1. Jul 25, 2005

### eljose

let be the product:

$$\prod_{\sigma}(1-s/\sigma)e^{s/\sigma}=g(s)$$ where the product is over the non-trivial zeroes of Riemann function

then we take Logarithms to both sides so we have the equality:

$$Lng(s)=\sum_{\sigma}Ln(1-s/\sigma)+s/\sigma$$

then we define the function M(x) in the way that gives the number of non trivial zeroes of riemann function up to x so how could we obtain using the same method that is applied to the product

$$\prod_p(1-p^{-s})$$ ot get the integral equation for M(x)?...

this can be useful as if Riemann hypothesis is true we will have that M(z) is only non zero when Re(z)=1/2

2. Jul 25, 2005

### matt grime

several things spring to mind.
namely the product series expansion of the zeta function isn't valid when Re(s)=1/2 nad even if it were we know there are other zeroes to the zeta fucntion, the riemann hypothesis only being abotu those that lie in the critical strip (the so called non-trivial ones)

3. Jul 26, 2005

### eljose

why is not valid?....

4. Jul 26, 2005

### matt grime

you do know the series and product expansions converge for Re(s)>1 (i hope i'm right on this one, the summation certainly only is valid for Re(s)>1), thuogh it depends on which expansion you use.

$$\prod_p(1-p^{-s})$$

exists iff

$$\sum_p -p^{-s}$$

exists which is iff Re(s)>1