How to evaluate a certain definite integral: $\int_{0}^{\infty}\frac{\log(x)}{e^{x}+1}dx$ How can I show that:
$$\int_{0}^{\infty}\frac{\log(x)}{e^{x}+1}dx=-\frac{\log^{2}(2)}{2}$$
EDIT: This is equivalent to showing that $\eta'(1)=-\ln2\gamma-\dfrac{\ln^2(2)}{2}$.
 A: Since for any $\alpha$ such that $\Re(\alpha)>-1$ we have:
$$f(\alpha)=\int_{0}^{+\infty}\frac{x^\alpha}{e^x+1}\,dx = \left(1-\frac{1}{2^{\alpha}}\right)\Gamma(1+\alpha)\zeta(1+\alpha)\tag{1}$$
by expanding the integrand function as a geometric series, we just need to find the limit of the derivative of the RHS of $(1)$ when $\alpha\to 0^+$. It is worth to consider logarithmic derivatives and exploit the identity $g'(z)=g(z)\cdot\frac{d}{dz}\log g(z)$, since the RHS of $(1)$ is a product and $f(0)=\log 2$. Now we have, in a neighbourhood of zero:
$$ 2^z-1 = z \log2 + \frac{z^2}{2}\log^2 2 + o(z^3),$$
$$ \Gamma(z+1) = 1-\gamma z + O(z^2),$$
$$ \zeta(z+1) = \frac{1}{z}+\gamma+O(z),$$
hence their product is
$$ \log 2 + \frac{z}{2}\log^2 2 + O(z^2)$$
and the value of the logarithmic derivative in zero of $\left(1-\frac{1}{2^{z}}\right)\Gamma(1+z)\zeta(1+z)$ is just $-\frac{\log 2}{2}$.
This gives:

