$\int_0^{\infty } \frac{\log (x)}{e^x+1} \, dx = -\frac{1}{2} \log ^2(2)$ How to show? $$
\int_0^{\infty } \frac{\log (x)}{e^x+1} \, dx = -\frac{1}{2} \log ^2(2)
$$
Anyone an idea on how to prove this?
 A: By the recursive relation $\Gamma(x+1)=x\Gamma(x)$, we get
$$
\small{\log(\Gamma(x))=\log(\Gamma(n+x))-\log(x)-\log(x+1)-\log(x+2)-\dots-\log(x+n-1)}\tag{1}
$$
Differentiating $(1)$ with respect to $x$, evaluating at $x=1$, and letting $n\to\infty$ yields
$$
\begin{align}
\frac{\Gamma'(1)}{\Gamma(1)}&=\log(n)+O\left(\frac1n\right)-\frac11-\frac12-\frac13-\dots-\frac1n\\
&\to-\gamma\tag{2}
\end{align}
$$
Next, apply $(2)$ to the following:
$$
\begin{align}
\int_0^\infty\log(t)\;e^{-t}\;\mathrm{d}t
&=\left.\frac{\mathrm{d}}{\mathrm{d}x}\int_0^\infty t^x\;e^{-t}\;\mathrm{d}t\right]_{x=0}\\
&=\Gamma'(1)\\
&=-\gamma\tag{3}
\end{align}
$$
Then, a simple change of variables yields
$$
\int_0^\infty\log(t)\;e^{-nt}\;\mathrm{d}t=-\frac{\gamma+\log(n)}{n}\tag{4}
$$
Since $\dfrac{1}{e^t+1}=e^{-t}-e^{-2t}+e^{-3t}-e^{-4t}+\dots$, by applying $(4)$ to this result, we have that
$$
\begin{align}
\int_0^\infty\frac{\log(t)}{e^t+1}\mathrm{d}t
&=\int_0^\infty\sum_{n=1}^\infty(-1)^{n-1}\log(t)\;e^{-nt}\;\mathrm{d}t\\
&=\sum_{n=1}^\infty(-1)^n\frac{\gamma+\log(n)}{n}\\
&=-\frac12\log(2)^2\tag{5}
\end{align}
$$

More about $\mathbf{(2)}$:
The fact that $\frac{\mathrm{d}}{\mathrm{d}x}\log(\Gamma(x))=\log(x)+O\left(\frac1x\right)$ relies on the log-convexity of $\Gamma(x)$; that is, $\frac{\mathrm{d}}{\mathrm{d}x}\log(\Gamma(x))$ is monotonically increasing.  By the recursive relation for $\Gamma(x)$, we have that
$$
\log(\Gamma(x))-\log(\Gamma(x-1))=\log(x-1)\tag{6}
$$
and that
$$
\log(\Gamma(x+1))-\log(\Gamma(x))=\log(x)\tag{7}
$$
The Mean Value Theorem and $(6)$ imply that $\frac{\mathrm{d}}{\mathrm{d}x}\log(\Gamma(\xi_1))=\log(x{-}1)$ for some $\xi_1{\in}(x{-}1,x)$.
The Mean Value Theorem and $(7)$ imply that $\frac{\mathrm{d}}{\mathrm{d}x}\log(\Gamma(\xi_2))=\log(x)$ for some $\xi_2{\in}(x,x{+}1)$.
