Integral $\int_0^{\infty}\frac{\ln x}{e^x+1}\,dx$ I know such integral: $\int_0^{\infty}\frac{\ln x}{e^x}\,dx=-\gamma$. What about the integral $\int_0^{\infty}\frac{\ln x}{e^x+1}\,dx$? 
The answer seems very nice: $-\frac{1}{2}{\ln}^22$ but how it could be calculated? I tried integration by parts but the limit $\displaystyle{\lim_{x\to 0}\ln x\ln(1+e^{-x})}$ doesn't exist. Or I can also write the following equality 
$$\int_0^{\infty}\frac{\ln x}{e^x+1}\,dx=\lim\limits_{t\to 0}\frac{d}{dt}\left(\int_0^{\infty}\frac{x^t}{e^x+1}\, dx\right)$$
but I don't know what to do next.  
 A: Since : $$ \left(\forall x\in\mathbb{R}_{+}^{*}\right),\ \frac{1}{\mathrm{e}^{x}+1}=\sum_{n=1}^{+\infty}{\left(-1\right)^{n-1}\mathrm{e}^{-nx}} $$
And, using the result you gave, we have : \begin{aligned} \int_{0}^{+\infty}{\mathrm{e}^{-nx}\ln{x}\,\mathrm{d}x}=\frac{1}{n}\int_{0}^{+\infty}{\mathrm{e}^{-x}\ln{\left(\frac{x}{n}\right)}\,\mathrm{d}x}&=\frac{1}{n}\int_{0}^{+\infty}{\mathrm{e}^{-x}\ln{x}\,\mathrm{d}x}-\frac{\ln{n}}{n}\int_{0}^{+\infty}{\mathrm{e}^{-x}\,\mathrm{d}x}\\&=-\frac{\gamma +\ln{n}}{n} \end{aligned}
Thus : \begin{aligned}\int_{0}^{+\infty}{\frac{\ln{x}}{\mathrm{e}^{x}+1}\,\mathrm{d}x}&=\gamma\sum_{n=1}^{+\infty}{\frac{\left(-1\right)^{n}}{n}}+\sum_{n=1}^{+\infty}{\left(-1\right)^{n}\frac{\ln{n}}{n}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(*\right)\end{aligned}
Since $ x\mapsto\frac{\ln{x}}{x} $ is positive and decreasing on $ \left]1,+\infty\right[ $, $ \left(\sum\limits_{k=1}^{n}{f\left(k\right)}-\int\limits_{1}^{n}{f\left(x\right)\mathrm{d}x}\right)_{n} $ converges to some constant $ \ell $, thus : \begin{aligned} \sum_{k=1}^{n}{\frac{\ln{k}}{k}}&=\int_{1}^{n}{\frac{\ln{x}}{x}\,\mathrm{d}x}+\ell+\underset{\overset{n\to +\infty}{}}{\mathcal{o}}\left(1\right)=\frac{\ln^{2}{n}}{2}+\ell+\underset{\overset{n\to +\infty}{}}{\mathcal{o}}\left(1\right) \end{aligned}
Now let $ n $ be a positive integer, we have :\begin{aligned} \sum_{k=1}^{2n}{\left(-1\right)^{k}\frac{\ln{k}}{k}}&=\sum_{k=1}^{n}{\frac{\ln{\left(2k\right)}}{2k}}-\sum_{k=0}^{n-1}{\frac{\ln{\left(2k+1\right)}}{2k+1}}\\&=\sum_{k=1}^{n}{\frac{\ln{\left(2k\right)}}{k}}-\sum_{k=1}^{2n}{\frac{\ln{k}}{k}}\\ &=\ln{2}\sum_{k=1}^{n}{\frac{1}{k}}+\sum_{k=1}^{n}{\frac{\ln{k}}{k}}-\sum_{k=1}^{2n}{\frac{\ln{k}}{k}}\\ &=\ln{2}\left(\ln{n}+\gamma +\underset{\overset{n\to +\infty}{}}{\mathcal{o}}\left(1\right)\right)+\left(\frac{\ln^{2}{n}}{2}+\ell+\underset{\overset{n\to +\infty}{}}{\mathcal{o}}\left(1\right)\right)-\left(\frac{\ln^{2}{\left(2n\right)}}{2}+\ell+\underset{\overset{n\to +\infty}{}}{\mathcal{o}}\left(1\right)\right)\\ \sum_{k=1}^{2n}{\left(-1\right)^{k}\frac{\ln{k}}{k}}&=-\frac{\ln^{2}{2}}{2}+\gamma\ln{2}+\underset{\overset{n\to +\infty}{}}{\mathcal{o}}\left(1\right) \end{aligned}
Meaning : $$ \sum_{n=1}^{+\infty}{\left(-1\right)^{n}\frac{\ln{n}}{n}}=-\frac{\ln^{2}{2}}{2}+\gamma\ln{2} $$
Hence, $ \left(*\right) $ becames : $$ \int_{0}^{+\infty}{\frac{\ln{x}}{\mathrm{e}^{x}+1}\,\mathrm{d}x}=-\gamma\ln{2}-\frac{\ln^{2}{2}}{2}+\gamma\ln{2}=-\frac{\ln^{2}{2}}{2} $$
A: Integrating by parts can obtain
\begin{align*}
I& :=\int_0^{\infty}\frac{\ln x}{e^x+1}\mathrm{d}x=\int_0^1+\int_1^{\infty}\\
& =\int_0^1\ln\left(\frac{1+e^{-x}}{2}\right)\frac{\mathrm{d}x}{x}\\
& \ \ \ +\int_1^{\infty}\ln\left(1+e^{-x}\right)\frac{\mathrm{d}x}{x}\\
& =\int_0^1\Bigl.\ln\left(\frac{1-e^{-xy}}{y}\right)\Bigr|_{y=1}^{y=2}\frac{\mathrm{d}x}{x}\\
& \ \ \ +\int_1^{\infty}\Bigl.\ln\left(1-e^{-xy}\right)\Bigr|_{y=1}^{y=2}\frac{\mathrm{d}x}{x}\\
& =\iint_{[0,1]\times [1,2]}\left(\frac{1}{e^{xy}-1}-\frac{1}{xy}\right)\mathrm{d}x\mathrm{d}y\\
& \ \ \ +\iint_{[1,\infty)\times [1,2]}\frac{\mathrm{d}x\mathrm{d}y}{e^{xy}-1}\\
& =\int_1^2\Bigl.\ln\left(\frac{1-e^{-xy}}{x}\right)\Bigr|_{x=0}^{x=1}\frac{\mathrm{d}y}{y}\\
& \ \ \ +\int_1^2\Bigl.\ln\left(1-e^{-xy}\right)\Bigr|_{x=1}^{x=\infty}\frac{\mathrm{d}y}{y}\\
& =-\int_1^2\frac{\ln y}{y}\mathrm{d}y=-\frac{\ln^2 2}{2}.
\end{align*}
