24 Nov 2021


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

\[\Lambda(n) = \begin{cases} \log p & \text{ if }n=p^m\text{ is a prime power} \\ 0 & \text{ otherwise} \end{cases}\]

Our goal is to elucidate the connection between $\Lambda$, 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:

\[f(s) = \sum_{n=1}^\infty \frac{\Lambda(n)}{n^s}\]

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:

\[\frac{d}{ds}\log \zeta(s) = \frac{\zeta'(s)}{\zeta(s)}\]

Main Theorem

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

\[-\frac{\zeta'(s)}{\zeta(s)} = \sum_{n=1}^\infty \frac{\Lambda(n)}{n^s}\]


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

\[\zeta(s) = \prod_{p\in \mathscr{P}} \frac{1}{1-p^{-s}}\]

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

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

\[\log \zeta(s) = \log \prod_{p\in \mathscr{P}} \frac{1}{1-p^{-s}} = -\sum_{p\in \mathscr{P}} \log(1-p^{-s})\]

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:

\[\begin{align*} -\frac{\zeta'(s)}{\zeta(s)} &= -\frac{d}{ds}\log \zeta(s) \\ &= \frac{d}{ds} \sum_{p\in \mathscr{P}} \log(1-p^{-s}) \\ &= \sum_{p\in \mathscr{P}} \frac{d}{ds} \log(1-p^{-s}) \\ &= \sum_{p\in \mathscr{P}} (1-p^{-s})^{-1}(- p^{-s})(-\log p) \\ &= \sum_{p\in \mathscr{P}} \log p \frac{p^{-s}}{1-p^{-s}} \end{align*}\]

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:

\[\frac{1}{1-p^{-\sigma}} = \sum_{m=0}^\infty (p^{-\sigma})^m = \sum_{m=0}^\infty (p^m)^{-\sigma}\]

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

\[\frac{1}{1-p^{-s}} = \sum_{m=0}^\infty (p^{-s})^m = \sum_{m=0}^\infty (p^m)^{-s}\]

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 $p^{-s}$:

\[\frac{p^{-s}}{1-p^{-s}} = \sum_{m=1}^\infty (p^m)^{-s}\]

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

\[\begin{align*} -\frac{\zeta'(s)}{\zeta(s)} &= \sum_{p\in \mathscr{P}} \log p \frac{p^{-s}}{1-p^{-s}} \\ &= \sum_{p\in \mathscr{P}} \log p \sum_{m=1}^\infty (p^m)^{-s} \\ &= \sum_{p\in \mathscr{P}} \sum_{m=1}^\infty \frac{\log p}{(p^m)^s} \end{align*}\]

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:

\[\sum_{p\in \mathscr{P}} \sum_{m=1}^\infty \frac{\log p}{(p^m)^s} = \sum_{n=p^m} \frac{\log p}{n^s}\]

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 $\Lambda(n)=0$ 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:

\[\sum_{n=p^m} \frac{\log p}{n^s} = \sum_{n=p^m} \frac{\Lambda(n)}{n^s} = \sum_{n=1}^\infty \frac{\Lambda(n)}{n^s}\]

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

$ \blacksquare $