aa r X i v : . [ m a t h . N T ] S e p AN EXPLICIT UPPER BOUND FOR | ζ (1 + it ) | DHIR PATEL
Abstract.
In this paper we provide an explicit bound for | ζ (1 + it ) | in the form of | ζ (1 + it ) | ≤ min (cid:0) log t, log t + 1 . , log t + 44 . (cid:1) . This improves on the current best-known explicit boundof | ζ (1 + it ) | ≤ . t ) / up until t of the magnitude 10 . introduction The study of the growth rate of ζ (1 + it ) has been of great interest because of its application inestimating S ( T ) as shown in [28] and computing zero free regions for the Riemann zeta function.In 1900 Mellin [20] was the first to obtain a result in this direction and showed that for real t bounded away from 0 that ζ (1 + it ) = O (log | t | ) (1)In 1921, Weyl improved (1) using Weyl’s sums in [31] to ζ (1 + it ) = O (cid:18) log t log log t (cid:19) (2)and was in turn improved upon by Vinogradov which can be found in [26, Theorem 6.14] to ζ (1 + it ) = O (log / t log / log t ) (3)Several authours namely Flett [8], Walfisz [30], and Korobv [14] between 1950-58 obtainedbounds of the form O (log / t log log / ǫ t ) , O (log / t log log / t ) , O (log / ǫ ) respectively. More-over, authors such as Vinogradov [29], Korobov [15] [13] in 1958 and Richert [23] in 1967 gave thebest known unconditional estimate ζ (1 + it ) = O (log / t ) (4)There are several conditional bounds known for ζ (1 + it ). One such was given by Littlewood in1912, assuming the Lindel¨of hypothesis and showed that ζ (1 + it ) = O (log log t log log log t ) (5)He further improved upon this result in 1928 [19] and provided the best known conditional boundassuming the Riemann Hypothesis that states ζ (1 + it ) = O (log log t ) . (6)In addition to the asymptotic behaviour of ζ (1 + it ), many explicit bounds of the form | ζ (1 + it ) | ≤ a log t, for t ≥ t (7)have also been derived for it. One of the earliest known results is given by Landau [18] in 1903where he shows a = 2, t = 10. Backlund [2] in 1918 improved this result to a = 1 and t >
50 andthis t was lowered by Trudgian in [27] to 2 . . . . and in the same paper he showed that a = and t = 3.The best known explicit bounds for large t are of the form | ζ (1 + it ) | ≤ A log / t, for t ≥ t (8) This explicit estimate with a = 3 / | ζ (1 + it ) | estimate in [27] no longer holds. This iselaborated further in section 2. In 1967, Richert first obtained (8) in [23] for unknown constant A which was computed in 1985 byEllison[7] to be 2100 with t = 3 and in 1995 Cheng [4] improved it to 175 with t = 2. More recentimprovements have been given by Ford in [9] where he showed A = 72 . A = 62 . t = 3 in [27].Moreover, bounds such as (8) seem to improve over bounds like (7) only when t is astronomicallylarge because of extremely large A value. Hence, it is worth obtaining good explict bounds of theform (7) for computational purposes when t is relatively small. Keeping this in mind we have thefollowing theorem Theorem 1.1. If t ≥ , then | ζ (1 + it ) | ≤ min (cid:18) log t,
12 log t + 1 . ,
15 log t + 44 . (cid:19) (9) In particular for t ≥ . . . . × | ζ (1 + it ) | ≤
15 log t + 44 .
02 (10)2.
Remark on Erroneous computational lemma in literature
An important tool used to obtain bounds such as (7) with a = is an explicit version of vander Corput’s second derivative test. This result can be found in the work of Cheng-Graham in [3,Lemma 3]. However a computational flaw was discovered by Kevin Ford [10] and Reyna in [6] withthis Cheng-Graham lemma. This affects many explicit estimates in the literature such as [9], [27],[1] to name a few. However, we note that even if (7) with a = were true, Theorem 1 . t ≥ . × .For the remainder of the section we record the errors in the work of Cheng-Graham [3, Lemma2 and 3]. We also correct another result in literature [1, Lemma 1.2] affected due to these errors.We begin by providing corrected version of flawed Cheng-Graham lemma in [3, Lemma 2]. To doso, we first define k x k := min n ∈ Z | x − n | and we observe that 0 ≤ k x k ≤ / . Lemma 2.1.
Suppose f is a continuously differentiable real-valued function with a monotonicderivative and k f ′ k ≥ U − for some positive real number U on the interval ( a, b ] . Then | S | = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X n ∈ ( a,b ] e πif ( n ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ π U. (11) Remark 2.2.
This result is often attributed to the works of Kuzmin-Landau [16, 17] in literature.Proof.
