Equality with Euler–Mascheroni constant While trying to prove $\int_0^{\infty } \frac{\log (x)}{e^x+1} \, dx = -\frac{1}{2} \log ^2(2)$ How to show? in an alternative way, I came to this solution:
$$\sum_{k=0}^{+\infty}(-1)^{k+1}\frac{\log (k+1)+\gamma }{(k+1)}.$$
As both solutions have to be the same, the following equality should be valid:
$$\sum_{k=0}^{+\infty}(-1)^{k+1}\frac{\log (k+1)+\gamma }{(k+1)}=- \frac{1}{2}{{\log }^2(2)}. $$
Can anyone give me some advice on how to prove this equality.
p.s. You can be sure that the equality is correct, as I checked it numerically.
 A: Note that
$$
\begin{align}
\sum_{k=1}^\infty(-1)^k\frac{\log(k)+\gamma}{k}
&=\lim_{n\to\infty}\left(2\sum_{k=1}^{n}\frac{\log(2k)+\gamma}{2k}-\sum_{k=1}^{2n}\frac{\log(k)+\gamma}{k}\right)\\
&=\lim_{n\to\infty}\left(\sum_{k=1}^{n}\frac{\log(2)+\log(k)+\gamma}{k}-\sum_{k=1}^{2n}\frac{\log(k)+\gamma}{k}\right)\tag{1}
\end{align}
$$
Using the Euler-Maclaurin Sum Formula, we get that
$$
\sum_{k=1}^n\frac{1}{k}=\log(n)+\gamma+\frac{1}{2n}-\frac{1}{12n^2}+O\left(\frac{1}{n^4}\right)\tag{2}
$$
and that
$$
\sum_{k=1}^n\frac{\log(k)}{k}=\frac12\log(n)^2+C+\frac{\log(n)}{2n}-\frac{\log(n)-1}{12n^2}+O\left(\frac{\log(n)}{n^4}\right)\tag{3}
$$
Applying $(2)$ and $(3)$ to $(1)$, leaving out the terms which vanish, we get
$$
\begin{align}
&\sum_{k=1}^\infty(-1)^k\frac{\log(k)+\gamma}{k}\\
&=\small{\lim_{n\to\infty}\left(\log(2)(\log(n){+}\gamma)+\left(\frac12\log(n)^2+C+\gamma(\log(n){+}\gamma)\right)-\left(\frac12\log(2n)^2+C+\gamma(\log(2n){+}\gamma)\right)\right)}\\
&=\lim_{n\to\infty}\left(\log(2)(\log(n)+\gamma)-\log(2)\log(n)-\frac12\log(2)^2-\gamma\log(2)\right)\\
&=-\frac12\log(2)^2\tag{4}
\end{align}
$$
A: This can be solved similarly to the original problem.  The Dirichlet eta function is defined by
$$
\eta(s):=\sum_{n\ge 1} \frac{(-1)^{n-1}}{n^s}.
$$
The given sum can be rewritten as 
$$
\sum_{n\ge 1} (-1)^n \frac{\log n}{n}+\sum_{n\ge 1} (-1)^n \frac{\gamma}{n}=
\eta'(1)-\gamma \log 2.\qquad (*)$$ We have
$$
\eta(s)=\sum_{n\ge 1} \frac{1}{n^s}-2\sum_{n\ge 1} \frac{1}{(2n)^s}=(1-2^{1-s})\zeta(s)
$$
so, using the expansions 
$$\zeta(s)=\frac{1}{s-1}+\gamma+O(s-1),$$ $$
2^{1-s}=e^{(1-s)\log 2}=1-(s-1)\log 2+\frac{1}{2}(\log 2)^2 (s-1)^2+O((s-1)^3),
$$
we get $\eta'(1)=\gamma \log 2 -\frac{1}{2}(\log 2)^2$, so (*) equals $-\frac{1}{2}(\log 2)^2$.
A: The problem is equivalent to showing
$$
\sum_{n\ge1}{(-1)^n\log n\over n}=\gamma\log2+\frac12\log^22
$$
Now, let's first consider the finite case:
$$
\begin{aligned}
\sum_{n\le N}{(-1)^n\log n\over n}
&=2\sum_{n\le N/2}{\log(2n)\over2n}-\sum_{n\le N}{\log n\over n}+\mathcal O\left(\log N\over N\right) \\
&=\sum_{n\le N/2}{\log2+\log n\over n}-\sum_{n\le N}{\log n\over n}+\mathcal O\left(\log N\over N\right) \\
&=\log2\sum_{n\le N/2}\frac1n-\sum_{N/2<n\le N}{\log n\over n}+\mathcal O\left(\log N\over N\right) \\
\end{aligned}
$$
In fact, using Riemann-Stieltjes integral, we can show
$$
\begin{aligned}
\sum_{N/2<n\le N}{\log n\over n}
&=\int_{N/2}^N{\log x\over x}\mathrm d\lfloor x\rfloor \\
&={N\log(N)-N\log(N/2)\over N}-\int_{N/2}^N[x-\{x\}]\mathrm d\left(\log x\over x\right)+\mathcal O\left(\frac1n\right) \\
&=\log2-\int_{N/2}^N\left({1-\log x\over x}\right)\mathrm dx+\mathcal O\left(\log N\over N\right) \\
&=\int_{N/2}^N{\log x\over x}\mathrm dx+\mathcal O\left(\log N\over N\right) \\
&=\frac12[\log^2N-\log^2(N/2)]+\mathcal O\left(\log N\over N\right) \\
&=\frac12[\log N+\log(N/2)][\log N-\log(N/2)]+\mathcal O\left(\log N\over N\right) \\
&=\frac12\log2[2\log N-\log2]+\mathcal O\left(\log N\over N\right) \\
&=\log2\log N-\frac12\log^22+\mathcal O\left(\log N\over N\right)
\end{aligned}
$$
Now, employing this obtained identity and the asymptotic formula for harmonic series yields:
$$
\begin{aligned}
\sum_{n\le N}{(-1)^n\log n\over n}
&=\log2(\log N+\gamma)-\log2\log N+\frac12\log^22+\mathcal O\left(\log N\over N\right) \\
&=\gamma\log2+\frac12\log^22+\mathcal O\left(\log N\over N\right)
\end{aligned}
$$
Now, take the limit $N\to\infty$ on both side gives
$$
\sum_{n\ge1}{(-1)^n\log n\over n}=\gamma\log2+\frac12\log^22
$$
