Mellin-Perron inverse transform.

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SUMMARY

The discussion focuses on the application of the Mellin-Perron inverse transform to derive an asymptotic formula for the Mertens function, utilizing the Laurent series of the Riemann zeta function. The key equations presented include the Laurent series representation of the zeta function and the Mellin-Perron inverse formula. The user seeks to establish an asymptotic formula for M(exp(t)) based on the derived series. A reference to a relevant resource is provided for further exploration of the topic.

PREREQUISITES
  • Understanding of the Riemann zeta function and its properties
  • Familiarity with Laurent series and their applications
  • Knowledge of the Mellin-Perron inverse transform
  • Basic concepts of asymptotic analysis in number theory
NEXT STEPS
  • Study the derivation of the Laurent series for 1/ζ(s)
  • Explore the application of the Mellin-Perron inverse formula in number theory
  • Investigate asymptotic formulas related to the Mertens function
  • Review Chapter 12 of the provided resource for deeper insights into the topic
USEFUL FOR

Mathematicians, number theorists, and researchers interested in analytic number theory, particularly those focusing on the properties of the Riemann zeta function and the Mertens function.

tpm
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Don't know if this can be done but taking the Luarent series for the Riemann zeta function converging for every s but Re (s) =1 we have:

[tex]\zeta (s) = \sum_{n=-\infty}^{\infty}\gamma _{n} (s-a)^{n}[/tex] (1)

Then the Mellin-Perron inverse formula for Mertens function:

[tex]M(exp(t))2\pi i = \int_{C}ds \frac{x^{s}}{s\zeta (s)}[/tex] (2)

From expression (1) we could use it to find a Laurent series for [tex]1/\zeta (s)[/tex] to put it into (2) hence we find tor Mertens function:

[tex]M(e^{t})= \sum_{n=0}^{\infty}\frac{a(n)t^{n}}{n!}[/tex] (3)

my problem is , assuming (3) is true then i would like to find an asymptotic formula for big t for M(exp(t)) thanx.
 
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