First we notice that 0 < U − ≤ /
2. Next, we make note of the error in the proof of Lemma2 in [3] and give a possible fix. This fix is based on the ideas found in [21]. Most of the proof ofLemma 2 in [3] is valid except we note a typo on Page 1266 where the equality for G ( n ) − G ( n − G ( n ) − G ( n −
1) = 12 i (cot( πg ( n − − cot( πg ( n ))) (12)However, there is a fatal flaw that originates in the first inequality on Page 1267 of [3] becauseof missing absolute values after the first two cotangent terms. That inequality should instead read: (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) M X n = L e πif ( n ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤
12 (cot( πg ( L )) − cot( πg ( M − (cid:12)(cid:12)(cid:12)(cid:12)
12 + i πg ( L )) (cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12) − i πg ( M − (cid:12)(cid:12)(cid:12)(cid:12) (13) N EXPLICIT UPPER BOUND FOR | ζ (1 + it ) | With this fix in mind, we now provide a possible way to finish Cheng-Graham proof correctly givingus (11). To do so, we note: (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) M X n = L e πif ( n ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:18) cot( πg ( L )) + 1sin( πg ( L )) (cid:19) + 12 (cid:18) πg ( M − − cot( πg ( M − (cid:19) (14)= cos( πg ( L )) + 12 sin( πg ( L )) + 1 − cos( πg ( M − πg ( M − (cid:18) πg ( L )2 (cid:19) + 12 tan (cid:18) πg ( M − (cid:19) (16) ≤
12 cot (cid:18) πU − (cid:19) + 12 tan (cid:18) π (1 − U − )2 (cid:19) (17) ≤ cot (cid:18) πU − (cid:19) (18) ≤ π U (19)For proof readability we make several remarks regarding the above inequalities here. First notethat going from (13) to (14) we use0 < U − ≤ g ( L ) , g ( M − ≤ − U − < , (20)1 + i cot( x ) = ie − ix sin x , sin( πx ) > x ∈ (0 , . (21)Next to go from (14) to (15) we write cot( x ) in terms of sin( x ) and cos( x ) and then gather liketerms. To pass from (15) to (16) we use the following relation valid for x = kπ where k ∈ Z :cot (cid:16) x (cid:17) = 1 + cos x sin x , tan (cid:16) x (cid:17) = 1 − cos x sin x . To go from (16) to (17) we use bound on g ( L ) , g ( M −
1) in (20) along with the fact that on (0 , π/ x is non-negative and decreasing function and tan x is non-negative and increasing function. Forinequality (17) to (18) we use the relation below valid for θ = kπ and k ∈ Z :tan (cid:16) π − θ (cid:17) = cot ( θ ) . And lastly inequality (19) follows from (18) from the fact that cot x ≤ /x for 0 < x < π/ (cid:3) Furthermore, for an alternate proof of Lemma 2.1 one can refer to [24, Lemma 6.6] albeit wenote a couple of typos in that proof , the estimate on | c n | in the beginning of the proof shouldinclude an equality because of equality in (21) and read: | c n | = | − c n | = 12 | sin πy n | ≤
12 sin πϑ Next, the inequality at the end of that proof should involve a negative sign between cotangentterms and read: (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X ≤ n ≤ N e ( x n ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤
12 cot( πy ) −
12 cot( πy N − ) + | c | + | − c N − | (22)Moreover, Landau showed in [17] that the constant 2 /π in (11) is the best possible. For historicalcontext behind this result we refer to Reyna’s work in [6] where additionally Reyna gives an alternateproof of Lemma 2.1 above. Although we make note of a typo in [6, Lemma 2(a)], where theinequality on b k should read: πθ ≤ b k ≤ π (1 − θ ). These errors were also pointed out by Kevin Ford in [10]
DHIR PATEL
Next, having corrected Lemma 2 in [3] we now correct Lemma 3 in [3] which is a crucial toolused in literature to obtain explicit estimates.
Lemma 2.3.
Assume that f is a real-valued function with two consecutive derivatives on [ N +1 , N + L ] . If there exists two real numbers
V < W with
W > such that W ≤ | f ′′ ( x ) | ≤ V (23) for x on [ N + 1 , N + L ] , then (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) N + L X n = N +1 e πif ( n ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:18) LV + 2 (cid:19) r Wπ + 1 ! (24) Remark 2.4.
Because of this correction, the leading term in the incorrect Cheng-Graham Lemma3 is off by a factor of √ .Proof. The proof in [3, Lemma 3] is modified as follows: The k − y j , x j +1 ] are bounded by π ∆ instead of 1 / ( π ∆) + 1. And we take ∆ = 1 / √ πW instead of1 / √ πW . We also make note of a couple of typos in the proof: First, the estimate on k should be k ≤ L/V + 3. Next, when estimating the sum trivially, the mean value theorem in this case shouldbe applied to ( f ′ ) − instead of f − . (cid:3) Moreover using Platt-Trudgian’s observation in [22, Lemma 1] we obtain a slight improvementto (24) in the form of (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) N + L X n = N +1 e πif ( n ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:18) L − V + 2 (cid:19) r Wπ + 12 ! + 1 (25)The corrected version (24) in turn gives us new constants in the explicit third dervivative test foundin [1, Lemma 1.2] and again for completeness we state the corrected version here. Lemma 2.5. Let f ( x ) be a real-valued function with four continuous derivatives on [ N + 1 , N + L ] .Suppose there are constants W > and λ ≥ such that W ≤ | f (3) ( x ) | ≤ λ W for N + 1 ≤ x ≤ N + L .If η > , then (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) N + L X n = N +1 e πif ( n ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ ( LW − / + η )( α L + β W / ) , where α = 1 η + 32 λ √ π q η + W − / + 2 λ η W / + 2 λ W / ,β = 643 √ π √ η + 4 W / . This corrected version of the explicit third derivative test in turn changes the explicit van der Corput boundderived by Hiary in [1, Theorem 1.1]. In that paper, Hiary obtained | ζ (1 / it ) | ≤ . t / log t at the time. Thisbound was an improvement to Platt-Trudgian’s result in [22, Theorem 1] that stated | ζ (1 / it ) | ≤ . t / log t .With the correction provided in Lemma 2.5, the stimate obtained by Hiary in [1] now becomes | ζ (1 / it ) | ≤ . t / log t (with t = 2 × in that paper). And since the Platt-Trudgian bound for | ζ (1 / it ) | uses theincorrect Cheng-Graham lemma, it may no longer be valid.Nonetheless the constant, 0.77, is currently being improved by the author and the result will be published soonalong with some additional estimates for | ζ (1 / it ) | . N EXPLICIT UPPER BOUND FOR | ζ (1 + it ) | Proof.
