# What is the Dirichlet Transform of $a(n)=\frac{\mu(n)}{n^2}\,\log\left(\frac{2\,\pi}{n}\right)$?

This question is related to my previous question at the following link.

Do the Prime Number Theorem and/or Riemann Hypothesis predict a limit on the accuracy of this formula for $$\gamma$$?

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)

## 1 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.

• I suspect part of my confusion with respect to your answers to my original question and my more specific question here may perhaps be related to a sign problem. I think $\frac{B}{\zeta (s+2)}+\frac{\zeta '(s+2)}{\zeta (s+2)^2}$ should perhaps be $\frac{B}{\zeta (s+2)}-\frac{\zeta '(s+2)}{\zeta (s+2)^2}$. – Steven Clark Mar 19 at 20:59
• right I missed a sign but it doesn't change anything. see the paragraph I added – reuns Mar 19 at 22:47