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From this paper page 514-515, we have known that

$$\eta_2(s,2)=\frac{1}{\Gamma(s)}\int_{0}^\infty\frac{x^{s-1}e^{-2x}}{(1+e^{-x})^2}dx $$ , where $\displaystyle\eta_2(s,2)=\sum_{n=1}^\infty\sum_{k=1}^\infty\frac{(-1)^{n+k}}{(n+k)^s}$.

By differentiating both sides with respect to $s$, we will get

$$-\sum_{n=1}^\infty\sum_{k=1}^\infty\frac{(-1)^{n+k}\ln(n+k)}{(n+k)^s}=\frac{1}{\Gamma(s)}\int_{0}^\infty\frac{x^{s-1}e^{-2x}\ln{x}}{(1+e^{-x})^2}dx-\frac{\Gamma'(s)}{(\Gamma(s))^2}\int_{0}^\infty\frac{x^{s-1}e^{-2x}}{(1+e^{-x})^2}dx$$ Then substitute $s=2$ : $$-\sum_{n=1}^\infty\sum_{k=1}^\infty\frac{(-1)^{n+k}\ln(n+k)}{(n+k)^2}=\int_{0}^\infty\frac{xe^{-2x}\ln{x}}{(1+e^{-x})^2}dx-(1-\gamma)(\frac{\pi^2}{12}-\ln2)$$ How to evaluate $\displaystyle\int_{0}^\infty\frac{xe^{-2x}\ln{x}}{(1+e^{-x})^2}dx$ ?

Thank in advances.

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    There is a confusion, we have $\partial_{\color{red}{s}}(x^s)=(\ln x)\cdot x^s$ whereas $\partial_{\color{red}{x}}(x^s)=s\cdot x^{s-1}$. Two distinct results. – 2017-01-01
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    Ok I get it thank you – 2017-01-01
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    I have edited my question. – 2017-01-01
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    Hint. By expanding the following integrand $\frac{x^{s-1}e^{-2x}}{(1+e^{-x})}$ on gets terms like $e^{-kx}x^{s-1}$, a termwise integration gives $\frac{1}{\Gamma(s)}\int_{0}^\infty\frac{x^{s-1}e^{-2x}}{(1+e^{-x})}dx=2^{-s} \left(\zeta(s)-\zeta\left(s,\frac{3}{2}\right)\right)$ then an IBP plus $\partial_s$ gives the latter integral in your question. – 2017-01-01
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    Does IBP mean Integration by part? What you choose $u$ and $dv$ – 2017-01-01
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    Yes, integration by parts. You can take $dv=\frac{e^{-x}}{(1+e^{-x})^2}dx$, $u=x^{s-1}e^{-x}$, this leads to relate $\int_{0}^\infty\frac{x^{s-1}e^{-2x}}{(1+e^{-x})^2}dx$ to $\int_{0}^\infty\frac{x^{s-1}e^{-2x}}{(1+e^{-x})}dx$ and its closed form (with the Hurwitz zeta function) in my comment above. Isn't it? – 2017-01-01
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    Ok I get it , thanks – 2017-01-01
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    You are welcome. – 2017-01-01

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