To obtain this corrected version, we use (24) above in the proof of Lemma 1.2 in [1] andreplace the estmiate for | S ′ m ( L ) | in equation (43) of that proof with | S ′ m ( L ) | ≤ λ L p m/W √ π + 2 λ LmW + 8 p W /m √ π + 4 . (26) (cid:3) We note that here since α , β are decreasing function in W and W > α ≤ f α = 1 η + 32 λ √ π p η + 1 + 2 λ η + 2 λ , β ≤ e β = 643 √ π √ η + 4 . Note that we will obtain our explicit result for k derivative test by using the k − . Preliminary Results
Lemma 3.1.
Let f ( x ) be a real-valued function with four continuous derivatives on [ N + 1 , N + L ] .Suppose there are constants W > and λ ≥ such that W ≤ | f (4) ( x ) | ≤ λ W for N + 1 ≤ x ≤ N + L .If η > , then | S ( L ) | := (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) N + L X n = N +1 e πif ( n ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ ( LW − / + η )( α L + γ √ LW / + β W / ) , where α = 1 η + 7291 pe α ( η + W − / ) / , γ = 7255 q e β η − / + p η e α W − / , β = 1820 q e β η η − / . Proof.
We will use the Weyl-van der Corput Lemma in Cheng and Graham [3, Lemma 5], but usethe form given at the bottom of page 1273 as well as a further refinement by Platt and Trudgian[22, Lemma 2]. In all, if M is a positive integer, then | S ( L ) | ≤ ( L + M − LM + 2 M M X m =1 (cid:16) − mM (cid:17) | S ′ m ( L ) | ! , (27)where S ′ m ( L ) = N + L − m X r = N +1 e πi ( f ( r + m ) − f ( r )) . (28)Here, we can assume that m < L and L > . Otherwise, the sum S ′ m ( L ) is empty and does notcontribute to the upper bound.Now, let g ( x ) := f ( x + m ) − f ( x ) where N +1 ≤ x ≤ N + L − m . Then g ′′′ ( x ) = f ′′′ ( x + m ) − f ′′′ ( x ) . Hence, using the mean value theorem we obtain g ′′′ ( x ) = mf (4) ( ξ ) for some ξ ∈ ( x, x + m ) ⊂ [ N + 1 , N + L ]. Next, using the given bound on f (4) ( x ), we deduce that mW ≤ | g ′′′ ( x ) | ≤ mλ W , ( N + 1 ≤ x ≤ N + L − m ) . (29)Applying Lemma 2 . | S ′ m ( L ) | , multiplying the terms out and then using the inequalities: √ x + x + . . . ≤ √ x + √ x + . . . , where x , x , . . . > α ≤ e α and β ≤ e β , L − m < L for m ≥ . (30) DHIR PATEL | S ′ m ( L ) | < pe α LW − / m / + q e β LW / m − / + p η e α L + q e β η W / m − / . (31)Next, let us bound M X m =1 (cid:16) − mM (cid:17) | S ′ m ( L ) | using (31) and the below estimate valid for − < q < M X m =1 (cid:16) − mM (cid:17) m q ≤ M q +1 ( q + 1)( q + 2) . (32)To prove the above estimate for q ≥ q < | S ( L ) | ≤ ( L + M − LM + 7291 pe α LW − / M / + 7255 q e β LW / M − / + p η e α L + 1810 q e β η W / M − / ! . (33)Now we would like to make the first two terms in (33) of the same magnitude to minimize therhs. This can be achieved if we choose M = ⌈ η W / ⌉ for some free parameter η > M we obtain the inequality η W / ≤ M ≤ η W / + 1. Using thisinequality and then factoring W / term from the first parenthesis and multiplying it in the secondwe deduce that | S ( L ) | ≤ ( LW − / + η ) W / η − W − / + 7291 pe α η W / + 1 W ! / ! L (34)+ q e β η − / W / + p η e α ! √ L + 1810 q e β η η − / W / ! (35) ≤ ( LW − / + η )( α L + γ √ LW / + β W / ) (36)where α , γ , and β are defined as in the statement of the lemma. (cid:3) Lemma 3.2.
