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Question Number 198156 by Erico last updated on 12/Oct/23

Prove that   ((2t−1)/(lnt−ln(1−t)))=∫^( 1) _( 0) t^x (1−t)^(1−x) dx  and    ∫^( 1) _( 0) ((2t−1)/(lnt−ln(1−t)))dt  =  (π/2)∫^( 1) _( 0) ((x(1−x))/(sin(πx)))dx

Provethat2t1lntln(1t)=01tx(1t)1xdxand012t1lntln(1t)dt=π201x(1x)sin(πx)dx

Answered by witcher3 last updated on 12/Oct/23

∫_0 ^1 e^((1−x)ln(1−t)+xln(t)) dx  =∫_0 ^1 (1−t)e^(x(ln((t/(1−t)))) dx  =(((1−t)((t/(1−t))−1))/(ln(t)−ln(1−t)))=((2t−1)/(ln(t)−ln(1−t)))  ∫_0 ^1 ((2t−1)/(ln(t)−ln(1−t)))dt=∫_0 ^1 ∫_0 ^1 t^x (1−t)^(1−x) dxdt  =∫_0 ^1 ∫_0 ^1 t^x (1−t)^(1−x) dxdt=∫_0 ^1 ∫_0 ^1 t^x (1−t)^(1−x) dtdx  =∫_0 ^1 β(x+1,2−x)dx  =∫_0 ^1 ((x(1−x))/(Γ(3))).Γ(x)Γ(1−x)dx=(1/2)∫_0 ^1 ((x(1−x)π)/(sin(πx)))  =(π/2)∫_0 ^1 ((x(1−x))/(sin(πx)))dx=(π/2).7((ζ(3))/π^3 )=((7ζ(3))/(2π^2 ))

01e(1x)ln(1t)+xln(t)dx=01(1t)ex(ln(t1t)dx=(1t)(t1t1)ln(t)ln(1t)=2t1ln(t)ln(1t)012t1ln(t)ln(1t)dt=0101tx(1t)1xdxdt=0101tx(1t)1xdxdt=0101tx(1t)1xdtdx=01β(x+1,2x)dx=01x(1x)Γ(3).Γ(x)Γ(1x)dx=1201x(1x)πsin(πx)=π201x(1x)sin(πx)dx=π2.7ζ(3)π3=7ζ(3)2π2

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