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Question Number 140885 by mnjuly1970 last updated on 13/May/21
.......Advanced::::::::::★★★::::::::::Calculus.......findthevalueoftheinfiniteseries::Θ:=∑∞n=1(−1)n−1H2n2n−1=???.......M.N.july.1970........
Answered by mindispower last updated on 15/May/21
H2n=H2n−1+12nHn=−n∫01xn−1ln(1−x)dx...⇔Θ=∑n⩾1(−1)n−1.(−∫01x2n−2ln(1−x)dx+1(2n)(2n−1))=−∫01∑n⩾1(−x2)n−1ln(1−x)dx+∑n⩾1(−1)n−12n−1+∑n⩾1(−1)n2n=∫01−ln(1−x)1+x2dx+∑n⩾0(−1)n2n+1−∑n⩾0(−1)nn+1=π4−ln(2)−∫01ln(1−x)1+x2dx=Ax=tg(t)⇔A=−∫0π4ln(cos(t)−sin(t)cos(t))dtA=−∫0π4ln(cos(π4+t))dt−∫0π4ln(cos(t))dt=−∫0π4ln(sin(t))dt−∫0π4ln(cos)t))dt=−∫0π4ln(sin(2t)2)dt=−12∫0π2ln(sin(t))dt+π4ln(2).
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