Editing Mertens conjecture (section)

== Connection to the Riemann hypothesis ==

The connection to the Riemann hypothesis is based on the [[Dirichlet series]]
for the reciprocal of the [[Riemann zeta function]],

:<math>\frac{1}{\zeta(s)} = \sum_{n=1}^\infty \frac{\mu(n)}{n^s},</math>

valid in the region <math>\mathcal{Re}(s) > 1</math>. We can rewrite this as a 
[[Stieltjes integral]]

:<math>\frac{1}{\zeta(s)} = \int_0^\infty x^{-s} dM(x)</math>

and after integrating by parts, obtain the reciprocal of the zeta function
as a [[Mellin transform]]

:<math>\frac{1}{s \zeta(s)} = \left\{ \mathcal{M} M \right\}(-s)
= \int_0^\infty x^{-s} M(x)\, \frac{dx}{x}.</math>

Using the [[Mellin inversion theorem]] we now can express {{mvar|M}} in terms of {{frac|1|{{mvar|&zeta;}}}} as

:<math>M(x) = \frac{1}{2 \pi i} \int_{\sigma-i\infty}^{\sigma+i\infty} \frac{x^s}{s \zeta(s)}\,ds</math>

which is valid for {{math|1 < &sigma; < 2}}, and valid for {{math|{{frac|1|2}} < &sigma; < 2}} on the Riemann hypothesis. 
From this, the Mellin transform integral must be convergent, and hence
{{math|''M''(''x'')}} must be {{math|''O''(''x''<sup>e</sup>)}} for every exponent ''e'' greater than {{sfrac|1|2}}. From this it follows that 
:<math>M(x) = O\Big(x^{\tfrac{1}{2} + \epsilon}\Big)</math>

for all positive {{mvar|&epsilon;}} is equivalent to the Riemann hypothesis, which therefore would have followed from the stronger Mertens hypothesis, and follows from the hypothesis of Stieltjes that
:<math>M(x) = O\Big(x^\tfrac{1}{2}\Big).</math>