TITLE: How to evaluate a certain definite integral: $\int_{0}^{\infty}\frac{\log(x)}{e^{x}+1}dx$
QUESTION [14 upvotes]: 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 $\eta'(1)=-\ln2\gamma-\dfrac{\ln^2(2)}{2}$.

REPLY [2 votes]: 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.