Hurwitz Zeta Values for RNS Normal Ordering

Fix an integer $N\ge 1$, put $x=N+a$, and take $f(t)=t^{-s}$ for $\mathrm{Re}s>1$ and $t\ge x$. Then $f$ is $C^4$, its derivatives through order three tend to zero at infinity, and $f^{(4)}$ is integrable. Four integrations by parts on each interval $[x+n,x+n+1]$ against the periodic Bernoulli polynomials, summed over $n\ge 0$, give the Euler–Maclaurin tail identity ${\sum }_{n=0}^{\mathrm{\infty }}f(x+n)={\int }_x^{\mathrm{\infty }}f(t)dt+\frac{1}{2}f(x)-\frac{B_2}{2!}{f}^{\mathrm{\prime }}(x)-\frac{B_4}{4!}{f}^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}(x)-\frac{1}{4!}{\int }_x^{\mathrm{\infty }}{\tilde{B}}_4(t-x)f^{(4)}(t)dt,$ where $B_2=1\mathrm{/}6$, $B_4=-1\mathrm{/}30$, and ${\tilde{B}}_4(t)=B_4(t-\lfloor t\rfloor )$. The endpoint terms telescope because the periodic Bernoulli polynomials agree at consecutive integer endpoints; the last integral is the remaining integration-by-parts term. Here ${\int }_x^{\mathrm{\infty }}{t}^{-s}dt=x^{1-s}\mathrm{/}(s-1)$, and substitution of the derivatives gives $\zeta (s,a)={\sum }_{n=0}^{N-1}(n+a{)}^{-s}+\frac{x^{1-s}}{s-1}+\frac{1}{2}{x}^{-s}+\frac{s}{12}{x}^{-s-1}-\frac{s(s+1)(s+2)}{720}{x}^{-s-3}+R_2(s,x),$ with $R_2(s,x)=-\frac{s(s+1)(s+2)(s+3)}{24}{\int }_x^{\mathrm{\infty }}{\tilde{B}}_4(t-x)t^{-s-4}dt$. The integral converges locally uniformly for $\mathrm{Re}s>-3$, so this expression continues the defining series there (apart from its pole at $s=1$). At $s=-1$, both the third-derivative correction and $R_2$ vanish because they contain a factor $s+1$.

At $s=-1$, the finite sum in step 1.1 is ${\sum }_{n=0}^{N-1}(n+a)=N(N-1)\mathrm{/}2+Na$, the integral term is $-x^2\mathrm{/}2$, the endpoint term is $x\mathrm{/}2$, and the $B_2$ correction is $-1\mathrm{/}12$. The $B_4$ correction and remainder vanish because they contain a factor $s+1$. Therefore $\zeta (-1,a)=\frac{N(N-1)}{2}+Na-\frac{(N+a{)}^2}{2}+\frac{N+a}{2}-\frac{1}{12}=-\frac{1}{2}(a^2-a+\frac{1}{6})=-\frac{1}{2}{B}_2(a)\mathrm{.}$ The right side is independent of $N$, as required for the continuation value.

Substitution gives $B_2(1)=1\mathrm{/}6$, so $\zeta (-1,1)=-1\mathrm{/}12$. Also $B_2(1\mathrm{/}2)=1\mathrm{/}4-1\mathrm{/}2+1\mathrm{/}6=-1\mathrm{/}12$, so $\zeta (-1,1\mathrm{/}2)=1\mathrm{/}24$. These are exactly the integer and half-integer vacuum sums used in the oscillator calculation.