$\zeta(1)=\frac12\ln(2)$? Did I do something wrong?

Question

I attempted to calculate $\zeta(1)$ and I got $\frac12\ln(2)$.

\begin{align}\zeta(1)&=\lim_{\epsilon\to0}\frac{\zeta(1+\epsilon)+\zeta(1-\epsilon)}2\\&=\lim_{\epsilon\to0}\frac{\frac1{1-2^{-\epsilon}}\eta(1+\epsilon)+\frac1{1-2^\epsilon}\eta(1-\epsilon)}2\\&=\frac12\eta(1)\lim_{\epsilon\to0}\frac1{1-2^{-\epsilon}}+\frac1{1-2^\epsilon}\\&=\frac12\eta(1)\lim_{\epsilon\to0}\frac{2-2^\epsilon-2^{-\epsilon}}{(1-2^{-\epsilon})(1-2^\epsilon)}\\&=\frac12\eta(1)\lim_{\epsilon\to0}1\\&=\frac12\ln(2)\end{align}

Usually, one has $\zeta(1)=\pm\infty,\gamma,$ or $\ln(2)$ as a possible result that replaces $\gamma$, but I've never seen someone get $\zeta(1)=\frac12\ln(2)$. Did I do anything wrong?

Answer 1

Since

$$\frac{1}{1-2^{-\epsilon}} = \frac{1}{1-\exp(-\epsilon\log 2)} = \frac{1}{\epsilon\log 2 + O(\epsilon^2)}$$

and similarly for $\frac{1}{1-2^{\epsilon}}$, the small difference between $\eta(1)$ and $\eta(1\pm \epsilon)$ matters. If we use a first order Taylor expansion of $\eta$ at $1$, namely $\eta(1\pm \epsilon) = \eta(1) \pm \epsilon \eta'(1) + O(\epsilon^2)$, we find

\begin{align} \frac{\zeta(1+\epsilon) + \zeta(1-\epsilon)}{2} &= \frac{1}{2}\biggl(\frac{\eta(1) + \epsilon \eta'(1) + O(\epsilon^2)}{1-2^{-\epsilon}} + \frac{\eta(1) -\epsilon\eta'(1) + O(\epsilon^2)}{1-2^{\epsilon}}\biggr) \\ &= \frac{1}{2}\Biggl(\eta(1)\biggl(\frac{2^{\epsilon}}{2^{\epsilon}-1} + \frac{1}{1-2^{\epsilon}}\biggr) + \\ &\qquad\quad\eta'(1)\biggl(\frac{\epsilon}{\epsilon\log 2 + O(\epsilon^2)} + \frac{-\epsilon}{-\epsilon\log 2 + O(\epsilon^2)}\biggr) + O(\epsilon)\Biggr) \\ &= \frac{\eta(1)}{2} + \frac{\eta'(1)}{\log 2} + O(\epsilon) \end{align}

and therefore

$$\gamma = \lim_{\epsilon \to 0}\frac{\zeta(1+\epsilon) + \zeta(1-\epsilon)}{2} = \frac{\log 2}{2} + \frac{\eta'(1)}{\log 2}.$$