Let f ( x ) be a real-valued function with five continuous derivatives on [ N + 1 , N + L ] .Suppose there are constants W > and λ ≥ such that W ≤ | f (5) ( x ) | ≤ λ W for N + 1 ≤ x ≤ N + L .If η > , then (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) N + L X n = N +1 e πif ( n ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ ( LW − / + η )( α L + τ L / W / + γ √ LW / + ω L / W / + β W / ); where α = 1 η + 392435 pe α (cid:16) η + W − / (cid:17) / , τ = 392351 pe γ η − / , γ = 9878 p β η − / + p η e α W − / ,ω = 9878 p η e γ η − / , β = 392275 p η β η − / . Proof.
The proof of this lemma is very similar to Lemma 3 . | S ′ m ( L ) | by letting g ( x ) := f ( x + m ) − f ( x ) where N + 1 ≤ x ≤ N + L − m . Then g (4) ( x ) = f (4) ( x + m ) − f (4) ( x ) fromwhich we can deduce using the mean value theorem and given bound on f (5) ( x ) that mW ≤ | g (4) ( x ) | ≤ mλ W , ( N + 1 ≤ x ≤ N + L − m ) . (37) N EXPLICIT UPPER BOUND FOR | ζ (1 + it ) | Applying the result for fourth derivative stated in Lemma 3 . α ≤ f α = 1 η + 7291 pe α ( η + 1) / , γ ≤ e γ = 7255 q e β η − / + p η e α (38)and inequalities similar to (30) to bound | S ′ m | , we get (cid:12)(cid:12) S ′ m ( L ) (cid:12)(cid:12) ≤ pf α LW − / m / + p e γ L / W / m − / + p β LW / m − / + p η f α L + p η e γ L / W / m − / + p η β W / m − / . (39)Now we bound M X m =1 (cid:16) − mM (cid:17) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) N + L − m X r = N +1 e πi ( g ( r )) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) using (39) and (32) and then substitute them in anexpression like (27) to get (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) N + L X n = N +1 e πif ( n ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ ( L + M − LM + 392435 pe α LW − / M / + 392351 pe γ L / W / M − / + 9878 p β LW / M − / + p η e α L + 9878 p η e γ L / W / M − / + 392275 p η β W / M − / ! (40)Now we would like to make the first two terms in (40) of the same magnitude to minimize the rhsin (40). This can be achieved if we choose M = ⌈ η W / ⌉ for some free parameter η > η W / ≤ M ≤ η W / + 1 obtained because of thechoice of M and using similar algebraic manipulations as in proof of Lemma 3 . (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) N + L X n = N +1 e πif ( n ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ ( LW − / + η ) η − + 392435 pe α (cid:16) η + W − / (cid:17) / ! L + 392351 pe γ η − / L / W / + p β η − / + p η e α W − / ! √ LW / + 9878 p η e γ η − / L / W / + 392275 p η β η − / W / ! . (41)This finally gives us: (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) N + L X n = N +1 e πif ( n ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ ( LW − / + η )( α L + τ L / W / + γ √ LW / + ω L / W / + β W / )(42)where α , τ , γ , ω and β are defined as in the statement of the lemma. (cid:3) Note that Lemmas 2.5, 3.1 and 3.2 are explicit versions of processes
AB, A B, A B in the theoryof exponent pairs respectively. For an introduction to the theory of exponent pairs the author refersthe reader to [11]. Moreover these lemmas give a saving of ≈ W n when compared to the trivialbound where n = − , −
17 and −
115 for Lemmas 2 . , . .
2. In application, it is often unclearon how to choose the correct derivative test to obtain an estimate. For instance as it will be seenlater, the choice of W , dictates the length of the interval over which the k -th derivative test isapplied while bounding the zeta function. This along with the method used to estimate the initialsum determines the derivative test to be applied. DHIR PATEL
Lemma 3.3. If s = σ + it where σ > and t > then we have √ πe − πt/ ξ ( σ, t ) t σ − / ≤ | Γ( s ) | ≤ √ πe − πt/ ξ ( σ, t ) t σ − / (43) where ξ ( σ, t ) = exp (cid:18) − σ t − π t − σ t (cid:19) ,ξ ( σ, t ) = exp (cid:18)(cid:18) σ − (cid:19) σ t + σ t + π t (cid:19) . Proof.
