This question is related to my previous question at the following link.
This question assumes the following definitions. My suspicion is $F(s)$ defined in (2) below converges (as $N\to\infty$) for $\Re(s)>-1$, but this depends on the definition of the Dirichlet Transform of $a(n)$.
(1) $\quad f(x)=\sum\limits_{n=1}^x\frac{\mu(n)}{n^2}\,\log\left(\frac{2\,\pi}{n}\right)$
(2) $\quad F(s)=s\int\limits_0^{\infty }f(x)\,x^{-s-1}\,dx=\sum\limits_{n=1}^N\frac{\mu(n)}{n^{s+2}}\,\log\left(\frac{2\,\pi}{n}\right),\,\quad N\to\infty$
I attempted to determine the Dirichlet Transform of $a(n)$ with the Mathematica evaluation illustrated in (3) below which produced the result illustrated in (4) and (5) below.
(3) $\quad\text{DirichletTransform}\left[\frac{\mu(n)}{n^2}\,\log\left(\frac{2\,\pi}{n}\right),n,s\right]$
(4) $\quad\frac{\log(2)+\log(\pi)+…\,\zeta(s+2)}{\zeta(s+2)}$
where
(5) $\quad…=\text{Hold}[\text{RuleCondition}[\text{Sum$\grave{ }$SumTableLookUpDump$\grave{ }$tabres},\text{FreeQ}[\text{Sum$\grave{ }$SumTableLookUpDump$\grave{ }$tabres},\text{$\$$Failed}]]]$
Question (1): What is the Dirichlet Transform of $a(n)=\frac{\mu(n)}{n^2}\,\log\left(\frac{2\,\pi}{n}\right)$?
The following figure illustrates $F(s)$ defined in (2) above evaluated at $N=101$ and $N=404$ in blue and orange respectively. The dashed-gray horizontal reference line is at $\log(2\,\pi)$. I believe $F(s)$ converges (as $N\to\infty$) to the Dirichlet Transform of $a(n)$ for $\Re(s)>-1$, but this depends on the definition of the Dirichlet Transform of $a(n)$.
Figure (1): Illustration of $F(s)$ evaluated at $N=101$ (blue curve) and $N=404$ (orange curve)
Best Answer
For $2+\Re(s) > \sup_\rho \Re(\rho)$ $$\sum_{n=1}^\infty\frac{\mu(n)}{n^2}(B-\log n) n^{-s} = \frac{B}{\zeta(s+2)}-\frac{\zeta'(s+2)}{\zeta(s+2)^2}$$
For $\Re(s) > 0$
$$= s \int_1^\infty (\sum_{n\le x}\frac{\mu(n)}{n^2}\log(\frac{e^B}{n})) x^{-s-1}dx$$ For $2+\Re(s) > \sup_\rho \Re(\rho)$
$$ \int_1^\infty (-C+\sum_{n\le x}\frac{\mu(n)}{n^2}\log(\frac{2\,\pi}{n})) x^{-s-1}dx= \frac{1}{s}(\frac{\log 2\pi}{\zeta(s+2)}-\frac{\zeta'(s+2)}{\zeta(s+2)^2}- C)$$
where $$C = \frac{\log 2\pi}{\zeta(2)}-\frac{\zeta'(2)}{\zeta(2)^2}$$
Also what should be clear to you is that if $\sum_{n=1}^\infty \nu(n) n^{-s}$ converges for $\Re(s) > \sigma$ and $c(n)-c(n+1) \ge 0$ for $n> N$ and $c(n) \to 0$ then there is a bound for $\Re(s) > \sigma$, $\forall n,|\sum_{m=1}^n \nu(m)m^{-s}|\le A(s) $ thus
$$|\sum_{n=1}^\infty \nu(n) n^{-s} c(n)| = |\sum_{n=1}^\infty (\sum_{m=1}^n \nu(m) m^{-s}) (c(n)-c(n+1))| \le \sum_{n=1}^\infty A(s) |c(n)-c(n+1)|\\ = A(s)\sum_{n=1}^N |c(n)-c(n+1)|+A(s)\sum_{n=N+1}^\infty (c(n)-c(n+1))\\=A(s)(c(N+1)+\sum_{n=1}^N |c(n)-c(n+1)|)$$ which implies the convergence and analyticity of $\sum_{n=1}^\infty \nu(n) n^{-s} c(n)$ and hence with $\nu(n) = \mu(n) n^{-2},c(n) = n^{-\epsilon}(-B+\log n)$ and $\nu(n) = \mu(n) n^{-2} \log(e^B/n), c(n) = \frac{-1}{B-\log n}$ you obtain the abscissa of convergence of $\sum_{n=1}^\infty\frac{\mu(n)}{n^2}(B-\log n) n^{-s} $ and $\sum_{n=1}^\infty\frac{\mu(n)}{n^2} n^{-s} $ are the same.