How to Derive the Exact Expression for Dirichlet Series g(s)?

[mitchell2007]

Let be the series in the form g(s)= \sum_{1 \le n } |\Lambda (n) |^{2} n^{-s} where lambda is Von Mangoldt function, my question is how could i get an exact or at least almost exact expresion for g(s) . My other question is how could i obtainthe Mellin transform of the function \Lambda (n+2) \Lambda (n+1) i have tried sum by parts but got no results.

 

[mitchell2007]

of course i tried (wrong ??) with the property, given a Dirchlet series g(s)= \sum_{1 \le n} f(n)n^{-s} then -g'(s)/g(s)= \sum_{1 \le n} f(n) \Lambda (n) n^{-s} but i think this does not work unless f(n) is multiplicative, however could it be used at least as a good approximation or modifying it a bit coul work??

 

[matt grime]

Welcome back, Jose.

 

[mitchell2007]

readng Tom Apostol's 'Introduction to Analytic number theory' the desired series is just AT^{-1} \int_{-T}^{T}dt | \frac{ \zeta ' (a+it)}{\zeta (a+it)} |^{2} = \sum_{1 \le n} | \Lambda (n) |^{2} n^{2a} , with a>1 and T tending to infinity. but i don't know the value of the integral, could someone provide an approximate value of the integral over t (-T,T) above ??, thank you.

 

[DeadWolfe]

How many accounts has this guy made on here?

 

[mitchell2007]

I have also tried using partial summation so:

\sum_{n \le x } \Lambda (n+2) \Lambda (n) = B(x)= \Lambda (x+2) \Psi (x) - \sum_{n \le x } \Psi (x) ( \Lambda (x+2) - \Lambda (x+1) )

so g(s)= \sum_{1 \le n} \Lambda (n+2) \Lambda (n) n^{-s}= s \int_{1}^{\infty}B(x) x^{- (s+1)}

to obtain 'g(s)' any hint please?? thanx.