Explicit Estimates for the Riemann Zeta Function

An improved explicit bound on|ζ(12+it)|

“…Proof. This is Lemma 5 in [1] with L + M changed to L + M − 1, a substitution that is clearly permitted as per the displayed equation at the bottom of [1, p. 1272]. This differs from Lemma 5.10 in [11] in three respects: there is no upper restriction on M , the coefficients are smaller (in [11] both terms in (3) have 4 as their leading coefficients), and the factor (1 − m/M ) is present.…”

“…This is Lemma 3 in [1] with three slight adjustments. First, when applying the mean-value theorem on the first line of page 1268 of [1] one obtains k ≤ (L − 1)/V + 2 instead of k ≤ L/V + 2. Second, when estimating the 2(k − 1) intervals trivially, one may note that there are two intervals of length W ∆ + 1, namely those intervals from (C k − ∆, C k ) and (C 1 , C 1 + ∆), whereas there are k − 2 intervals of length 2W ∆ + 1.…”

“…One of the advantages of using Lemma 1 over Lemma 3 in [1] is that, according to (9), L − 1 generates only one term. The displayed formulae on page 1277 of [1] show that…”

Explicit upper bounds for the remainder term in the divisor problem

“…Going below x = 3 does not make much sense: if we extend the range to cover [2,3], the constant 0.9 when M = 4 becomes 1.7, but we cannot reach x = 1, because our upper bound vanishes (since log 1 = 0), but not the difference. A similar remark applies to the case M = 1.…”

Counting zeros of the Riemann zeta function

“…Recall that Patel [10] showed that |ζ(1 + it)| ≤ log t for t ≥ 3 (i.e., in (1.2), (c 1 , c 2 , t 0 ) = (1, 1, 3) is admissible) and that by the work of Hiary [8], (k 1 , k 2 , k 3 , t 1 ) = (0.77, 1 6 , 1, 3) is admissible for (1.3). 4 In addition, for (c 1 , c 2 , t 0 ) = (1, 1, 3) and (k 1 , k 2 , k 3 , t 1 ) = (0.77, 1 6 , 1, 3), we may take…”