Just playing around considering some older hint from some answer here in MSE (a couple of years ago) I came to the following observation.
Assume a small $\epsilon >0 $ then the arithmetic mean $$ {\zeta(1+\epsilon) + \zeta(1-\epsilon) \over 2} \approx \gamma = 0.577... \tag 1$$ and there is even the well known equality in the limit for $\epsilon \to zero$ which I've seen some times stated here and elsewhere for to explain the (Ramanujan's) evaluation/replacement of $\zeta(1)\underset{\mathfrak R}= \gamma$.

Well, for $\epsilon \to 0$ this approximates nicely, but of course, if $\epsilon$ increases, say $\epsilon = \frac12$ or even greater $\epsilon = 6 $ then this approximation worsens rapidly (and of course follows the oscillatory divergent values for $\zeta$ at the negative integers).

Now I tried what would happen, if I do not take $\pm \epsilon$ which means $\epsilon \cdot (1,-1)$ but higher complex unit-roots $\omega_k$ say, for order $d$, the vector $\epsilon \cdot [ 1, \omega, \omega^2, ... \omega^{d-1}]$ and compute the arithmetic mean and compare to the Eulerconstant $\gamma$.

Interestingly this approximates $\gamma$ again, and taking $d=5$, $d=50$, or even $d=150$ we can even have $\epsilon =1 $ to have approximation to more than 100 digits (if $d=150$ for instance) such that

$$\lim_{d \to \infty} \frac 1d \sum_{k=0}^{d-1} \zeta(1+ r \exp(2 i \pi \frac kd)) = \gamma \tag 2$$ where $r$ may be non-infinitesimal and for instance $r=1$ or $r=13$.

This is much surprising for me - and I'd like to know

a) is this true?
b) how can this be generalized/extended to $d \to \infty$ : does this mean to introduce an integral instead of the sum/mean? (I've no idea yet how this should actually be written - this is, perhaps, "contour-integration" ?)

  • $\begingroup$ You seem to have forgotten a factor of $1/d$ in your definition. Surely you need $\lim_{d \to \infty} \frac{1}{d}\sum_{k=0}^{d-1} \zeta(1+ r \exp(2 i \pi \frac kd)) = \gamma \tag 2$ $\endgroup$
    – sharding4
    Jun 21 '17 at 12:35
  • $\begingroup$ Upps - thank you. Corrected. $\endgroup$ Jun 21 '17 at 13:09

Presumably there should be a $\frac{1}{d}$ in front of your sum (so that it is an average, as in your first example). Then, viewing your integral as a Riemann sum, the limit as $d \to \infty$ will be $$\lim_{d \to \infty} \frac{1}{d}\sum_{k=0}^{d-1}\zeta(1+re^{2\pi i k/d}) \to \int_{0}^{1}\zeta(1+re^{2\pi i x})d x \\ = \frac{1}{2\pi}\int_{0}^{2\pi}\zeta(1+re^{i \theta})d \theta$$ By making the substitution $z=re^{i \theta}$, $0 \le \theta \le 2\pi$ this integral may be naturally viewed as the contour integral $$\frac{1}{2\pi i}\int_{\gamma}\frac{\zeta(1+z)}{z}dz$$ where $\gamma$ is the circle of radius $r$ centred at the origin. By the residue theorem, this is equal to $$\sum_{i} Res(f,z_{i})$$ where the sum is over all poles of $f(z)=\frac{\zeta(1+z)}{z}$ inside $\gamma$. For any value of $r$, there is exactly one pole in this region, and it has order $2$. It follows that the residue of the pole (and hence the contour integral) is the principal part of $\zeta(1)$, which is the constant $\gamma$.

I get the impression you have not seen contour integration before, so this argument might not be as helpful as it could be - I'd be happy to explain, but if I give a pre-emptive explanation of all of it then this answer would be far too long. All the theorems used would be found in any book/course on complex analysis.

  • $\begingroup$ Thank you, Daniel. Well, I'm rather illiterate with integration, and sometimes, when some problem gets me and it requires an advanced understanding I might be able to get familiar with this (I'm a hobbyist, learner-on-my-own): for all understanding of the formalia I need vitally the relation/foundation with/on an understandable problem. Perhaps I'm beginning now to understand the idea of a contour-integral - having now an own application... Your answer is surely helpful. $\endgroup$ Jun 21 '17 at 13:16
  • $\begingroup$ I'm glad it was helpful! I hope this motivates you to look into complex analysis - it really is a beautiful subject. $\endgroup$ Jun 21 '17 at 13:55

You should read a course on complex analysis.

$F(s)=(s-1)\zeta(s)$ is entire (analytic on $\mathbb{C}$), in particular it is holomorphic at $s=1$ with $F(1) = 1, F'(1) = \gamma$.

  • Thus for any complex number $a \ne 0$ $$\lim_{h \to 0} \frac{F(1+ah)-F(1)}{ah} = F'(1) = \gamma$$

  • Also note we have the Taylor series $F(s) = \sum_{n=0}^\infty \frac{F^{(n)}(1)}{n!} (s-1)^n$ so that $$\sum_{k=0}^{d-1} \zeta(1+ a \exp(2 i \pi \frac kd))= \sum_{k=0}^{d-1} \sum_{n=0}^\infty \frac{F^{(n)}(1)}{n!} a^{n-1} \exp( i (n-1) 2\pi \frac kd) \\= \sum_{n=0}^\infty \frac{F^{(n)}(1)}{n!} a^{n-1} \underbrace{(\sum_{k=0}^{d-1} \exp( i (n-1) 2\pi \frac kd)}_{=0 \text{ if } \frac{n-1}{d} \not \in \mathbb{Z}} = d \sum_{n=0}^\infty \frac{F^{(nd+1)}}{(nd+1)!} a^{nd}$$

  • Dividing by $d$ and letting $d \to \infty$ we obtain the Fourier series theorem $$\frac{1}{2\pi} \int_0^{2\pi} \frac{F(1+a e^{it})}{a e^{it}} dt = \frac{1}{2\pi } \int_0^{2\pi} (\sum_{n=0}^\infty \frac{F^{(n)}(1)}{n!} a^{n-1} e^{int}) e^{-it} dt \\ =\sum_{n=0}^\infty \frac{F^{(n)}(1)}{n!}\frac{1}{2\pi} a^{n-1}\underbrace{\int_0^{2\pi} e^{i(n-1) t} dt}_{= 0 \text{ if } n - 1 \ne 0}= F'(1)$$

  • $\begingroup$ Thanks, since I've worked with the Taylor-series for the zeta I can follow the two first bullet-paragraphs. The last series-representation there looks like a nice idea - I need to chew on that a bit. The third one is something "out-of-the-blue" after that (I've also never had an introduction in Fourier series and besides its definition no familiarity/no trained experience with its applications and formulae) so I do not yet see, what the final result "$F'(1) $ equal ...things..." does give me at all. But thank you anyway so far! $\endgroup$ Jun 22 '17 at 7:04
  • $\begingroup$ @GottfriedHelms You need to read also a course on Fourier series. The last bullet says that since $f(t) = F(1+a e^{it})$ has a (uniformly convergent) Fourier series, we can recover its coefficients from $\int_0^{2\pi} f(t) e^{-int}dt$ $\endgroup$
    – reuns
    Jun 22 '17 at 7:26

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