The proof follows a similar strategy as in [12]. By Stirling’s formula for complex values s such that − π + δ ≤ arg s ≤ π − δ given in [25, Page 151] we havelog Γ( s ) = (cid:18) s − (cid:19) log s − s + log(2 π )2 − Z ∞ { x } − / x + s dx. (44)This giveslog | Γ( s ) | = (cid:18) σ − (cid:19) log | s | + t (cid:18) − arctan (cid:18) tσ (cid:19) − σt (cid:19) + log(2 π )2 − ℜ (cid:18)Z ∞ { x } − / x + s dx (cid:19) . (45)Using integration by parts we have Z ∞ { x } − / x + s dx = − s + 12 Z ∞ { x } − { x } + 1 / x + s ) Hence we have using triangle inequality and σ ≥ (cid:12)(cid:12)(cid:12)(cid:12) ℜ (cid:18)Z ∞ { x } − / x + s dx (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) ≤ σ | s | + 112 Z ∞ dx ( x + σ ) + t ≤ σ t + π t (46)Also, since 0 ≤ log(1 + x ) ≤ x for x ≥ ≤ log | s | − log t = 12 log (cid:18) σ t (cid:19) ≤ σ t , for σ ≥ σ > − π − σ t ≤ − arctan (cid:18) tσ (cid:19) − σt = arctan (cid:16) σt (cid:17) − π − σt ≤ − π x ≥ x − x ≤ arctan( x ) ≤ x and the equality in the middle is becausearctan( x ) + arctan(1 /x ) = π , for x > , (48) , (46) in (45) we obtainlog | Γ( s ) | ≤ log √ π + (cid:18) σ − (cid:19) log t − π t + (cid:18) σ − (cid:19) σ t + σ t + π t (49)log | Γ( s ) | ≥ log(2 π )2 + (cid:18) σ − (cid:19) log t − π t − σ t − σ t − π t (50)Therefore exponentiating on both sides we get √ πe − πt/ ξ ( σ, t ) t σ − / ≤ Γ( s ) ≤ √ πe − πt / ξ ( σ, t ) t σ − / where ξ ( σ, t ) = exp (cid:18) − σ t − π t − σ t (cid:19) (51) ξ ( σ, t ) = exp (cid:18)(cid:18) σ − (cid:19) σ t + σ t + π t (cid:19) (52) N EXPLICIT UPPER BOUND FOR | ζ (1 + it ) | Here we remark that for σ > , t ≥ t > ξ ( σ, t ) ≥ ξ ( σ, t ) , ξ ( σ, t ) ≤ σ t + π t for 0 < σ < ξ ( σ, t ) for σ ≥ (53) (cid:3) We also remark that (43) is also valid for t < t is replace with | t | since Γ( s ) = Γ( s )and | Γ( s ) | = | Γ( s ) | Corollary 3.4.
We have for σ ≤ and t > ξ ( σ, t, n ) | Γ(( σ + n ) + it ) | ≤ | Γ( σ + it ) | ≤ t n | Γ(( σ + n ) + it ) | where n is the smallest positive integer such that ℜ ( σ + n ) > and ξ ( σ, t, n ) = 1( − σ + t )( − σ − t ) . . . ( − σ − n + 1 + t ) . Proof.
The functional equation of gamma function statesΓ( z ) = 1 z Γ( z + 1) , z ∈ C \ { , − , − , . . . } (54)Now let n ∈ Z > be the smallest integer such that ℜ ( z + n ) > | Γ( z ) | = 1 | z ( z + 1)( z + 2) · · · ( z + n − | | Γ( z + n ) | . Hence substituting z = σ + it and using Lemma 3.3 with p ( σ + l ) + t ≥ t for l ∈ { , , . . . , n − } we get that | Γ( σ + it ) | ≤ t n | Γ( σ + n + it ) | And since σ + l < , we can find a lower bound using p ( σ + l ) + t ≤ − ( σ + l ) + t , for l ∈{ , , . . . n − } and thus | Γ( σ + it ) | ≥ − σ + t )( − σ − t ) . . . ( − σ − n + 1 + t ) | Γ(( σ + n ) + it ) | . Hence the result follows. (cid:3)
Corollary 3.5.
For t > we have | χ (1 + it ) | ≤ g ( t ) t / (55) where g ( t ) = √ π exp (cid:18) t + π t (cid:19) (56) Proof.
We have the following definition for χ (1 + it ) as stated in [26, Page 16] when s = 1 + it issubstituted χ (1 + it ) = π / it Γ (cid:0) − it (cid:1) Γ (cid:0) it (cid:1) . (57)In order to bound | χ (1+ it ) | using (57) we will first bound (cid:12)(cid:12)(cid:12)(cid:12) Γ (cid:18) − it (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) from above and (cid:12)(cid:12)(cid:12)(cid:12) Γ (cid:18) it (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) from below. First using Corollary 3 . . have that (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Γ (cid:18) − it (cid:19) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ √ πt / e − πt/ ξ (1 , t/ , (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Γ (cid:18) it (cid:19) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≥ √ πe − πt/ ξ (1 / , t/
2) (58)which in turn gives us the result | χ (1 + it ) | ≤ √ π exp (cid:18) t + π t (cid:19) t / where we used n = 1 and (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ξ (1 , t/ ξ (1 / , t/ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ exp (cid:18) t + π t (cid:19) exp (cid:18) − t − π t (cid:19) = exp (cid:18) t + π t (cid:19) , for t > (cid:3) Theorem 3.6.
For t > and n = ⌊ p t/ π ⌋ we have | ζ (1 + it ) | ≤ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n X n =1 n it (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + g ( t ) t / (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n X n =1 n − it (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + R (60) where R = R ( t ) := 1 t / (cid:18)r π g ( t )2 (cid:19) + 1 t r π g ( t ) p π (3 − ! + 1 t / π / + g ( t )242 π ! . and g ( t ) is given by (56) .Proof. Note that Siegel had obtained the following expression for ζ ( s ) : ζ ( s ) = R ( s ) + χ ( s ) R (1 − s ) , (61)where R ( s ) is defined by an integral as given in [5] and R ( s ) = R ( s ) , χ ( s ) = π s − / Γ( − s )Γ( s )Reyna showed in [5] that with σ and t real and t >
0, and an integer K ≥ R ( s ) = N X n =1 n s + ( − N − U a − σ ( K X k =0 C k ( p ) a k + RS K ) (62)where a := r t π , N := ⌊ a ⌋ , p := 1 − a − N ) U := exp (cid:18) − i (cid:18) t t π − t − π (cid:19)(cid:19) and RS K and C k ( p ) are defined in [5, Page 999].In particular substituting s = 1 + it in (61) we obtain | ζ (1 + it ) | ≤ |R (1 + it ) | + | χ (1 + it ) ||R ( − it ) | (63)Thus to bound | ζ (1 + it ) | we will first bound |R (1 + it ) | , |R ( − it ) | using (62), triangle inequalityand [5] and combine it with the bound | χ (1 + it ) | found in Corollary 3 .
