>>874 の件

・x/(e^x-1)　= { x + x^2/2! + x^3/3! +... -(...) }/(e^x-1)
= 1- x^2*{1/2!+x/3!+...}/(e^x-1)

・∫[ε,2ε] log(x)/(1+e^x) dx
= ∫[ε,2ε] log(x)/x * x/(1+e^x) dx
= ∫[ε,2ε] log(x)/x dx - ∫[ε,2ε] log(x)*x*(1/2!+x/3!+...)/(e^x-1)
= [(1/2)log(x)^2][ε,2ε] + o(1)
= (1/2)*{(log(2)+log(ε))^2 - log(ε)^2 } + o(1)
= (1/2)log(2)^2 + log(2)log(ε) + o(1)

・∫[2ε,∞] 1/(e^x-1) dx
= ∫[2ε,∞] (1-e^{-x})'/(1-e^{-x}) dx
= [ log(1-e^{-x}) ][2ε,∞]
= -log(1-e^{-2ε}) = -log(1+e^{-ε}) -log(1-e^{-ε})
= -log(2-1+e^{-2ε}) - log(ε) - log(1 -ε/2! +ε^2/3! -...)
= -log(2) - log(ε) + o(1)

・1/(e^x+1) = 1/(e^x-1) - 2/(e^{2x}-1)

以上をまとめて
∫[ε,∞]log(x)/(1+e^x)dx
= ∫[ε,∞]log(x)/(e^x-1) dx -2∫[ε,∞]log(x)/(e^{2x}-1) dx
= ∫[ε,∞]log(x)/(e^x-1) dx -∫[2ε,∞]log(x/2)/(e^x-1) dx
= ∫[ε,2ε]log(x)/(e^x-1) dx + log(2)∫[2ε,∞] 1/(e^x-1) dx
= (1/2)log(2)^2 + log(2)log(ε) +log(2)*(-log(2) - log(ε)) + o(1)
= -(1/2)log(2)^2 + o(1)

参考
https://math.stackexchange.com/questions/2585960/evaluate-int-0-infty-frac-log-x1ex-dx
少し補足しただけでほぼそのまま頂いた.
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