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 want to find following residue :
$$Res_{s=\frac{1}{2}} \frac{\int F(s)ds+(\frac{1}{4e}-\frac{1}{2√e})\ln(s-\frac{1}{2})}{(s-\frac{1}{2})²}$$
I think the value of this residue is zero but it's under Riemann hypothesis. Need more insight into this
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.
Is $G(s)=\int F(s)ds+(\frac{1}{4e}-\frac{1}{2√e})\log(s-\frac{1}{2})$ analytic at $s=1/2$ ? If so $\frac{G(s)-G(1/2)}{s-1/2}$ is analytic at $1/2$ so $$Res(\frac{G(s)-G(1/2)}{s-1/2},1/2)=0, \qquad Res(\frac{G(s)}{s-1/2},1/2)=Res(\frac{G(1/2)}{s-1/2},1/2)=G(1/2)$$ Note that $G(1/2)$ depends on the constant of integration in $\int F(s)ds$ (as well as the branch of $\log(s-1/2)$)