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Question Number 63232 by mathmax by abdo last updated on 01/Jul/19

let B(x,y) =∫_0 ^1 (1−t)^(x−1) t^(y−1) dt 1) study the convergence of B(x,y) 1) prove that B(x,y)=B(y,x) prove that B(x,y) =∫_0 ^∞ (t^(x−1) /((1+t)^(x+y) )) dt 2) prove that B(x,y) =((Γ(x).Γ(y))/(Γ(x+y))) 3) prove that Γ(x).Γ(1−x) =(π/(sin(πx))) for allx ∈]0,1[

letB(x,y)=∫01(1−t)x−1ty−1dt1)studytheconvergenceofB(x,y)1)provethatB(x,y)=B(y,x)provethatB(x,y)=∫0∞tx−1(1+t)x+ydt2)provethatB(x,y)=Γ(x).Γ(y)Γ(x+y)3)provethatΓ(x).Γ(1−x)=πsin(πx)forallx∈]0,1[

Commented by mathmax by abdo last updated on 03/Jul/19

1)at V(0) (1−t)^(x−1) t^(y−1) ∼t^(y−1) ⇒ B(x,y)∼∫_0 ^1 (dt/t^(1−y) ) this integral converges ⇔1−y<1 ⇔y>0 ar V(1) (1−t)^(x−1) t^(y−1) ∼ (1−t)^(x−1) ⇒B(x,y)∼∫_0 ^1 (dx/((1−t)^(1−x) )) wich converges ⇔1−x<1 ⇔x>0 so B(x,y) converges ⇔ x>0 and y>0 2)we have B(x,y) =∫_0 ^1 (1−t)^(x−1) t^(y−1) dt =_(1−t=u) ∫_1 ^o u^(x−1) (1−u)^(y−1) (−du) =∫_0 ^1 (1−u)^(y−1) u^(x−1) du =B(y,x) we say that B(x,y) is symetric

1)atV(0)(1−t)x−1ty−1∼ty−1⇒B(x,y)∼∫01dtt1−ythisintegralconverges⇔1−y<1⇔y>0arV(1)(1−t)x−1ty−1∼(1−t)x−1⇒B(x,y)∼∫01dx(1−t)1−xwichconverges⇔1−x<1⇔x>0soB(x,y)converges⇔x>0andy>02)wehaveB(x,y)=∫01(1−t)x−1ty−1dt=1−t=u∫1oux−1(1−u)y−1(−du)=∫01(1−u)y−1ux−1du=B(y,x)wesaythatB(x,y)issymetric

Commented by mathmax by abdo last updated on 04/Jul/19

we have B(x,y) =∫_0 ^1 (1−t)^(x−1) t^(y−1) dt let use the changement t =(1/(u+1)) ⇒ B(x,y) =−∫_0 ^∞ (1−(1/(u+1)))^(x−1) (1/((u+1)^(y−1) )) ((−du)/((u+1)^2 )) = ∫_0 ^∞ (u^(x−1) /((1+u)^(x−1) )) (1/((1+u)^(y+1) )) du =∫_0 ^∞ (u^(x−1) /((1+u)^(x+y) )) du so the result is proved .

wehaveB(x,y)=∫01(1−t)x−1ty−1dtletusethechangementt=1u+1⇒B(x,y)=−∫0∞(1−1u+1)x−11(u+1)y−1−du(u+1)2=∫0∞ux−1(1+u)x−11(1+u)y+1du=∫0∞ux−1(1+u)x+ydusotheresultisproved.

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