# Evaluation of inverse transform:

1. Mar 25, 2006

### eljose

hello my question (now) is how to evaluate the inverse Laplace transform of the function: (a>0)

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

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

$$\frac{d\zeta(as)}{\zeta(as)}=\sum_{n=1}^{\infty}c(n)n^{-as}$$

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

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

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

$$\sum_{n=1}^{\infty}d(n)\theta(t-aln(n))$$

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 $$d(n)=ac(n)/ln(n)$$ n=2,3,4,5,6,7,........