In many online sources, you can find $$ \zeta(s) \overset{C.P.}{=} \lim_{\epsilon \to 0} \left(\frac{\zeta(s+\epsilon)+\zeta(s-\epsilon)}{2}\right). $$ This seems quite logical, but I neither know how to derive it nor how to derive the second equality $$ \zeta(1) \overset{C.P.}{=}\lim_{\epsilon \to 0} \left(\frac{\zeta(1+\epsilon)+\zeta(1-\epsilon)}{2}\right)=\gamma $$ (where $\gamma$ is the Euler-Mascheroni-constant). I always knew the Cauchy principal value as a a method for assigning values to improper integrals, and although I know many integral representations of the Riemann Zeta function, I dont't really know how to show this specific result. Many thanks in advance!
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The asymptotic
$$ \zeta(s) = \frac{1}{s-1} + \gamma + O(|s - 1|) $$
for the Riemann zeta function near $ s = 1 $ is well known, and it immediately implies your result, since the $ 1/(s-1) $ terms cancel when computing the limit in your principal value.
For a proof of this asymptotic, see this post.
Ege Erdil
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Thank you very much! – cxrlo May 04 '20 at 14:20
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(a little remark: I think you meant to say $\zeta(s)$ instead of $\zeta(s-1)$.) – cxrlo May 04 '20 at 14:25
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1@cxrlo You're right, of course. – Ege Erdil May 04 '20 at 14:25