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Question Number 166180 by mnjuly1970 last updated on 15/Feb/22
provethatϕ=∫01ln2(1−x)x2dx=2ζ(2)−−−proof−−−ϕ=[−1xln2(1−x)]01−∫012ln(1−x)x(1−x)dx=−limξ→1−1ξln2(1−ξ)−2{∫01ln(1−x)1−xdx+∫01ln(1−x)xdx}=−limξ→1−{1ξln2(1−ξ)+ln2(1−ξ)}+2ζ(2)=limξ→1−(ξ−1ξ)ln2(1−ξ)+2ζ(2)=1−ξ=δξ→1−,δ→0+[limδ→0+(−δ1−δ)ln2(δ)=0]+2ζ(2)◼m.n∴ϕ=2ζ(2)
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