# Taking the logarithm of the euler product

1. Dec 18, 2003

### MathematicalPhysicist

can some one explain to me how is taking the logarithm of euler product gives you -sum(p)[log(1-p^s)]+log(s-1)=log[(s-1)z(s)]?

my question is coming after encoutering this equation in this text in page number 2: http://claymath.org/Millennium_Prize_Problems/Riemann_Hypothesis/_objects/Official_Problem_Description.pdf [Broken]

btw, does taking a logarithm out of a product gives you the summation or what?

Last edited by a moderator: May 1, 2017
2. Dec 18, 2003

### Muzza

I don't know if this is of any use to you... But the logarithm of a product is the sum of the logarithms of the factors. That is, log(ab) = log(a) + log(b).

3. Dec 18, 2003

### MathematicalPhysicist

yes, i know this but how can you interpret the euler product like this.
lets see it as an example the product of p is 1*2*3...*n so you take the logarithm and you get log1*2*3=log1+log2+log3 which is the summation on this i understand but why adding the term log(s-1)?