$$\int_{0}^{+\infty}\frac{\log x}{e^x+1}\,dx = -\frac{\log^2 2}{2}$$

as wanted.
A: \begin{align}
\int^\infty_0\frac{\color\red{\log{x}}}{e^x+1}dx
&=\color\red{\lim_{a \to 1}\frac{d}{da}}\int^\infty_0\frac{\color\red{x^{a-1}}e^{-x}}{1+e^{-x}}dx\tag1\\
&=\lim_{a \to 1}\frac{d}{da}\sum_{n \ge 0}(-1)^n\int^\infty_0x^{a-1}e^{-(n+1)x}dx\tag2\\
&=\lim_{a \to 1}\sum_{n \ge 0}(-1)^n\frac{d}{da}\frac{\Gamma(a)}{(n+1)^a}\tag3\\
&=\lim_{a \to 1}\sum_{n \ge 0}(-1)^n\frac{\Gamma(a)\psi(a)-\Gamma(a)\log(n+1)}{(n+1)^a}\tag4\\
&=\psi(1)\sum_{n \ge 0}\frac{(-1)^n}{n+1}-\sum_{n \ge 0}\frac{(-1)^n\log(n+1)}{n+1}\tag5\\
&=-\gamma\log{2}-\left(\frac{1}{2}\log^22-\gamma\log{2}\right)\tag6\\
&=-\frac{1}{2}\log^22\\
\end{align}
Explanation 
$(1)$: Divide numerator and denominator by $e^x$ 
$(2)$: Expand the integrand as a geometric series, swap the order of integration and summation 
$(3)$: Recognise the gamma function 
$(4)$: Quotient rule 
$(5)$: Apply the limit and split the sum into two. 
$(6)$: For the first sum, $\psi(1)=-\gamma$, and $\displaystyle\ln(1+1)=\sum_{n \ge 0}\frac{(-1)^n}{n+1}1^{n+1}$. I will now show how to evaluate the second sum.
\begin{align}
\sum_{n \ge 0}\frac{(-1)^n\log(n+1)}{n+1}
&=\sum_{n \ge 1}\frac{(-1)^{n-1}\log{n}}{n}\tag7\\
&=\lim_{s \to 1}\frac{d}{ds}\sum_{n \ge 1}\frac{(-1)^{n}}{n^s}\\
&=-\eta'(1)\tag8\\
&=\frac{1}{2}\log^22-\gamma\log{2}\tag9\\
\end{align}
$(7)$: Index shift 
$(8),(9)$: See Dirichlet eta function, Stieltjes Constants
A: So we would like to find $\eta'(1)$ which in essence mean computing $\sum_{k=1}^{\infty} \frac{(-1)^{k-1}\log k}{k}$.
We will present a solution which is in some sense reminscent of one of the proofs of the fact $\sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k}=\log 2$. Namely rewriting $\sum_{k=1}^{2n} \frac{(-1)^{k-1}}{k}=\sum_{k=1}^{2n} \frac{1}{k}-2\sum_{k=1}^{n} \frac{1}{2k}=H_{2n}-H_{n}=(H_{2n}-\log 2n)-(H_n-\log n)+\log 2$ ($H_n$ represents the $n$-th harmonic number) and using the fact $\lim_{n\to \infty}H_n-\log n$ exists (and is in fact equal to Euler-Mascheroni $\gamma$ constant).
First of all we prove that the following sequence is convergent - $b_n=\sum_{k=1}^{n}\frac{\log k}{k}-\frac{\log(n)^2}{2}$ (analog of $H_n-\log n$, see here). Rewrite $b_n=\frac{\log 2}{2}-\frac{\log^2 3}{2}+\sum_{k=3}^{n}\frac{\log k}{k}-\int_{3}^{n}\frac{\log x}{x}\mathrm{d}x$ and denote the last two summands as $c_n$ so it is equivalent to prove $c_n$ is convergent. Observe that $\frac{\log x}{x}$ is decreasing for $x\ge 3$. Now $c_{n}-c_{n-1}=\frac{\log n}{n}-\int_{n-1}^n \frac{\log x}{x}\mathrm{d}x\le \frac{\log n}{n}-\frac{\log n}{n}=0$, so the sequence is decreasing. Moreover $c_n=\sum_{k=3}^{n}\frac{\log k}{k}-\int_{3}^{n}\frac{\log x}{x}\mathrm{d}x=\sum_{k=3}^{n}\frac{\log k}{k}-\sum_{k=3}^{n-1}\int_{k}^{k+1}\frac{\log x}{x}\mathrm{d}x\ge \sum_{k=3}^{n}\frac{\log k}{k}-\sum_{k=3}^{n-1}\frac{\log k}{k}=\frac{\log n}{n}\ge 0$,
so the sequence is bounded from below, and it is convergent. (the proof is the same as that of the general fact here.)
Now $$a_{2n}=\sum_{k=1}^{2n}\frac{(-1)^{k-1}\log k}{k}=\sum_{k=1}^{2n}\frac{\log k}{k}-2\sum_{k=1}^{n}\frac{\log 2k}{2k}=\sum_{k=1}^{2n}\frac{\log k}{k}-\sum_{k=1}^{n}\frac{\log k}{k}-H_n\log 2=\left(\sum_{k=1}^{2n}\frac{\log k}{k}-\frac{\log^2 2n}{2}\right)-\left(\sum_{k=1}^{n}\frac{\log k}{k}-\frac{\log^2 n}{2}\right)-(H_n-\log n)\log 2+\left(-\log n\log2+\frac{\log^2 2n}{2}-\frac{\log^2 n}{2}\right)\to -\gamma \log 2 + \frac{\log^2 2}{2}$$ as $n$ goes to $\infty$.
In a similar manner the series $$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}\log^d n}{n}$$ for integer $d$ (i.e. the $d$-th derivative of eta at $1$) could be evaluated, however the Stieltjes constants (which we referred to earlier) will be present in the answer.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$$
\int_{0}^{\infty}{\ln\pars{x} \over \expo{x} + 1}\,\dd x
=\lim_{\mu\to 0}\partiald{}{\mu}\int_{0}^{\infty}
{x^{\mu}\expo{-x} \over 1 + \expo{-x}}\,\dd x
$$

