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Let $\mu$ denote the Möbius function, and define the the Mertens function $M(x) = \sum_{n \leq x} \mu(n)$. By Person's formula, one can express $M(x)$ as a sum over the nontrivial zeros of the Riemann zeta function. However, in every book or paper I have come across, this is done assuming the simplicity of the zeta zeros.

My question: is there an expression for $M(x)$ involving a sum over the Riemann zeros, that can be obtained without assuming the simplicity of the zeros ?

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    $\begingroup$ Sure; you just write $\mathrm{Res}_{s = \rho} \frac{1}{s\zeta(s)}$ instead of $\frac{1}{\rho \zeta'(\rho)}$ in the sum over zeroes. $\endgroup$
    – Peter Humphries
    Commented Jun 1, 2021 at 20:55
  • $\begingroup$ @Peter Humpries...thanks...but what is the expression of the residue at a multiple zero ? $\endgroup$
    – user257465
    Commented Jun 1, 2021 at 21:04
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    $\begingroup$ It depends on the multiplicity, but it's just basic complex analysis; if the zero is of order $m \geq 2$, then write out $\frac{1}{s\zeta(s)}$ as a Laurent series of the form $a_{-m} (s - \rho)^{-m} + a_{-m + 1} (s - \rho)^{-m + 1} + \cdots$. The residue is $a_{-1}$, which can be explicitly written down in terms of $\zeta^{(j)}(\rho)$ by writing $\frac{1}{s} = \sum_{j=0}^{\infty} (-1)^j \rho^{-j-1} (s - \rho)^j$ and $\zeta(s) = \sum_{j = m}^{\infty} \frac{\zeta^{(j)}(\rho)}{j!} (s - \rho)^j$ in terms of their Taylor series expansions... $\endgroup$
    – Peter Humphries
    Commented Jun 1, 2021 at 21:18
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    $\begingroup$ ...then invert the Taylor series to get the Laurent series expansion for $\frac{1}{\zeta(s)}$, multiply these two expansions together, and extract out the coefficient of $(s - \rho)^{-1}$. I don't see why you would ever want to do this though! Best just to leave it as $\mathrm{Res}_{s = \rho} \frac{1}{s\zeta(s)}$. $\endgroup$
    – Peter Humphries
    Commented Jun 1, 2021 at 21:20
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    $\begingroup$ Since we are talking about residues at poles and not essential singularities, there is a "closed form" version of Peter's comment: If $\rho$ is a zero of $\zeta(s)$ with order $m\geq 1$, then $\mathop{\mathrm{Res}}_{s=\rho}\frac{1}{s\zeta(s)} = \frac{1}{(m-1)!} \lim_{s\to\rho} \frac{d^{m-1}}{ds^{m-1}} \frac{(s-\rho)^m}{s\zeta(s)}$. These derivatives start to become unpleasant at $m=2$. I can't think of anyone who would want to see them computed in their entirety. $\endgroup$
    – 2734364041
    Commented Jun 2, 2021 at 1:26

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