Deriving the Riemann non-trivial zeros from $\zeta_{H}(s,a) + \zeta_{H}(s,1-a)$ - MathOverflow

Deriving the Riemann non-trivial zeros from $\zeta_{H}(s,a) + \zeta_{H}(s,1-a)$

2012-06-05T18:55:47Z

The Hurwitz zeta function:

$$\zeta_{H}(s,a)$$

reduces to $\zeta(s)$ when $a=1$ and to $(2^s-1)\zeta(s)$ when $a=\frac12$.

However, I stumbled upon a peculiar third connection:

$$\zeta_{H}(s,a) + \zeta_{H}(s,1-a)$$

that seems to exactly produce the non-trivial zeros of $\zeta(s)$,

when $a=\frac12$ (obviously), but also (and apparently only) when $a=\frac13, \frac14$ or $\frac16.$

Why does it only work for these values? Is there any reference to this in the literature?

Thanks.

2012-06-05T22:23:02Z

Found the answer, the question can be closed. answer

It boils down to:

$$\zeta_{H}(s,a) + \zeta_{H}(s,1-a) = \frac{4}{(2\pi)^{1-s}}\Gamma(1-s)C(1-s,a)$$

$$C(s,a)=\sum_{n=1}^\infty \frac{\cos(2n\pi a)}{n^s}$$

And $C(s,a)$ reducing to:

$a=\frac12 \rightarrow$ $(2^{1-s}-1)\zeta(s)$

$a=\frac13 \rightarrow$$\dfrac12(3^{1-s}-1)\zeta(s)$

$a=\frac14 \rightarrow$$2^{-s}(2^{1-s}-1)\zeta(s)$

$a=\frac16 \rightarrow$$\dfrac12(1-2^{1-s})(1-3^{1-s})\zeta(s)$

hence the non-trivial zeros.