Constant term in the Euler-Maclaurin expansion of $s_n=\sum_{k=1}^n \tfrac{1}{k+1/2}$

Question

This question is a followup from my previous post based on the Euler-Maclaurin formula: How to find the correct constant term with Euler-Maclaurin formula, $\sum_{j=1}^n j\log j$.

This time I am working with $s_n=\sum_{k=1}^n \tfrac{1}{k+1/2}$. By the Euler-Malcuarin formula we have $$\begin{align*} s_n & = \int_1^n \frac{dt}{t+1/2} +\frac{1}{2n+1}+\frac{1}{3}-\int_1^n \frac{B_1(t-\lfloor t\rfloor)}{(t+1/2)^2}\ dt \\[3mm] & = \log\left(n + {1 \over 2}\right) - \log\left(3 \over 2\right) + \frac{1}{2n+1}+\frac{1}{3} \\[3mm] & \phantom{=\,\,}-\int _1^{+\infty} \frac{B_1(t-\lfloor t\rfloor)}{(t+1/2)^2}\ dt +\int _n^{+\infty} \frac{B_1(t-\lfloor t\rfloor)}{(t+1/2)^2}\ dt.\label{1}\tag{1} \end{align*}$$ Since the periodic Bernoulli polynomial is a continuous and bounded function $$ \int_n^{+\infty}\frac{B_1(t-\lfloor t\rfloor)}{(t+1/2)^2}\ dt = O(1/n)$$ as $n\to +\infty$. Now applying the Euler-Maclaurin formula to $\zeta (s,1/2)=\sum_{k=1}^\infty \frac{1}{(k+1/2)^s}$ we have \begin{align*} \zeta(s,1/2) & = \int _1^{+\infty} \frac{dt}{(t+1/2)^s}+\frac{1}{2}\left(\frac{2}{3}\right)^{s} \\[3mm] & -s\int_1^{+\infty}\frac{B_1(t-\lfloor t\rfloor)}{(t+1/2)^{s+1}}\ dt \\[5mm] & = \frac{(3/2)^{1-s}}{s-1}+\frac{1}{2}\left(\frac{2}{3}\right)^s \\[3mm] & -s\int_1^{+\infty}\frac{B_1(t-\lfloor t\rfloor)}{(t+1/2)^{s+1}}\ dt\label{2}\tag{2} \end{align*} for $\text{Re}(s)>1$. So the value of the $\int_1^{+\infty}$ integral in (\ref{1}) can be found via setting $s=1$ in (\ref{2}), but there is a singularity there. What do I do?

Answer 1

First note that $$ \sum\limits_{k = 1}^\infty {\frac{1}{{(k + 1/2)^s }}} = - 2^s + \sum\limits_{k = 0}^\infty {\frac{1}{{(k + 1/2)^s }}} = \zeta (s,1/2)\color{red}{ - 2^s}. $$ From (the corrected version of) your last equation $$ \mathop {\lim }\limits_{s \to 1} \left( {\zeta (s,1/2) - \frac{1}{{s - 1}}} \right) = 2- \log \left( {\tfrac{3}{2}} \right) + \tfrac{1}{3} - \int_1^{ + \infty } {\frac{{B_1 (t - \left\lfloor t \right\rfloor )}}{{(t + 1/2)^2 }}{\rm d}t} . $$ (Note that the corrected equation holds for $\operatorname{Re}(s)>0$, $s\ne 1$ by analytic continuation.) By the Laurent series expansion of the Hurwitz zeta function, $$ \mathop {\lim }\limits_{s \to 1} \left( {\zeta (s,1/2) - \frac{1}{{s - 1}}} \right) = \gamma _0\! \left( {\tfrac{1}{2}} \right) = - \psi\! \left( {\tfrac{1}{2}} \right) = \gamma + 2\log 2, $$ where $\gamma_0(a)$ is a generalised Steiltjes constant and $\psi$ is the digamma function. Thus, $$ s_n = \log \left( {n + \tfrac{1}{2}} \right) + \gamma + 2\log 2 - 2 + \mathcal{O}\!\left( {\frac{1}{n}} \right) = \log n + \gamma + 2\log 2 - 2 + \mathcal{O}\!\left( {\frac{1}{n}} \right), $$ as $n\to+\infty$.