Prove that Σ prime p≤x p=3 (mod 10) 1 Р = for some constant A. log log x + A+O 1 log x ,

Let Σ 1 and g(x) = Σ logp. f(x) = prime p≤x p=3 (mod 10) prime p≤x p=3 (mod 10) g(x) = f(x) logx - Ր _☑ t¯¹ƒ(t) dt. Assuming that f(x) ~ 1½π(x), prove that g(x) ~ 1x. 米 (You may assume the Prime Number Theorem: 7(x) ~ x/log x.) *

Let Σ logp. f(x) = Σ 1 and g(x) = Σ prime p≤x p=3 (mod 10) (i) Find ƒ(40) and g(40). prime p≤x p=3 (mod 10) (ii) Prove that g(x) = f(x) logx – [*t^¹ƒ(t) dt. 2