Marcel Riesz defined a function :

$R(x) = \sum_{n=1}^\infty \frac {(-1)^n x^n} {\zeta(2n)\Gamma(n)}$

The Riemann hypothesis holds if $R(x)= O( x^{1/4 + {\varepsilon}}$) For any $\varepsilon$

We have a relatively trivial bound $O(x^{1/2})$

Also various standard properties can be extracted from basic zeta function theory .


(1) Does exponent between (1/4) and (1/2) for bound exists. i.e is there a bound better than $x^{1/2}$ but weaker than $x^{1/4 + {\varepsilon}}$?

(2) If such intermediate bound is achieved , what equivalent property about zeroes of Zeta function would be proved ?

(3) $R(x)$ looks lot like Taylor or any other symmetrical functional expansion . Is there any possible compact form for $R(x)$?

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    $\begingroup$ Online on MO after long time , anything new on the question ? $\endgroup$
    – TPC
    Mar 31, 2020 at 21:24
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    $\begingroup$ The formula 2.4 ($\frac{\Gamma(1-s/2)}{\zeta(s)}=\int_0^{\infty}x^{-s/2-1}R(x)dx$) in Broughan Equivalents of the Riemann Hypothesis II clearly shows that any estimate $R(x)= O( x^{c + {\varepsilon}}), 1/4 \le c <1/2$ gives a non zero halfplane $\Re s >2c$ for the RZ $\endgroup$
    – Conrad
    Feb 14, 2021 at 16:43
  • $\begingroup$ @Conrad The integral associated with the Mellin transform of $x^n$ doesn't converge (the result is actually a distribution). Was a different formula than the OP's for $R(x)$ used to evaluate the Mellin transform integral? $\endgroup$ Feb 14, 2021 at 21:32
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    $\begingroup$ @Steven not sure I understand the question; it is known (and easy to prove by Mobius inversion) that $R(x)=O(x^{1/2+\epsilon})$ for large $x$ and $R(x) << x$ for small $0<x<1$ which means that $x^{-s/2-1}R(x)$ is integrable at both ends for $\Re s \in [1+\delta, 2-\delta], \epsilon < \delta <1/2$ and it is a matter of computation to show that the Mellin transform is $\Gamma (1-s/2)/\zeta(s)$ while if we have a better estimate for large $x$ we extend the left integrability to $2c+\delta, 1/4 \le c < 1/2$ so extend the zero free region of $\zeta$ to $\Re s >2c$ as $\delta$ is arbitrary $\endgroup$
    – Conrad
    Feb 14, 2021 at 21:46
  • $\begingroup$ @Conrad I don't believe the Mellin transform can be derived term-wise from the OP's formula, but I see now the Mellin transform can be derived term-wise from the alternate formula $R(x)=x\sum\limits_{n=1}^\infty\frac{\mu(n)}{n^2} e^{-\frac{x}{n^2}}$. $\endgroup$ Feb 14, 2021 at 22:24


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