Deriving closed form of $\sum_{n=1}^\infty\frac{(-1)^{n+1}\ln(n+1)}{n} $ I wanted to derive closed form of $\sum_{n=1}^\infty\frac{(-1)^{n+1}\ln(n+1)}{n}$ which I converted into a integral,
$$\sum_{n=1}^\infty\frac{(-1)^{n+1}\ln(n+1)}{n}=\int_0^1\int_0^1\frac{u^v}{1+u^v}\text{d}u\text{d}v$$
after some more manipulations I got,
$$\int_0^1\int_0^1\frac{u^v}{1+u^v}\text{d}u\text{d}v=1-\int_0^1\frac{\psi^{(0)}\left(\frac{1}{v}\right)-\psi^{(0)}\left(\frac{1}{2v}\right)-\ln(2)}{v}\text{d}v$$
not sure how to proceed further...
 A: As you did
$$\int_0^1\frac{u^v}{u^v+1}\,du=\frac{H_{\frac{1}{2 v}}-H_{\frac{1}{2 v}-\frac12}}{2 v}$$
At this point, I am stuck with formal but we can notice that the rhs (our new integrand)
$$f(v)=\frac{H_{\frac{1}{2 v}}-H_{\frac{1}{2 v}-\frac12}}{2 v}\sim \frac{1}{2}- \left(\log (2)-\frac{1}{2}\right)v^{4/5}$$ This approximation matches the function value at the end points and corresponds to the minimum of the infinite norm.
This makes
$$\int_0^1f(v)\,dv \sim
\frac{1}{9} (7-5 \log (2))\approx 0.392696$$ while the exact value is $0.392259$.
For improvement, I think that we need an approximation at least able to reproduce the following values
$$\left(
\begin{array}{ccc}
v & f(v) & f'(v) \\
 0 & \frac{1}{2} & -\frac{1}{4} \\
 \frac{1}{2} & -1+2\log (2) & \frac{2}{3} \left(\pi ^2-6 (1+\log (2))\right) \\
 1 & 1-\log (2) & -\frac{\pi ^2}{12}+\log (2)
\end{array}
\right)$$
Being lazy, for the time being, I used a quintic polynomial matching all the above conditions and obtained as an approximation
$$\frac{1}{720} \left(\pi ^2+588 \log (2)-135\right)\approx 0.392278$$ which is better but not enough.
What is amazing is
$$\frac{1}{10} \left(\psi \left(\frac{4}{9}\right)+\psi
   \left(\frac{3}{5}\right)-\psi \left(\frac{1}{4}\right)-\psi
  \left(\frac{3}{10}\right)\right)\approx 0.392259435$$ while the exact value is                          $0.392259418$
