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The following equation is for the reciprocal of the zeta function at $Re(s) > 1$:

$$ \frac{1}{\zeta(s)} = \sum_{n=1}^{\infty} \frac{\mu(n)}{n^{s}} = s \int_{1}^{\infty} \frac{M(x)}{x^{1+s}}dx, \ M(x) = \sum_{n \leq x} \mu(n) $$

Is there a similar equation for the Hurwitz zeta function $Re(s) > 1$:

$$ \frac{1}{\zeta(s,a)} $$

jcin
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  • First, I don't think you're using Riemann hypothesis, your formulas make sense when $Re(s)>1$. Second, what do you mean by equation of $F(x)$? – Julian Mejia May 28 '19 at 00:49
  • When $Re(s) < 1$ there is a formula for M(x) using the zeros of the zeta function as the residue theorem is applied to the inverse mellin transform. link M(x), Mertens function, for $ Re(s) > 1 $ involves $\mu(n)$. Is there a similar function for the reciprocal of the Hurwitz zeta function? I removed the reference to the Riemann hypothesis to avoid confusion and added clarification for s. – jcin May 28 '19 at 00:55
  • Perhaps OP means that it’s not known whether or not the right side converges for $1/2\leq Re(s)\leq 1$. – Alex R. May 28 '19 at 00:56
  • It seems like there’s no obvious analogue for the Hurwitz Zeta function. In this paper: http://www.jams.or.jp/scm/contents/e-2009-2/2009-13.pdf

    An analogue is defined via $R(s,x)$. See section 3.

     – Alex R. May 28 '19 at 01:03
  • $\mu$ appears because $\mu$ is the inverse of $1$ in the "algebra" of arithmetic functions with the convolution as operation, i.e. $\mu\ast 1=\delta$. But in the Hurwitz zeta function, you lose arithmetic information, this function is no longer a usual Dirichlet series. – Julian Mejia May 28 '19 at 04:47
  • Would it be possible to represent the Hurwitz zeta function as a general Dirichlet series, find the reciprocal and use the inverse Mellin transform? – jcin May 28 '19 at 05:07

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For $q \ge 1,a \ge 1$ integers $$q^{-s}\zeta(s,a/q) = \sum_{n=0}^\infty (qn+a)^{-s}= \sum_{n=1}^\infty c_n n^{-s}$$ is a Dirichlet series but it has no (Euler-like) product so there is no simple expression for its $\log$ and inverse.

Iff $a = 1$ then $c_1 =1\ne 0$ so that $$\frac{1}{q^{-s} \zeta(s,a/q)}= \frac{1}{1-(-\sum_{n=2}^\infty c_n n^{-s})}=1+\sum_{k=1}^\infty (-\sum_{n=2}^\infty c_n n^{-s})^k= 1+\sum_{k=1}^\infty\sum_{n= 2^k}^\infty b_{k,n} n^{-s}$$ is a Dirichlet series.

For $a \ne 1$ the inverse is a generalized Dirichlet series.

If $a/q \not \in \Bbb{Q}$ then both $\zeta(s,a/q), \frac{1}{\zeta(s,a/q)}$ are generalized Dirichlet series.

reuns
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