Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Mellin-Perron inverse transform.

  1. Apr 29, 2007 #1


    User Avatar

    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.
  2. jcsd
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted