 Solution

Each finite-interval integral $f_n(s)$ is entire in $s$: the integrand is $(x-n)e^{-(s+1)\log x}$ and its complex derivatives may be integrated on the compact interval. On a compact subset of $\mathrm{Re}s>0$, choose $\delta >0$ with $\mathrm{Re}s\ge \delta$. Then $|f_n(s)|\le n^{-1-\delta }$. The Weierstrass uniform-convergence theorem gives convergence and holomorphy of ${\sum }_n{f}_n(s)$ throughout that half-plane.

Integration by parts givesFor $\mathrm{Re}s>1$, sum absolutely and use the integral over $[1,\infty )$ to obtain $s{\sum }_n{f}_n(s)=1-\zeta (s)+1/(s-1)$. Rearranging gives the first requested identity.

On $[n,n+1]$, the magnitude of $(x-n)(x-n-1)$ is at most $1/4$. On a compact subset of $\mathrm{Re}s>-1$, its integrals are bounded by a convergent constant multiple of $n^{-1-\delta }$ for some $\delta >0$. Thus ${\sum }_n{g}_n(s)$ is holomorphic there. The prefactor $s(s+1)$ is entire, leaving only the explicit simple pole at one in $\tilde{\zeta }$, with residue one.

Differentiate $(x-n)(x-n-1)x^{-s-1}$ and integrate. Its endpoint values both vanish, givingAt $s=0$ the last quotient has its continuous value $-\log ((n+1)/n)$. For $\mathrm{Re}s>1$, telescoping over $n$ gives $(s+1){\sum }_n{g}_n-2{\sum }_n{f}_n=-1/s$. Substitution into $\tilde{\zeta }$ yieldsThis supplies the analytic continuation to $\mathrm{Re}s>-1$ apart from the pole. Evaluating its holomorphic expression at zero makes the series term vanish and gives $\boxed{\zeta (0)=-1/2}$.