TITLE: Euler constant as limit of zeta function
QUESTION [5 upvotes]: I want to prove that $ \lim \limits_{s \to 0} (\zeta(1+s)+\zeta(1-s))= 2\gamma$ , I divide it to two limits, $\lim \limits_{s \to 1^{+}} (\zeta(s)-\frac{1}{s-1}) = \gamma$ which I proved using the definition for $\gamma = H(\infty)-\ln (\infty)$  and $\zeta(s) = \sum \limits_{n=1}^{\infty} n^{-s} $ for $s>1$ but the definition of $\zeta(s)$ fails when evaluating $\lim \limits_{s\to 1^{-}} (\zeta(s)-\frac{1}{s-1})$, so I want to use $\zeta(s) = \frac{1}{1-2^{-s}} \sum \limits_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^s}$ for $s>0$, the problem for me now is that this new definition does not easily transform to the definition of $\gamma = \lim \limits_{n \to \infty } (H_n -\ln n)$, I want to see a proof of the case when $s\to 1^{-}$ ?
Thanks

REPLY [2 votes]: We can express both situations using Euler-Maclaurin formula:
$$
\sum_{n=1}^N\frac1n=\log N+\frac12+{1\over2N}-\int_1^N{\overline B_1(x)\over x^2}\mathrm dx
$$
Using the definition of Euler-Mascheroni constant, we can see that
$$
\gamma=\frac12-\int_1^\infty{\overline B_1(x)\over x^2}\mathrm dx
$$
Now, let's apply Euler-Maclaurin formula to $\zeta(s)$
$$
\zeta(s)=\sum_{n=1}^\infty{1\over n^s}={1\over s-1}+\frac12-s\int_1^\infty{\overline B_1(x)\over x^{s+1}}\mathrm dx
$$
Since $|\overline B_1(x)|$ is bounded, we see that the right-most integral converges for $\Re(s)>1$, so the RHS is an analytic continuation. Consequently, we have
$$
\lim_{s\to1}\left\{\zeta(s)-{1\over s-1}\right\}=\frac12-\int_1^\infty{\overline B_1(x)\over x^2}\mathrm dx=\gamma
$$