5. Hence, after using (62)and triangle inequality we get |R (1 + it ) | ≤ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) N X n =1 n it (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + (cid:18) πt (cid:19) / (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) K X k =0 C k ( p ) a k (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + | RS K | ! . (64)It remains to bound the last two terms to the right of the inequality above. First, we will boundthe second sum on the right hand side of (64) using K = 1, triangle inequality after explanding the N EXPLICIT UPPER BOUND FOR | ζ (1 + it ) | sum and [5, Theorem 4.1] with σ = 1 > , and Γ(1 /
2) = √ π . We thus have (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X k =0 C k ( p ) a k (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ | C ( p ) | + 92 t / ≤
12 + 92 t / . (65)The last inequality above follows from bounding | C ( p ) | using (5 .
2) and Theorem 6.1 in [5].Next to bound | RS K | we use [5, Theorem 4.2], again with K = 1 and σ = 1 , t > | RS | ≤ π t / . (66)Now plugging in (65) and (66) in (64) we get |R (1 + it ) | ≤ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) N X n =1 n it (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + 1 t / (2 π ) / π / t + 9(2 π ) / t / ! (67)Similarly we can a bound for |R ( − it ) | by first observing |R ( − it ) | = |R ( it ) | = |R ( it ) | and usingtriangle inequality and (62) we get |R ( − it ) | ≤ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) N X n =1 n it (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) K X k =0 C k ( p ) a k (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + | RS K | ! (68)Now to bound the second sum and | RS K | above we follow similar steps used to derive (65) and(66) with σ = 0 and t > |R ( − it ) | ≤ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) N X n =1 n it (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + 12 + 1 p π (3 − t / + 242 π t (69)Lastly combining (61), (67), (69), and the bound for | χ (1 + it ) | using Corollary 3 . (cid:3) Lemma 3.7 (Partial Summation) . Let b ≥ b ≥ . . . ≥ b n ≥ , and s m = a + a + . . . + a m wherethe a ′ s are any real or complex numbers. Then if | s m | ≤ M ( m = 1 , , . . . ) , | a b + a b + . . . a n b n | ≤ b M. (70)This lemma can be found in [26, Page 96]. In this paper we will use it to remove the n − weightfrom the sum P n − − it . In order to do so, we let b n = n , a n = n − it and M = max ≤ m ≤ L (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) N + m X k = N +1 n − it (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) giving us (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) N + L X n = N +1 e − it log n n (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ N + 1 max ≤ ∆ ≤ L (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) N +∆ X N +1 e − it log n (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . Proof of Theorem 1.1
First using Theorem 3 . ⌊ t / / √ π ⌋ ≥ t / / √ π − t > π : | ζ (1 + it ) | ≤
12 log t + γ + √ πt / − √ π −
12 log(2 π ) + exp (cid:18) t + π t (cid:19) + R (71) To improve upon (71) we make use of Lemma 3 .
2. With this in mind, we first split the sums onthe right of the inequality in (60) as follows: | ζ (1 + it ) | ≤ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ⌊ jt / ⌋− δ X n =1 n it (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ⌊ √ t π ⌋ X n = ⌈ jt / ⌉ n it (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + g ( t ) t / (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ⌊ √ t π ⌋ X n =1 n − it (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + R . (72)where j ∈ Z > to be chosen later and δ := ( jt / ∈ Z ≤ n ≤ ⌊ jt / ⌋ − δ is computed using triangle inequality andexplicit bound on harmonic sum obtained using partial summation and stated in [27] X n ≤ N n ≤ log N + γ + 1 N . (73)The second sum in the range ⌈ jt / ⌉ ≤ n ≤ ⌊ p t/ (2 π ) ⌋ is divided into dyadic pieces where each ofthese pieces is estimated using Lemma 3 .
