EXERCISE SET 3 (2 MARCH 2021) (1) (a) For any x > 1, note that | Ap(x)| ? p-1. By the partial summation formula, for x ? 1, X an (p) = Ap(x)x-$ + s | Ap(u)u-5-1 du. ns nx Since | Ap(u)| ? p-1, the integral converges absolutely for Rt(s) > 0, and Ap(x)x-8 -> 0 as x > co, so taking x > co, we get Fp(s) = s 00 Ap(u)u-8-1 du. 1 Applying to the above formula an(p) nx ns = Ap(x) x -$ + s 00 x Ap(u)u-8-1 du -s | Ap(u)u-8-1 du. X | 1 =Fp(s) } (b) Set ®(s) = > 0. By (1)(a), for o € [1/2, 2], an (p) 00. |Ap(u)|u-0-1 du |Fp(s)| ? > nº + Ap(x) \x+|s| nx x ? (p-1) > 1 x= + | s | x = \ . nx 12 For X n<x ni <1+ 1 už du= - 1+ 2x 2, by taking x = |}(s)|, we get Fp(s) ? (p -1){ - 1 +2x2 +x+|s|x == >< 6x= = 6/S(s). (2) (a) In the proof of Lemma 2.5: We have 0 < 4(x) - 0(x) = > A(n) - > log p = 2 - Clogp p?x,k>1 px nx psx pk<x = X `log p = 2 2 logp p?x,k?2 p?x p<Vx,k>2 px pk <x pk <x log x log p log p ? log x PVx > 15 Vxlog x. 5 PVx So we get, for all x ? 2, ?(x)-0(x) = x + O(xe-cvloga) - 0(x) = Vxlogx. 1 1
Note that xe-cvlogx > x zelog loga for all x ? 2, we have 0(x)= x + O(xe-cvlogx) with the same constant c > 0. (b) By the partial summation formula, for T(t) = 0 for t < 2, log p log x 0(x) + 12 T(x) = log p psx 8 0(t) t logªt dt x x e = log x + O -cy/log x + 1. log x 2 tlog2t dt = log x x + O(xe-c vlogx) + 2 logª t 1 dt + O x. e-cvlogt dt ) . 2 logª t Now, by integration by parts, Li(x) = 12 logt dt t x logt + 1, dt = log x x 12 log2 t 1 2 + 1. dt log 2 12 logª t 1 and we get TT (x) = x log x + O(xe- log) + Li(x) - log x x + 2 log 2 + O - 1, e-cylogt 12 log2 t dt e-cylogt = Li(x) + O(xe-cvlogz) + ( / 2 log2 t dt ) . Now we estimate the integral: change the variable u = vlogt 20 e-CVlogt Vlogx eu2-cu du = 2e vlog 2 ·?log x e(u-f)2 12 log2 t dt = 2 vlog 2 u3 u3 du vlog 2 u-3 du = 2xe-Vlog x vlog x vlog 2 u-3 du < 2e-4 e 4 e (Vlogx- )2 vlog x K xe-Vlog x . So we get TT (x) = Li(x)