103 last updated on 22/Jul/22
If f(x)=x^(a−1) and g(x)=x^(b−1) also h(t)=(f∗g)(t)=∫_0 ^( t) f(x)g(t−x)dx is the convolution ⇒h(t)=∫_0 ^( t) x^(a−1) (t−x)^(b−1) dx x=ty⇒dx=t dy ⇒h(t)=∫_0 ^( 1) t^(a−1) y^(a−1) t^(b−1) (1−y)^(b−1) tdy= =t^(a+b−1) ∫_0 ^( 1) t^(a−1) (1−y)^(b−1) dy= =B(a,b)t^(a+b−1) for convolution: H(s)=F(s)G(s) where F,G,H are the laplace transforms of f,g,h respectively. Since L{x^p }(s)=((Γ(p+1))/s^(p+1) ), then H(s)=B(a,b)((Γ(a+b))/s^(a+b) ) F(s)=((Γ(a))/s^a ) and G(s)=((Γ(b))/s^b ) ⇒((Γ(a)Γ(b))/s^(a+b) )=B(a,b)((Γ(a+b))/s^(a+b) ) ⇒B(a,b)=((Γ(a)Γ(b))/(Γ(a+b)))
Iff(x)=xa−1andg(x)=xb−1alsoh(t)=(f∗g)(t)=∫0tf(x)g(t−x)dxistheconvolution⇒h(t)=∫0txa−1(t−x)b−1dxx=ty⇒dx=tdy⇒h(t)=∫01ta−1ya−1tb−1(1−y)b−1tdy==ta+b−1∫01ta−1(1−y)b−1dy==B(a,b)ta+b−1forconvolution:H(s)=F(s)G(s)whereF,G,Harethelaplacetransformsoff,g,hrespectively.SinceL{xp}(s)=Γ(p+1)sp+1,thenH(s)=B(a,b)Γ(a+b)sa+bF(s)=Γ(a)saandG(s)=Γ(b)sb⇒Γ(a)Γ(b)sa+b=B(a,b)Γ(a+b)sa+b⇒B(a,b)=Γ(a)Γ(b)Γ(a+b)
Commented by savitar last updated on 22/Jul/22
Very nice sir
Verynicesir

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