2. Lastly, the third sum is bounded trivially using triangleinequality.The dyadic subdivision for this second sum in (72) is carried out in the following manner andthen using triangle inequality subject to ǫ > jt / > ⌊ √ t/ π ⌋ X n = ⌈ jt / ⌉ n it = R ( ǫ ) − X r =0 ⌊ (1+ ǫ ) r +1 jt / ⌋− δ r X n = ⌈ (1+ ǫ ) r jt / ⌉ n it (74)where, δ r := ( ǫ ) r +1 jt / ∈ Z R ( ǫ ) ≤ $ log t − log √ πj log(1 + ǫ ) % + 1 (75)along with R ( ǫ ) ≥ . Here, R ( ǫ ) gives the number of dyadic pieces of our main sum obtained usingthe parameter ǫ . This bound is obtained using the following inequality and solving for R ( ǫ ):(1 + ǫ ) R ( ǫ ) − jt / ≤ ⌈ (1 + ǫ ) R ( ǫ ) − jt / ⌉ ≤ $r t π % ≤ r t π . Also note that that since ǫ > jt / we have ⌈ (1 + ǫ ) r jt / ⌉ < ⌊ (1 + ǫ ) r +1 jt / ⌋ − δ r .To bound the sum on the right hand side in (74) we first use Lemma 3.7 and thus we get S := (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) R ( ǫ ) − X r =0 ⌊ (1+ ǫ ) r +1 jt / ⌋− δ r X n = ⌈ (1+ ǫ ) r jt / ⌉ n it (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ R ( ǫ ) − X r =0 ⌈ (1 + ǫ ) r jt / ⌉ max ≤ ∆ ≤ L (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ⌈ (1+ ǫ ) r jt / ⌉ +∆ − X n = ⌈ (1+ ǫ ) r jt / ⌉ n − it (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (76)where L := L r = ⌊ (1 + ǫ ) r +1 jt / ⌋ − δ r − ⌈ (1 + ǫ ) r jt / ⌉ + 1 where L > ǫ > jt / .We will apply Lemma 3 . f ( x ) := − t π log x, ⌈ (1 + ǫ ) r jt / ⌉ ≤ x ≤ ⌈ (1 + ǫ ) r jt / ⌉ + (∆ − , N EXPLICIT UPPER BOUND FOR | ζ (1 + it ) | then f (4) ( x ) := − tπx , ⌈ (1 + ǫ ) r jt / ⌉ ≤ x ≤ ⌈ (1 + ǫ ) r jt / ⌉ + (∆ − . Now using 1 ≤ ∆ ≤ L we have, 12 tπ ( ⌈ (1 + ǫ ) r jt / ⌉ + ∆ − ≤ | f (5) ( x ) | ≤ tπ ⌈ (1 + ǫ ) r jt / ⌉ (77)12 tπ ( ⌊ (1 + ǫ ) r +1 jt / ⌋ ) ≤ | f (5) ( x ) | ≤ tπ ⌈ (1 + ǫ ) r jt / ⌉ (78)1 W ≤ | f (5) ( x ) | ≤ λ W (79)Thus, we set W := W r = π ( ⌊ (1 + ǫ ) r +1 jt / ⌋ ) t , λ := λ r = ( ⌊ (1 + ǫ ) r +1 jt / ⌋ ) ( ⌈ (1 + ǫ ) r jt / ⌉ ) Using ⌈ (1 + ǫ ) r jt / ⌉ ≤ ⌊ (1 + ǫ ) r +1 jt / ⌋ − δ r ≤ ⌊ (1 + ǫ ) r +1 jt / ⌋ and r ≥ , j > W > λ ≥
1. Hence we can apply Lemma 3 . S ≤ R ( ǫ ) − X r =0 ⌈ (1 + ǫ ) r jt / ⌉ ( α L W − / + τ L / W / + γ L / W / + ω L / W / + β LW / + η α L + η τ L / W / + η γ L / W / + η ω L / W / + η β W / ) / Factoring L W − / gives us that S ≤ R ( ǫ ) − X r =0 √ L W − / ⌈ (1 + ǫ ) r jt / ⌉ ( α + τ L − / W / + γ L − / W / + ω L − / W / + β L − W / + η α L − W / + η τ L − / W / + η γ L − / W / + η ω L − / W / + η β L − W / ) / (80)Next let us bound L W − / using the definition of L and W . With this in mind and using ⌊ x ⌋ ≤ x , ⌈ x ⌉ ≥ x and δ r ≥ L = ⌊ (1 + ǫ ) r +1 jt / ⌋ − δ r − ⌈ (1 + ǫ ) r jt / ⌉ + 1 (81) ≤ ǫ (1 + ǫ ) r jt / ψ ǫ,r,t (82)where ψ ǫ,r,t := 1 + 1 ǫ (1 + ǫ ) r jt / (83)With this at hand we now bound LW − / from above using ⌊ (1+ ǫ ) r +1 jt / ⌋ ≥ ⌊ (1+ ǫ ) r +1 jt / ⌋− δ r ≥ ⌈ (1 + ǫ ) r jt / ⌉ ≥ (1 + ǫ ) r jt / and get LW − / ≤ ǫ (1 + ǫ ) r jt / ψ ǫ,r,t π ( ⌊ (1 + ǫ ) r +1 jt / ⌋ ) t ! − / (84) ≤ ǫψ ǫ,r,t (cid:18) π (cid:19) / j / (1 + ǫ ) r/ t / . (85)Substituting (85) and (82) in (80) we have S ≤ (cid:18) π (cid:19) / ǫ R ( ǫ ) − X r =0 ψ ǫ,r,t j / (1 + ǫ ) r/ B r (86) where B r = α + τ W / L / + γ W / L / + ω W / L / + β W / L + η α W / L + η τ W / L / + η γ W / L / + η ω W / L / + η β W / L ! / (87)Now let us focus on bounding B r using upper and lower bounds for W and L respectively.We have the following upper bound for W : W = π ⌊ (1 + ǫ ) r +1 jt / ⌋ t ≤ π
