On infinite sum containing logarithmic derivative of Zeta function and Möbius function:

Question

Consider the following function:

$$F(s)= \sum_m \mu(m) \sum_n \frac{e^{-n/2}\zeta^\prime (mns)}{n \zeta(mns)}$$

Now, we can see, that function has simple poles ${\left[\frac{1}{n}\right]}_{n=1}^\infty$ due to each log derivative of Zeta factor and singularities ${\left[\frac{\alpha_t+i\beta_t}{n}\right]}_{n=1}^\infty$ belonging to each Zeta at denominator where $\alpha_t+i\beta_t$ is zero of $\zeta(s)$.i.e. imaginary axis is 'natural boundary'.

Also I can see some other elementary properties of this.

I'd like to compute the following:

$$Res_{s=\frac{1}{2}} \frac{\int F(s)ds+(\frac{1}{4e}-\frac{1}{2√e})\ln(s-\frac{1}{2})}{(s-\frac{1}{2})²}$$

I think this is equal to zero . I need more insight into this. Any comments regarding this are welcome.

As we can see $\int F(s)ds$ has logarithmic singularity at s=1/2 so, the log term in the numerator is to cancel that singularity.

Used Laurent series of Log derivative of Zeta in $ F(s) $:

i.e.

$$-\frac{\zeta^\prime}{\zeta}(s)=\frac{1}{s-1}+\sum_{j=0}^\infty \frac{(-1)^j}{j!}\eta_j (s-1)^j.$$

Now,

$$F(s)= \left [\sum_{j=0}^\infty \frac{(-1)^{j+1}}{j!}\eta_j \sum_m{\mu(m)} m^j \sum_n e^{-n/2}n^{k-1}\left(s-\frac{1}{mn}\right)^j\right]_1$$

$$-\left[\sum_m\frac{\mu(m)}{m}\sum_n \frac{e^{-n/2}}{n^2\left(s-\frac{1}{mn}\right)}\right]_2 $$

References:

[1] W.T.Ross, H.S.Shapiro Generalized Analytic Continuation

[2] Kimoto,Wakayama Remarks on zeta regularized products

Answer 1

Your question doesn't make much sense. Let $a_d =\sum_{m|d} \mu(m) \frac{e^{-(d/m)/2}}{d/m}=O(1)$. As $\frac{\zeta'(s)}{\zeta(s)}=O(2^{-s})$ then for $\Re(s) >0$, $F(s) = \sum_{d\ge 1} a_d\frac{\zeta'(ds)}{\zeta(ds)}$ converges locally uniformly, it is meromorphic, with simple poles at $1/d$ and $\rho/d$ accumulating to the right of $\Re(s)=0$ (we need to know there are infinitely many non-trivial zeros) making it a natural boundary.