New Bound on the Chebyshev function and the Riemann Hypothesis

Under the assumption that the Riemann hypothesis is true, von Koch deduced the improved asymptotic formula θ ( x ) = x + O ( √ x × log 2 x ), where θ ( x ) is the Chebyshev function. We prove if there exists some real number x ≥ 10 8 such that θ ( x ) > x + 1 logloglog x × √ x × log 2 x , then the Riemann hypothesis should be false. In this way, we show that under the assumption that the Riemann hypothesis is true, then θ ( x ) < x + 1 logloglog x × √ x × log 2 x for all x ≥ 10 8 .


Introduction
The Riemann hypothesis is a conjecture that the Riemann zeta function has its zeros only at the negative even integers and complex numbers with real part 1 2 [1]. The Riemann hypothesis belongs to the David Hilbert's list of 23 unsolved problems [1]. Besides, it is one of the Clay Mathematics Institute's Millennium Prize Problems [1]. This problem has remained unsolved for many years [1]. In mathematics, the Chebyshev function θ(x) is given by where p ≤ x means all the prime numbers p that are less than or equal to x. Say Nicolas(p n ) holds provided q≤p n q q − 1 > e γ × log θ(p n ).
The constant γ ≈ 0.57721 is the Euler-Mascheroni constant, log is the natural logarithm, and p n is the n th prime number. The importance of this property is: [3]. Nicolas(p n ) holds for all prime numbers p n > 2 if and only if the Riemann hypothesis is true.
We also know that Theorem 1.4. [6]. If the Riemann hypothesis holds, then for all numbers x ≥ 13.1.
Let's define H = γ − B such that B ≈ 0.2614972128 is the Meissel-Mertens constant [7]. We know from the constant H, the following formula: For x ≥ 2, the function u(x) is defined as follows We use the following theorems: Theorem 1.6. [9]. For x > −1: Theorem 1.7. [10]. For x ≥ 1: Let's define: Definition 1.8. We define another function: Putting all together yields the proof that the inequality ϖ(x) > u(x) is satisfied for a number x ≥ 3 if and only if Nicolas(p) holds, where p is the greatest prime number such that p ≤ x. In this way, we introduce another criterion for the Riemann hypothesis based on the Nicolas criterion and deduce some of its consequences. 2

Results
Theorem 2.1. The Riemann hypothesis is true if and only if the inequality ϖ(x) > u(x) is satisfied for all numbers x ≥ 3.
Proof. In the paper [3] is defined the function: We know that f (x) is lesser than 1 when Nicolas(p) holds, where p is the greatest prime number such that 2 < p ≤ x. In the same paper, we found that which is the same as ϖ(x) > u(x). Therefore, this is a consequence of the theorem 1.1.
Theorem 2.2. If the Riemann hypothesis holds, then for all numbers x ≥ 13.1.
Proof. Under the assumption that the Riemann hypothesis is true, then we would have after of distributing the terms based on the theorem 1.4 for all numbers x ≥ 13.1. If we apply the logarithm to the both sides of the previous inequality, then we obtain that That would be equivalent to where we know that according to theorem 1.7 since 8×π× √ x 3×log x+5 ≥ 1 for all numbers x ≥ 13.1. We use the theorem 1.5 to show that which is the same as We eliminate the value of H and thus, Under the assumption that the Riemann hypothesis is true, we know from the theorem 2.1 that ϖ(x) > u(x) for all numbers x ≥ 13.1 and therefore, Hence, Suppose that θ(x) = ϵ × x for some constant ϵ > 1. Then, In addition, we know that using the theorem 1.6 since log ϵ log x > −1 when ϵ > 1. Certainly, we will have that Thus, .
If we add the following value of log x log θ(x) to the both sides of the inequality, then We know this inequality is satisfied when 0 < ϵ ≤ 1 since we would obtain that log x log θ(x) ≥ 1. Therefore, the proof is done. Theorem 2.3. If there exists some real number x ≥ 10 8 such that then the Riemann hypothesis is false.
Proof. If the Riemann hypothesis holds, then for all x ≥ 10 8 due to the theorem 1.2. Now, suppose there is a real number x ≥ 10 8 such that θ(x) > x + 1 log log log x × √ x × log 2 x. That would be equivalent to and so, 1 log θ(x) for all numbers x ≥ 10 8 . Hence, If the Riemann hypothesis holds, then for those values of x that complies with due to the theorem 2.2. By contraposition, if there exists some number y ≥ 10 8 such that for all x ≥ y the inequality is satisfied, then the Riemann hypothesis should be false. Let's define the function The Riemann hypothesis would be false when there exists some number y ≥ 10 8 such that for all x ≥ y the inequality υ(x) ≤ 0 is always satisfied. We ignore when 2 ≤ x ≤ 10 8 since θ(x) < x according to the theorem 1.3. We know that the function υ(x) is monotonically decreasing for every number x ≥ 10 8 . The derivative of υ(x) is negative for all x ≥ 10 8 . Indeed, a function υ(x) of a real variable x is monotonically decreasing in some interval if the derivative of υ(x) is lesser than zero and the function υ(x) is continuous over that interval [11]. It is enough to find a value of y ≥ 10 8 such that υ(y) ≤ 0 since for all x ≥ y we would have that υ(x) ≤ υ(y) ≤ 0, because of υ(x) is monotonically decreasing. We found the value y = 10 8 complies with υ(y) ≤ 0. In this way, we obtain that υ(x) ≤ 0 for every number x ≥ 10 8 . Hence, the proof is complete.
Theorem 2.4. Under the assumption that the Riemann hypothesis is true, then for all x ≥ 10 8 .
Proof. This is a direct consequence of the theorem 2.3.