Mertens function

Using the Euler product, one finds that

${\displaystyle {\frac {1}{\zeta (s)}}=\prod _{p}(1-p^{-s})=\sum _{n=1}^{\infty }{\frac {\mu (n)}{n^{s}}},}$

where ${\displaystyle \zeta (s)}$ is the Riemann zeta function, and the product is taken over primes. Then, using this Dirichlet series with Perron's formula, one obtains

${\displaystyle {\frac {1}{2\pi i}}\int _{c-i\infty }^{c+i\infty }{\frac {x^{s}}{s\zeta (s)}}\,ds=M(x),}$

where c > 1.

Conversely, one has the Mellin transform

${\displaystyle {\frac {1}{\zeta (s)}}=s\int _{1}^{\infty }{\frac {M(x)}{x^{s+1}}}\,dx,}$

which holds for ${\displaystyle \operatorname {Re} (s)>1}$.

A curious relation given by Mertens himself involving the second Chebyshev function is

${\displaystyle \psi (x)=M\left({\frac {x}{2}}\right)\log 2+M\left({\frac {x}{3}}\right)\log 3+M\left({\frac {x}{4}}\right)\log 4+\cdots .}$

Assuming that the Riemann zeta function has no multiple non-trivial zeros, one has the "exact formula" by the residue theorem:

${\displaystyle M(x)=\sum _{\rho }{\frac {x^{\rho }}{\rho \zeta '(\rho )}}-2+\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}(2\pi )^{2n}}{(2n)!n\zeta (2n+1)x^{2n}}}.}$

Weyl conjectured that the Mertens function satisfied the approximate functional-differential equation

${\displaystyle {\frac {y(x)}{2}}-\sum _{r=1}^{N}{\frac {B_{2r}}{(2r)!}}D_{t}^{2r-1}y\left({\frac {x}{t+1}}\right)+x\int _{0}^{x}{\frac {y(u)}{u^{2}}}\,du=x^{-1}H(\log x),}$

where H(x) is the Heaviside step function, B are Bernoulli numbers, and all derivatives with respect to t are evaluated at t = 0.

There is also a trace formula involving a sum over the Möbius function and zeros of the Riemann zeta function in the form

${\displaystyle \sum _{n=1}^{\infty }{\frac {\mu (n)}{\sqrt {n}}}g(\log n)=\sum _{\gamma }{\frac {h(\gamma )}{\zeta '(1/2+i\gamma )}}+2\sum _{n=1}^{\infty }{\frac {(-1)^{n}(2\pi )^{2n}}{(2n)!\zeta (2n+1)}}\int _{-\infty }^{\infty }g(x)e^{-x(2n+1/2)}\,dx,}$

where the first sum on the right-hand side is taken over the non-trivial zeros of the Riemann zeta function, and (g, h) are related by the Fourier transform, such that

${\displaystyle 2\pi g(x)=\int _{-\infty }^{\infty }h(u)e^{iux}\,du.}$

Another formula for the Mertens function is

${\displaystyle M(n)=-1+\sum _{a\in {\mathcal {F}}_{n}}e^{2\pi ia},}$

where ${\displaystyle {\mathcal {F}}_{n}}$ is the Farey sequence of order n.

This formula is used in the proof of the Franel–Landau theorem.^[3]

M(n) is the determinant of the n × n Redheffer matrix, a (0, 1) matrix in which a_ij is 1 if either j is 1 or i divides j.

${\displaystyle M(x)=1-\sum _{2\leq a\leq x}1+{\underset {ab\leq x}{\sum _{a\geq 2}\sum _{b\geq 2}}}1-{\underset {abc\leq x}{\sum _{a\geq 2}\sum _{b\geq 2}\sum _{c\geq 2}}}1+{\underset {abcd\leq x}{\sum _{a\geq 2}\sum _{b\geq 2}\sum _{c\geq 2}\sum _{d\geq 2}}}1-\cdots }$

This formulation^{[citation needed]} expanding the Mertens function suggests asymptotic bounds obtained by considering the Piltz divisor problem, which generalizes the Dirichlet divisor problem of computing asymptotic estimates for the summatory function of the divisor function.

From ^[4] we have

${\displaystyle \sum _{d=1}^{n}M(\lfloor n/d\rfloor )=1\ .}$

Furthermore, from ^[5]

${\displaystyle \sum _{d=1}^{n}M(\lfloor n/d\rfloor )d=\Phi (n)\ ,}$

where ${\displaystyle \Phi (n)}$ is the totient summatory function.

Neither of the methods mentioned previously leads to practical algorithms to calculate the Mertens function. Using sieve methods similar to those used in prime counting, the Mertens function has been computed for all integers up to an increasing range of x.^[6][7]

The Mertens function for all integer values up to x may be computed in O(x log log x) time. A combinatorial algorithm has been developed incrementally starting in 1870 by Ernst Meissel,^[9] Lehmer,^[10] Lagarias-Miller-Odlyzko,^[11] and Deléglise-Rivat^[12] that computes isolated values of M(x) in O(x^2/3(log log x)^1/3) time; a further improvement by Harald Helfgott and Lola Thompson in 2021 improves this to O(x^3/5(log x)^3/5+ε),^[13] and an algorithm by Lagarias and Odlyzko based on integrals of the Riemann zeta function achieves a running time of O(x^1/2+ε).^[14]

See OEIS: A084237 for values of M(x) at powers of 10.

Ng notes that the Riemann hypothesis (RH) is equivalent to

${\displaystyle M(x)=O\left({\sqrt {x}}\exp \left({\frac {C\cdot \log x}{\log \log x}}\right)\right),}$

for some positive constant ${\displaystyle C>0}$. Other upper bounds have been obtained by Maier, Montgomery, and Soundarajan assuming the RH including

${\displaystyle {\begin{aligned}|M(x)|&\ll {\sqrt {x}}\exp \left(C_{2}\cdot (\log x)^{\frac {39}{61}}\right)\\|M(x)|&\ll {\sqrt {x}}\exp \left({\sqrt {\log x}}(\log \log x)^{14}\right).\end{aligned}}}$

Known explicit upper bounds without assuming the RH are given by:^[15]

${\displaystyle {\begin{aligned}|M(x)|&<{\frac {12590292\cdot x}{\log ^{236/75}(x)}},\ {\text{ for }}x>\exp(12282.3)\\|M(x)|&<{\frac {0.6437752\cdot x}{\log x}},\ {\text{ for }}x>1.\end{aligned}}}$

It is possible to simplify the above expression into a less restrictive but illustrative form as:

${\displaystyle {\begin{aligned}M(x)=O\left({\frac {x}{\log ^{\pi }(x)}}\right).\end{aligned}}}$

• Perron's formula
• Liouville's function