Problem
Show that $$ \gamma = \tfrac 12 \log 2 + {1 \over \log 2} \sum_{n=2}^\infty (-1)^n {\log n \over n}. $$
Progress
I tried writing the terms $1/k$ of the harmonic sum in the definition of $\gamma$ as $\int_0^1 x^{k-1} dx$, and interchanging the order of summation and integration. This gives $$ \gamma = \lim_{n \to \infty} \int_0^1 {1 - x^n - \log n + x \log n \over 1 - x} dx, $$ which might just be a dead end. I also know the alternate form $$ \gamma = \lim_{s \to 1} {\zeta'(s) \over \zeta(s)} + {1 \over s-1} = \lim_{s \to 1} -\sum_{n=1}^\infty \Lambda(n) n^{-s} + {1 \over s-1}, $$ but I don't see how it could help.
Notes: $\log x$ is the natural logarithm, $\Lambda(n)$ is the von Mangoldt function.
