TITLE: Closed form of $\sum_{n=1}^\infty(-1)^{n+1} \frac xn \ln\left(1+\frac xn\right), \quad x \in (0,1)$
QUESTION [9 upvotes]: Is there a known closed form of the series below?
$$\sum_{n=1}^\infty(-1)^{n+1} \frac xn \ln\left(1+\frac xn\right), \quad x \in (0,1]$$

REPLY [3 votes]: Denote the sum by $S(x)$. It is straightforward to check by direct differentiation that
$$S'(x)-\frac{S(x)}{x}=x\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2+nx}=
\frac{\psi\left(\frac{x+1}{2}\right) -\psi\left(\frac{x+2}{2}\right)}{2}+\ln 2.$$
The solution of this differential equation satisfying appropriate initial conditions at $x=0$ is
$$S(x)=\frac x2 \int_0^x\left[\psi\left(\frac{s+1}{2}\right) -\psi\left(\frac{s+2}{2}\right)+2\ln 2\right]\frac{ds}{s}.$$
However it does not seem to me that this antiderivative has any chances to be computed in closed form.