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Question Number 198156 by Erico last updated on 12/Oct/23
Provethat2t−1lnt−ln(1−t)=∫01tx(1−t)1−xdxand∫012t−1lnt−ln(1−t)dt=π2∫01x(1−x)sin(πx)dx
Answered by witcher3 last updated on 12/Oct/23
∫01e(1−x)ln(1−t)+xln(t)dx=∫01(1−t)ex(ln(t1−t)dx=(1−t)(t1−t−1)ln(t)−ln(1−t)=2t−1ln(t)−ln(1−t)∫012t−1ln(t)−ln(1−t)dt=∫01∫01tx(1−t)1−xdxdt=∫01∫01tx(1−t)1−xdxdt=∫01∫01tx(1−t)1−xdtdx=∫01β(x+1,2−x)dx=∫01x(1−x)Γ(3).Γ(x)Γ(1−x)dx=12∫01x(1−x)πsin(πx)=π2∫01x(1−x)sin(πx)dx=π2.7ζ(3)π3=7ζ(3)2π2
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