[tex] \ln[\zeta(sa)]/s [/tex] (1)

firs from number theory and dirichlet function will be exist a Dirichlet series so we have that:

[tex] \frac{d\zeta(as)}{\zeta(as)}=\sum_{n=1}^{\infty}c(n)n^{-as} [/tex]

So integrating respect to the variable s we have the Dirichlet representation in the form:

[tex] \ln[\zeta(sa)]= \sum_{n=1}^{infty}d(n)e^{-asln(n)} [/tex]

Now using the table of Laplace transforms we will have that the inverse transform of Log of riemann zeta function (1) is:

[tex] \sum_{n=1}^{\infty}d(n)\theta(t-aln(n)) [/tex]

of course i know that Riemann got it in a different way..i would like to know this way... here theta means the Heaviside,s step function and [tex] d(n)=ac(n)/ln(n) [/tex] n=2,3,4,5,6,7,........