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Question Number 162099 by mnjuly1970 last updated on 26/Dec/21
PROVETHATΩ=∫01Li2(x).ln(x)xdx=?−π490−−−−−−−−−−Ω=∫01ln(x)∑∞n=1xn−1n2dx=∑∞n=11n2∫01xn−1.ln(x)dx=∑∞n=11n2{[xnnln(x)]01−1n∫01xn−1dx}=∑∞n=1−1n4=−ζ(4)=−π490◼m.n−−−M.N−−−
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