# Logarithmic derivative of Riemann zeta, is this derivation correct?

Let matrix $T_2$ be defined below as the Dirichlet inverse of the Euler totient function as a function of the Greatest Common Divisor (GCD) of row index $n$ and column index $k$;

$$T_2(n,k) = a(GCD(n,k))$$

where:

$$a(n) = \lim\limits_{s \rightarrow 1} \zeta(s)\sum\limits_{d|n} \mu(d)(e^{d})^{(s-1)}$$

starts:

$$\displaystyle T_2 = \begin{bmatrix} +1&+1&+1&+1&+1&+1&+1&\cdots \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&-2&+1&+1&-2&+1 \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&+1&+1&-4&+1&+1 \\ +1&-1&-2&-1&+1&+2&+1 \\ +1&+1&+1&+1&+1&+1&-6 \\ \vdots&&&&&&&\ddots \end{bmatrix}$$

By the answer at Mathematics Stackexchange given by joriki, we know that the Dirichlet series in the rows and columns tend to the von Mangoldt function $\Lambda(n)$ and $\Lambda(k)$:

$$\displaystyle \begin{bmatrix} \frac{T_2(1,1)}{1 \cdot 1}&+\frac{T_2(1,2)}{1 \cdot 2}&+\frac{T_2(1,3)}{1 \cdot 3}+&\cdots&+\frac{T_2(1,k)}{1 \cdot k} \\ \frac{T_2(2,1)}{2 \cdot 1}&+\frac{T_2(2,2)}{2 \cdot 2}&+\frac{T_2(2,3)}{2 \cdot 3}+&\cdots&+\frac{T_2(2,k)}{2 \cdot k} \\ \frac{T_2(3,1)}{3 \cdot 1}&+\frac{T_2(3,2)}{3 \cdot 2}&+\frac{T_2(3,3)}{3 \cdot 3}+&\cdots&+\frac{T_2(3,k)}{3 \cdot k} \\ \vdots&\vdots&\vdots&\ddots&\vdots \\ \frac{T_2(n,1)}{n \cdot 1}&+\frac{T_2(n,2)}{n \cdot 2}&+\frac{T_2(n,3)}{n \cdot 3}+&\cdots&+\frac{T_2(n,k)}{n \cdot k} \end{bmatrix} = \begin{bmatrix} \frac{\infty}{1} \\ +\frac{\Lambda(2)}{2} \\ +\frac{\Lambda(3)}{3} \\ \vdots \\ +\frac{\Lambda(n)}{n} \end{bmatrix}$$ $$=\;\;\;\;\;\;\;\;\;\;\;$$ $$\displaystyle \begin{bmatrix} \frac{\infty}{1}&+\frac{\Lambda(2)}{2}&+\frac{\Lambda(3)}{3}+&\cdots&+\frac{\Lambda(k)}{k} \end{bmatrix}\;\;\;\;\;\;\;\;\;\;\;$$

Also by this other answer by GH from MO we know that the von Mangoldt function can written as a Dirichlet generating function when the $s$ goes towards the value $1$. Numerically for non complex values of $s$ the following seems to be true:

