Proof of the Riemann Hypothesis from a power-sum result under the density hypothesis

Details Proof of the Riemann Hypothesis from a power-sum result under the density hypothesis Proof of the Riemann Hypothesis from a power-sum result under the density hypothesis

2008-10-12

arxiv.org/abs/0810.2102v7

Mathematics

General Mathematics

20 pages, submitted on September 26, 2011

Scientific paper

The Riemann hypothesis is equivalent to the $\varpi$-form of the prime number theorem as $\varpi(x) =O(x\sp{1/2} \log\sp{2} x)$, where $\varpi(x) =\sum\sb{n\le x}\ \bigl(\Lambda(n) -1\big)$ with the sum running through the set of all natual integers. Let ${\mathsf Z}(s) = -\tfrac{\zeta\sp{\prime}(s)}{\zeta(s)} -\zeta(s)$. We use the classical integral formula for the Heaveside function in the form of ${\mathsf H}(x) =\int\sb{m -i\infty} \sp{m +i\infty} \tfrac{x\sp{s}}{s} \dd s$ where $m >0$, and ${\mathsf H}(x)$ is 0 when $x <1$, $\tfrac{1}{2}$ when x=1, and 1 when $x >1$. However, we diverge from the literature by applying Cauchy's residue theorem to the funtion ${\mathsf Z}(s) \cdot \tfrac{x\sp{s}} {s}$, rather than $-\tfrac{\zeta\sp{\prime}(s)} {\zeta(s)} \cdot \tfrac{x\sp{s}}{s}$, so that we may utilize the formula for $m <1$, under certain conditions. Starting with the estimate on $\varpi(x)$ from the trivial zero-free region $\sigma >1$ of ${\mathsf Z}(s)$, we use induction to reduce the size of the exponent $\theta$ in $\varpi(x) =O(x\sp{\theta} \log\sp{2} x)$, while we also use induction on $x$ when $\theta$ is fixed. We prove that the Riemann hypothesis is valid under the assumption of the density hypothesis.

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