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Question Number 127420 by liberty last updated on 29/Dec/20
∫0∞x−12x−1ln(2x−1)dx?
Answered by mnjuly1970 last updated on 29/Dec/20
solution:t=2x−1⇒t2+1=2xΩ=1ln2(2)∫0∞(ln(t2+1)−ln(2)2tln(t))2tdt(t2+1)=1ln2(2)∫0∞ln(t2+1)−ln(2)ln(t)(t2+1)dtΩ=t=1y1ln2(2)∫0∞ln(y2+1)−2ln(y)−ln(2)−ln(y)(1+y2)dy=−Ω+2ln2(2)∫0∞dy1+y22Ω=2ln2(2)(π2)⇒Ω=π2ln2(2)✓
Commented by liberty last updated on 30/Dec/20
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