\[\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(n+a)^{s}}=\frac{1}{\Gamma\left(s\right)}%
\int_{0}^{\infty}\frac{x^{s-1}e^{-ax}}{1+e^{-x}}\,\mathrm{d}x=2^{-s}\left(%
\zeta\left(s,\tfrac{1}{2}a\right)-\zeta\left(s,\tfrac{1}{2}(1+a)\right)\right),\]