By the monotonicity of $\frac{\mathrm{d}}{\mathrm{d}x}\log(\Gamma(x))$, we get that
$$
\log(x-1)\le\frac{\mathrm{d}}{\mathrm{d}x}\log(\Gamma(x))\le\log(x)\tag{8}
$$
Since $\log(x)-\log(x-1)=O\left(\frac1x\right)$, $(8)$ implies that
$$
\frac{\mathrm{d}}{\mathrm{d}x}\log(\Gamma(x))=\log(x)+O\left(\frac1x\right)\tag{9}
$$

Log-Convexity of $\mathbf{\Gamma(x)}$:
If $\frac{\mathrm{d}^2}{\mathrm{d}x^2}f(x)\ge0$, then $f$ is convex at $x$. Thus, if $\dfrac{f(x)f''(x)-f'(x)^2}{f(x)^2}=\frac{\mathrm{d}^2}{\mathrm{d}x^2}\log(f(x))\ge0$, then $f$ is log-convex. So we need to show that $\Gamma(x)\Gamma''(x)\ge\Gamma'(x)^2$.  That is,
$$
\int_0^\infty t^{x-1}\;e^{-t}\;\mathrm{d}t \int_0^\infty\log(t)^2\;t^{x-1}\;e^{-t}\;\mathrm{d}t \ge \left(\int_0^\infty\log(t)\;t^{x-1}\;e^{-t}\;\mathrm{d}t\right)^2\tag{10}
$$
Dividing both sides of $(10)$ by $\int_0^\infty t^{x-1}\;e^{-t}\;\mathrm{d}t$, $(10)$ becomes
$$
\int\log(t)^2\;\mathrm{d}\mu \ge \left(\int\log(t)\;\mathrm{d}\mu\right)^2\tag{11}
$$
where $\mathrm{d}\mu=\dfrac{t^{x-1}\;e^{-t}\;\mathrm{d}t}{\int_0^\infty t^{x-1}\;e^{-t}\;\mathrm{d}t}$ is a unit measure on $[0,\infty)$. Thus, $(11)$ is simply Jensen's inequality.
Strictly speaking:
Note that
$$
\log(t)^2 + a^2 \ge 2a\log(t)\tag{12}
$$
with equality if and only if $\log(t)=a$. Integrating $(12)$ w.r.t. the unit measure $\mathrm{d}\mu$, yields
$$
\int\log(t)^2\;\mathrm{d}\mu + a^2 \ge 2a\int\log(t)\;\mathrm{d}\mu\tag{13}
$$
with equality in $(13)$ if and only if $\log(t)=a$ a.e. $\mathrm{d}\mu$.  Let $a=\int\log(t)\;\mathrm{d}\mu$, then $(13)$ becomes
$$
\int\log(t)^2\;\mathrm{d}\mu \ge \left(\int\log(t)\;\mathrm{d}\mu\right)^2\tag{14}
$$
with equality if and only if $\log(t)$ is constant a.e. $\mathrm{d}\mu$. Since the $\mathrm{d}\mu$ in $(11)$ is absolutely continuous and $\log(t)$ is strictly increasing on $(0,\infty)$, the inequality in $(11)$ is strict. Therefore, $\Gamma$ is strictly log-convex.
A: Start with $J(s)$ given by
$$ J(s) = \int_0^\infty \frac{x^s}{1+e^x}dx. $$
Expand the denominator using geometric series, like so:
$$ J(s) = \sum_{k\geq0}\int_0^\infty (-1)^k x^s e^{-(1+k)x}dx$$
$$ = \sum_{k\geq1} \frac{(-1)^{k+1}}{k^{s+1}} \int_0^\infty x^s e^{-x}dx$$
Now, the sum is the Dirichlet eta function, related to the Riemann zeta function like so,
$$ \sum_{k\geq1}\frac{(-1)^{k+1}}{k^{s+1}} = (1-2^{-s})\zeta(s+1), $$
and the integral is $\Gamma(1+s)$. Thus
$$ J(s) = (1-2^{-s})\zeta(1+s) \Gamma(1+s). $$
To find the derivative at $s=0$ we need the Laurent series for each of these functions at $s=0$, ($\zeta(1+s)$ is singular at $s=0$, but $1-2^{-s}$ has a zero there, so $J$ is regular), they are
$$ (1-2^{-s})\zeta(1+s) = \log2 + (\gamma \log 2 - \frac{(\log 2)^2}{2})s + O(s^2), $$
$$ \Gamma(1+s) = 1 - \gamma s + O(s^2), $$
where $\gamma$ is Euler's constant. Multiplying the two series and taking the coefficient of $s$, we get
$$ \frac{d J}{ds}(0) = -\frac12 (\log 2)^2, $$
which is the integral you were looking for.
