# Mellin-Perron inverse transform.

1. Apr 29, 2007

### tpm

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:

$$\zeta (s) = \sum_{n=-\infty}^{\infty}\gamma _{n} (s-a)^{n}$$ (1)

Then the Mellin-Perron inverse formula for Mertens function:

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

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

$$M(e^{t})= \sum_{n=0}^{\infty}\frac{a(n)t^{n}}{n!}$$ (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.