$$\displaystyle \begin{bmatrix} \frac{T_2(1,1)}{1 \cdot 1^s}&+\frac{T_2(1,2)}{1 \cdot 2^s}&+\frac{T_2(1,3)}{1 \cdot 3^s}+&\cdots&+\frac{T_2(1,k)}{1 \cdot k^s} \\ \frac{T_2(2,1)}{2 \cdot 1^s}&+\frac{T_2(2,2)}{2 \cdot 2^s}&+\frac{T_2(2,3)}{2 \cdot 3^s}+&\cdots&+\frac{T_2(2,k)}{2 \cdot k^s} \\ \frac{T_2(3,1)}{3 \cdot 1^s}&+\frac{T_2(3,2)}{3 \cdot 2^s}&+\frac{T_2(3,3)}{3 \cdot 3^s}+&\cdots&+\frac{T_2(3,k)}{3 \cdot k^s} \\ \vdots&\vdots&\vdots&\ddots&\vdots \\ \frac{T_2(n,1)}{n \cdot 1^s}&+\frac{T_2(n,2)}{n \cdot 2^s}&+\frac{T_2(n,3)}{n \cdot 3^s}+&\cdots&+\frac{T_2(n,k)}{n \cdot k^s} \end{bmatrix} = \begin{bmatrix} \frac{\zeta(s)}{1} \\ +\frac{\zeta(s)\sum\limits_{d|2} \frac{\mu(d)}{d^{(s-1)}}}{2} \\ +\frac{\zeta(s)\sum\limits_{d|3} \frac{\mu(d)}{d^{(s-1)}}}{3} \\ \vdots \\ +\frac{\zeta(s)\sum\limits_{d|n} \frac{\mu(d)}{d^{(s-1)}}}{n} \end{bmatrix}$$ $$=\;\;\;\;\;\;\;\;\;\;\;$$ $$\displaystyle \begin{bmatrix} \frac{\infty}{1^s}&+\frac{\Lambda(2)}{2^s}&+\frac{\Lambda(3)}{3^s}+&\cdots&+\frac{\Lambda(k)}{k^s} \end{bmatrix}\;\;\;\;\;\;\;\;\;\;\;$$

but I don't know for sure. In order for the sum of columns sums to converge at all, the first term can not be allowed to be $\frac{\infty}{1}$. Therefore I set the terms in the first column equal to zero, and the matrix becomes:

$$\displaystyle \begin{bmatrix} \frac{0}{1 \cdot 1^s}&+\frac{T_2(1,2)}{1 \cdot 2^s}&+\frac{T_2(1,3)}{1 \cdot 3^s}+&\cdots&+\frac{T_2(1,k)}{1 \cdot k^s} \\ \frac{0}{2 \cdot 1^s}&+\frac{T_2(2,2)}{2 \cdot 2^s}&+\frac{T_2(2,3)}{2 \cdot 3^s}+&\cdots&+\frac{T_2(2,k)}{2 \cdot k^s} \\ \frac{0}{3 \cdot 1^s}&+\frac{T_2(3,2)}{3 \cdot 2^s}&+\frac{T_2(3,3)}{3 \cdot 3^s}+&\cdots&+\frac{T_2(3,k)}{3 \cdot k^s} \\ \vdots&\vdots&\vdots&\ddots&\vdots \\ \frac{0}{n \cdot 1^s}&+\frac{T_2(n,2)}{n \cdot 2^s}&+\frac{T_2(n,3)}{n \cdot 3^s}+&\cdots&+\frac{T_2(n,k)}{n \cdot k^s} \end{bmatrix} = \begin{bmatrix} \frac{\zeta(s)}{1}-\frac{1}{1} \\ +\frac{\zeta(s)\sum\limits_{d|2} \frac{\mu(d)}{d^{(s-1)}}}{2} -\frac{1}{2} \\ +\frac{\zeta(s)\sum\limits_{d|3} \frac{\mu(d)}{d^{(s-1)}}}{3} -\frac{1}{3} \\ \vdots \\ +\frac{\zeta(s)\sum\limits_{d|n} \frac{\mu(d)}{d^{(s-1)}}}{n} -\frac{1}{n} \end{bmatrix}$$

$$=\;\;\;\;\;\;\;\;\;\;\;$$ $$\displaystyle \begin{bmatrix} \frac{\Lambda(1)}{1^s}&+\frac{\Lambda(2)}{2^s}&+\frac{\Lambda(3)}{3^s}+&\cdots&+\frac{\Lambda(k)}{k^s} \end{bmatrix}\;\;\;\;\;\;\;\;\;\;\;$$

So if this is correct then the sum of row sums and the sum of column sums would be:

$$-\frac{\zeta'(s)}{\zeta(s)}=\sum\limits_{k=1}^{\infty} \frac{\Lambda(k)}{k^s} = \sum\limits_{n=1}^{\infty} \left(\zeta(s)\sum\limits_{d|n} \frac{\mu(d)}{d^{(s-1)}}-\frac{1}{n}\right)$$

as in wikipedia about the von Mangoldt function.

Is this correct?