24 Nov 2021

Introduction

In what follows, let $p$ denote a prime number. The von Mangoldt function $\mathrm{\Lambda }$ is defined by the following:

$$\mathrm{\Lambda }(n)=\{\begin{array}{ll}\log p& \text{ if }n=p^m\text{ is a prime power}\\ 0& \text{ otherwise}\end{array}$$

Our goal is to elucidate the connection between $\mathrm{\Lambda }$, the Riemann zeta function $\zeta$, and the prime numbers. Today, we will consider the classical Dirichlet series whose coefficients are given by the von Mangoldt function:

$$f(s)=\sum _{n=1}^{\mathrm{\infty }}\frac{\mathrm{\Lambda }(n)}{n^s}$$

We will show that this series is exactly given by (the negative of) the logarithmic derivative of the Riemann zeta function. This logarithmic derivative is given by:

$$\frac{d}{ds}\log \zeta (s)=\frac{{\zeta }^{\prime }(s)}{\zeta (s)}$$

Main Theorem

The Dirichlet series whose coefficients are prescribed by the von Mangoldt function $\mathrm{\Lambda }(n)$ is precisely the opposite of the logarithmic derivative of the Riemann zeta function $\zeta$. To wit:

$$-\frac{{\zeta }^{\prime }(s)}{\zeta (s)}=\sum _{n=1}^{\mathrm{\infty }}\frac{\mathrm{\Lambda }(n)}{n^s}$$

Proof

By the Euler product formula for the Riemann zeta function, we have:

$$\zeta (s)=\prod _{p\in \mathcal{P}}\frac{1}{1-p^{-s}}$$

where the product is taken over all primes $p$, and absolutely converges when $\sigma =\mathrm{\Re }(s)>1$.

Because the product is absolutely convergent, the following equivalent identity holds for the logarithm of the product:

$$\log \zeta (s)=\log \prod _{p\in \mathcal{P}}\frac{1}{1-p^{-s}}=-\sum _{p\in \mathcal{P}}\log (1-p^{-s})$$

Further, this latter series is absolutely convergent when $\sigma >1$. In particular, it follows that this sum is holomorphic in this domain since it is an absolutely convergent sum comprised of functions holomorphic in said half-plane. (Absolute convergence of the series is enough to ensure that the partial summands converge locally uniformly to a function which is holomorphic by a theorem of Weierstrauss in complex analysis.)

Therefore, term-by-term differentiation is valid for such an absolutely convergent holomorphic series. Thus we compute:

$$\begin{array}{rl}-\frac{{\zeta }^{\prime }(s)}{\zeta (s)}& =-\frac{d}{ds}\log \zeta (s)\\ & =\frac{d}{ds}\sum _{p\in \mathcal{P}}\log (1-p^{-s})\\ & =\sum _{p\in \mathcal{P}}\frac{d}{ds}\log (1-p^{-s})\\ & =\sum _{p\in \mathcal{P}}(1-p^{-s}{)}^{-1}(-p^{-s})(-\log p)\\ & =\sum _{p\in \mathcal{P}}\log p\frac{p^{-s}}{1-p^{-s}}\end{array}$$

Next, we shall note the fact that $\frac{1}{p^{\sigma }}<1$ for any prime $p$ and real number $\sigma >1$. In particular, these ensure that the following series is absolutely convergent:

$$\frac{1}{1-p^{-\sigma }}=\sum _{m=0}^{\mathrm{\infty }}(p^{-\sigma }{)}^m=\sum _{m=0}^{\mathrm{\infty }}(p^m{)}^{-\sigma }$$

Therefore, the complex valued series identity also holds when $\sigma =\mathrm{\Re }(s)>1$:

$$\frac{1}{1-p^{-s}}=\sum _{m=0}^{\mathrm{\infty }}(p^{-s}{)}^m=\sum _{m=0}^{\mathrm{\infty }}(p^m{)}^{-s}$$

The corresponding sum of absolute values of its terms is exactly the previous identity. We tweak the identity slightly by multiplying with the extra factor of $p^{-s}$:

$$\frac{p^{-s}}{1-p^{-s}}=\sum _{m=1}^{\mathrm{\infty }}(p^m{)}^{-s}$$

Using this fact, we now return to our previous calculation. We find:

$$\begin{array}{rl}-\frac{{\zeta }^{\prime }(s)}{\zeta (s)}& =\sum _{p\in \mathcal{P}}\log p\frac{p^{-s}}{1-p^{-s}}\\ & =\sum _{p\in \mathcal{P}}\log p\sum _{m=1}^{\mathrm{\infty }}(p^m{)}^{-s}\\ & =\sum _{p\in \mathcal{P}}\sum _{m=1}^{\mathrm{\infty }}\frac{\log p}{(p^m{)}^s}\end{array}$$

Now, this double sum has the precise effect of summing over every prime power of the form $p^m$ for some prime $p$ and positive integer $m$. This double sum can be interpreted as a particular ordering of a more general unordered sum over the countable set of all prime powers. As is proved here on another post, absolute convergence of one ordering of a sum ensures that any other rearrangement is also absolutely convergent. Using this fact, we may write:

$$\sum _{p\in \mathcal{P}}\sum _{m=1}^{\mathrm{\infty }}\frac{\log p}{(p^m{)}^s}=\sum _{n=p^m}\frac{\log p}{n^s}$$

where the latter sum is over natural numbers which are prime powers of the form $p^m$, with positive integer $m$.

By using the von Mangoldt function, we can convert this sum to a Dirichlet series which sums over the natural numbers. As $\mathrm{\Lambda }(n)=0$ for any term which is not a prime power, the two series will have exactly the same terms with the same weights. Thus, we may write:

$$\sum _{n=p^m}\frac{\log p}{n^s}=\sum _{n=p^m}\frac{\mathrm{\Lambda }(n)}{n^s}=\sum _{n=1}^{\mathrm{\infty }}\frac{\mathrm{\Lambda }(n)}{n^s}$$

Therfore, we may complete the proof of the identity by combining our previous calculations with these two identities.

$◼$