Mellin-Perron inverse transform.

  • Thread starter tpm
  • Start date

tpm

73
0
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.
 

Related Threads for: Mellin-Perron inverse transform.

  • Last Post
Replies
0
Views
1K
Replies
0
Views
3K
  • Last Post
Replies
0
Views
1K
Replies
0
Views
1K
Replies
0
Views
1K

Hot Threads

Top