\begin{align}
\int_{0}^{\infty}
{x^{\mu}\expo{-x} \over 1 + \expo{-x}}\,\dd x
&=\sum_{n = 0}^{\infty}\pars{-1}^{n}\int_{0}^{\infty}
x^{\mu}\expo{-\pars{n + 1}x}\,\dd x
=\sum_{n = 0}^{\infty}{\pars{-1}^{n} \over \pars{n + 1}^{\mu + 1}}
\int_{0}^{\infty}x^{\mu}\expo{-x}\,\dd x
\\[3mm]&=\Gamma\pars{\mu + 1}\sum_{n = 1}^{\infty}
{\pars{-1}^{n + 1} \over n^{\mu + 1}}
=-\Gamma\pars{\mu + 1}{\rm Li}_{\mu + 1}\pars{-1}
\\[3mm]&=\Gamma\pars{\mu + 1}\pars{1 - 2^{-\mu}}\zeta\pars{\mu + 1}
\end{align}

\begin{align}
&\color{#66f}{\large\int_{0}^{\infty}{\ln\pars{x} \over \expo{x} + 1}\,\dd x}
=\lim_{\mu\ \to\ 0}
\partiald{\bracks{\Gamma\pars{\mu + 1}\pars{1 - 2^{-\mu}}\zeta\pars{\mu + 1}}}{\mu}
\tag{1}
\\[3mm]&=\color{#66f}{\large -\,\half\,\ln^{2}\pars{2}} \approx -0.2402
\end{align}

See the Dirichlet Eta Function link and the PolyLogarithm Function Link.

ADDENDA
The limit, in expression $\pars{1}$ has to be handled carefully because $\ds{\zeta\pars{\mu + 1} \sim {1 \over \mu}}$ when $\ds{\mu \sim 0}$. It has the
Laurent Expansion
$$
\zeta\pars{\mu + 1}={1 \over \mu} + \sum_{n = 0}^{\infty}{\pars{-1}^{n} \over n!}\,
\gamma_{n}\,\mu^{n}
$$
where $\ds{\gamma_{n}}$'s are the Stieltjes Constants.
Let's consider the limit evaluation. In order to evaluate the limit we just need the expansion, up to order one, of each  factor in
( $\ds{\gamma}$ is the Euler-Mascheroni Constant ):
$$
\Gamma\pars{\mu + 1}\,{1 - 2^{-\mu} \over \mu}\,\bracks{\mu\zeta\pars{\mu + 1}}\,,
\qquad\left\vert\begin{array}{ccl}
\ \Gamma\pars{\mu + 1} & \sim & 1 - \gamma\mu
\\[2mm]
\ {1 - 2^{-\mu} \over \mu} & \sim & \ln\pars{2} - \half\,\ln^{2}\pars{2}\mu
\\[2mm]
\ \mu\zeta\pars{\mu + 1} & \sim & 1 + \gamma\mu\,,\quad\gamma_{0} = \gamma
\end{array}\right.
$$
From these expression we see that the product
$\ds{\Gamma\pars{\mu + 1}\bracks{\mu\zeta\pars{\mu + 1}} \sim
1 - \gamma^{2}\mu^{2}}$ is already of order $\ds{\mu^{2}}$. We are left with the'middle factor' $\ds{1 - 2^{-\mu} \over \mu}$ such that
\begin{align}&\lim_{\mu\ \to\ 0}
\partiald{\bracks{\Gamma\pars{\mu + 1}\pars{1 - 2^{-\mu}}\zeta\pars{\mu + 1}}}{\mu}
=\lim_{\mu\ \to\ 0}
\partiald{\bracks{\ln\pars{2} - \ln^{2}\pars{2}\mu/2}}{\mu}
\\[3mm]&=\color{#66f}{\large -\,\half\,\ln^{2}\pars{2}}
\end{align}
