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In a comment under this answer, a user boldly asserts that there is ONLY ONE way to prove that $$ \left. \frac d {ds}\,\frac 1 {\zeta(s)} \right|_{s=1} = 1 $$ where $\zeta$ is Riemann's zeta function.

So my question is: what examples of elementary ways of proving this best illustrate whatever diversity of such methods exists?

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Michael Hardy
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    First insight: the ways to prove must be countable, unless you allow infinite proofs. Hence the elementary ways must be countable as well. –  Feb 20 '16 at 19:26
  • What counts as elementary? The easiest way, I think, is to show that $\zeta$ has a simple pole with residue $1$ at $1$. – Daniel Fischer Feb 20 '16 at 19:30
  • @DanielFischer : I'd certainly count that as elementary if it can be shown in a way that will be understood by those who know the definitions of the terms used in your comment. $\qquad$ – Michael Hardy Feb 20 '16 at 19:32
  • This comment shows the integral. One would add a few more details in an answer of course. This answer linked by robjohn is IMO more complicated, but can still be counted as elementary. – Daniel Fischer Feb 20 '16 at 19:42

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this is more an explanation of intuitively how do I think to it than an answer.

this is mainly a question on how do we define $\mu(n)$. there are only two ways to define $\mu(n) $ (there are many others but equivalent to these) :

  • 1) as the Dirichlet convolution inverse of the constant sequence $1$ $$\left(\sum_{n=1}^\infty \mu(n) n^{-s} \right) \left(\sum_{m=1}^\infty m^{-s} \right) = 1$$

$$\text{or equivalently } \quad\sum_{d \mid n} \mu(d) = \delta_{n=1}$$

  • 2) as the function being $\pm 1$ for square-free integers $$\mu(1) = 1, \qquad \mu(n) = \begin{cases} (-1)^k & \text{ if } n = \prod_{i=1}^k p_i, \text{ the $p_i$ being different primes}, \\[4pt] 0 & \text{otherwise} \end{cases}$$

of course these definitions are shown equivalent by the Euler product $$\zeta(s) = \sum_{n=1}^\infty n^{-s} =\prod_p \frac{1}{1-p^{-s}}$$ the question now is : if we define $\mu(n)$ with 1) or 2) how can we prove that $$\left.\frac{d \ 1/\zeta(s)}{ds}\right|_{s=1} = 1 \tag{§} $$

  • 1) defines $\mu(n)$ as the Dirichlet convolution inverse of $1$ so we will have to prove that $$\zeta(s) = \frac{1}{s-1}+\mathcal{O}(1)\qquad \text{when }s\to 1 \tag{§§}$$

  • now if we define $\mu(n)$ with 2), can we prove $(§)$ without relying also on 1) and hence on $(§§)$?

    I think that for proving $(§)$ only from 2) we would need something like the prime number theorem. the original (analytic) proof of the PNT clearly relies on 1) and $(§§)$.

    so our only hope is with the ''elementary'' proof of the PNT, but it also relies on the definition 1), hence you cannot prove $(§)$ without using 1) in your proof, and your proof of $(§)$ will be somewhat equivalent to proving $(§§)$.

Michael Hardy
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