On a Sum Involving the Sum-of-Divisors Function

Sφ(x) � 6 π x log x + O x(log x) log2 x ( 􏼁 (1/3) 􏼐 􏼑, (4) and also proved that the error term in (4) isΩ(x), where log2 denotes the iterated logarithm. Some related works can be found in [4, 5]. Since the sum-of-divisors function σ(n) ≔ 􏽐d|nd has similar properties as the Euler function φ(n) in many cases, it seems natural and interesting to consider its analogy of (3). Our result is as follows.


Introduction
As usual, denote by φ(n) the Euler function and by [t] the integral part of real t, respectively. Recently, Bordellès et al. [1] studied the asymptotic behaviour of the quantity for x ⟶ ∞. By exponential sum technique, they proved that and conjectured that Very recently, Wu [2] improved (2) and Zhai [3] resolved conjecture (3) by showing (4) and also proved that the error term in (4) is Ω(x), where log 2 denotes the iterated logarithm. Some related works can be found in [4,5]. Since the sum-of-divisors function σ(n) ≔ d|n d has similar properties as the Euler function φ(n) in many cases, it seems natural and interesting to consider its analogy of (3). Our result is as follows.

Theorem 1 (i) For x ⟶ ∞, we have
(ii) Let E(x) be the error term in (5). en, for x ⟶ ∞, we have plays a key role, where c > 0 is a positive constant. Clearly, such a bound is not true for 1. By refining Zhai's approach, we shall prove our result.

Preliminary Lemmas
As in [3], we need some bounds on exponential sums of the en, for each positive integer P 1 not exceeding P, we have where e(t) ≔ e 2πit and A > 0, c > 0 are absolute constants. e next two lemmas are essentially a special case of [7, Lemmas 2.5 and 2.6] with a � 1. e only difference is that the ranges of T and N here are slightly larger than those of [7, Lemmas 2.5 and 2.6] (T ≥ N 2 in place of T ≥ N (3/2) and N ≤ x (2/3) in place of N ≤ x (1/2) , respectively). Although the proof is completely similar, for the convenience of readers, we still reproduce a proof here.
en, there exists an absolute positive constant c 5 such that where the implied constant is absolute.
Proof. We apply Lemma 1 to For this, we choose and take the s j to be all integers s such that Obviously the number r of s j is between c 0 k and k. Next we shall verify that f(t) satisfies the conditions (i) and (ii) of Lemma 1 with the parameters chosen above. For where Similarly for N ≤ t ≤ 2N, we find the inequality For the lower bound of (ii), we have where From Lemma 1, there exist two positive constants c and A such that 2 Journal of Mathematics with c 5 ≔ 10 − 4 c. is completes the proof of Lemma 2. □ Proof. By invoking a classical result on ψ(t) (see 8, page 39]) we can write, for any H ≥ 1, Taking e first term can be absorbed by the second, since c 5 can be chosen small enough to ensure that c 6 < 1 and since exp c * (log x) ( and an Abel summation produces the required result. □ en, where the implied constant is absolute. Since |ψ(t)| ≤ 1 for all t, we have On the other hand, since ψ(u) is of period 1, we have Inserting these two bounds into (24), we obtain the required result.

A Formula on the Mean Value of σ(n)
where Proof. Using σ(n) � dm�n m, the hyperbole principle of Dirichlet allows us to write where Journal of Mathematics (31) Firstly we have Secondly we can write where Δ(x, z) is as in (28). Inserting (32), (33), and (34) into (30) and using z 2 ≤ (x/z), we get (27). Taking z � 1 in (27) and noticing that we obtain the required bound. is completes the proof. □
(41) us, which implies that S k (d) ≪ (log x) − 5 . Inserting this into the expression of Δ † 1 (x, z), we get Next we bound Δ ♯ 1 (x, z). Let F(t) be a function of bounded variation on [n, n + 1] for each integer n and let V F [n, n + 1] be the total variation of F on [n, n + 1]. Integrating by parts, we have 4 Journal of Mathematics From this, we can derive that for n ≥ 1. Summing over n, we find that We apply this formula to where we have used the fact that z ≤ ��� N 0 and d ≤ (x/(N 0 z))⇒z ≤ (x/d) (1/3) and the bound Using (48), a simple partial integration allows us to derive that Combining (43) and (50), it follows that Similarly, we can prove the same bound for |Δ 2 (x, z)|. is completes the proof.
��� N 0 ] be a parameter to be chosen later.

(54)
With the help of the bound σ(n) ≪ n log 2 n, we can derive that For evaluating S 1 (x, σ), we write With the help of Lemma 5 (ii), a simple partial integration gives us Journal of Mathematics