12 (1 + ǫ ) j (1 + ǫ ) r (88)and using ⌊ x ⌋ ≥ x − ⌈ x ⌉ ≤ x + 1 and δ r ≤ L = ⌊ (1 + ǫ ) r +1 jt / ⌋ − δ r − ⌈ (1 + ǫ ) r jt / ⌉ + 1 ≥ (1 + ǫ ) r jt / (cid:18) ǫ − ǫ ) r jt / (cid:19) ≥ φ ǫ,j,t (1 + ǫ ) r jt / (89)where φ ǫ,j,t = (cid:18) ǫ − jt / (cid:19) and recall that ǫ > jt / .Now using (88) and (89) we can bound for a quantity in the form W a/ L b ≤ (cid:16) π (cid:17) a/ (1 + ǫ ) a/ j a/ − b φ bǫ,j,t (1 + ǫ ) ( a/ − b ) r t b/ (90)Using (90) we can bound the following quantities: W / L / , W / L / , W / L / , W / L , W / L , W / L / , W / L / , W / L / , W / L by substituting a = 1 , , , , b = , , , , , , , a, b values and the inequality √ x + x + . . . ≤ √ x + √ x + . . . for x , x , . . . >
0, we finally obtain the following complicated upper bound for B r B r ≤ √ α + τ (cid:16) π (cid:17) / (1 + ǫ ) / j / φ / ǫ,j,t (1 + ǫ ) r/ t / + γ (cid:16) π (cid:17) / (1 + ǫ ) j / φ / ǫ,j,t (1 + ǫ ) r/ t / + ω (cid:16) π (cid:17) / (1 + ǫ ) j / φ / ǫ,j,t (1 + ǫ ) r/ t / + β (cid:16) π (cid:17) / (1 + ǫ ) / j / φ ǫ,j,t (1 + ǫ ) r/ t / ! / + η α (cid:16) π (cid:17) / (1 + ǫ ) / j / φ ǫ,j,t ǫ ) r/ t / + η τ (cid:16) π (cid:17) / (1 + ǫ ) j / φ / ǫ,j,t ǫ ) r/ t / + η γ (cid:16) π (cid:17) / (1 + ǫ ) / j / φ / ǫ,j,t ǫ ) r/ t / + η ω (cid:16) π (cid:17) / (1 + ǫ ) / j / φ / ǫ,j,t ǫ ) r/ t / η β (cid:16) π (cid:17) / (1 + ǫ ) / φ ǫ,j,t j / ǫ ) r/ t / ! / (91)At this stage we can factor (1 + ǫ ) r/ t / and 1(1 + ǫ ) r/ t / from the second and third square-rootterms above and plugging the bound obtained for B r into inequality (86) we obtain N EXPLICIT UPPER BOUND FOR | ζ (1 + it ) | S ≤ (cid:18) π (cid:19) / ǫj / R ( ǫ ) − X r =0 ψ ǫ,r,t √ α (1 + ǫ ) r/ + C (1 + ǫ ) r/ t / + C ǫ ) r/ t / ! (92)where C := C ( t ) = τ (cid:16) π (cid:17) / (1 + ǫ ) / j / φ / ǫ,j,t + γ (cid:16) π (cid:17) / (1 + ǫ ) j / φ / ǫ,j,t t / + ω (cid:16) π (cid:17) / (1 + ǫ ) j / φ / ǫ,j,t t / + β (cid:16) π (cid:17) / (1 + ǫ ) / j / φ ǫ,j,t t / ! / C := C ( t ) = η α (cid:16) π (cid:17) / (1 + ǫ ) / j / φ ǫ,j,t + η τ (cid:16) π (cid:17) / (1 + ǫ ) j / φ / ǫ,j,t t / + η γ (cid:16) π (cid:17) / (1 + ǫ ) / j / φ / ǫ,j,t t / + η ω (cid:16) π (cid:17) / (1 + ǫ ) / j / φ / ǫ,j,t t / + η β (cid:16) π (cid:17) / (1 + ǫ ) / φ ǫ,j,t j / t / ! / Next using the definition of ψ ǫ,r,t from (83) we can write (92) as S ≤ (cid:18) π (cid:19) / ǫj / R ( ǫ ) − X r =0 √ α (1 + ǫ ) r/ + C (1 + ǫ ) r/ t / + C (1 + ǫ ) r/ t / ! + (cid:18) π (cid:19) / j / R ( ǫ ) − X r =0 √ α (1 + ǫ ) r/ t / + C (1 + ǫ ) r/ t / + C (1 + ǫ ) r/ t / ! (93)Let us estimate each of the above sums using the following inequality that is valid for c, d > R ( ǫ ) − X r =0 ǫ ) cr ≤ (1 + ǫ ) c (1 + ǫ ) c − , R ( ǫ ) − X r =0 (1 + ǫ ) dr ≤ (1 + ǫ ) d (2 π ) d/ j d ((1 + ǫ ) d − t d/ (94)Finally we can obtain a bound for S using (93) and (94) : S ≤ d √ α (1 + ǫ ) / (1 + ǫ ) / − C (1 + ǫ ) / (2 π ) / j / ((1 + ǫ ) / −
1) + C (1 + ǫ ) / (1 + ǫ ) / − t − / ! + d √ α (1 + ǫ ) / (1 + ǫ ) / − t − / + C (1 + ǫ ) / (1 + ǫ ) / − t − / + C (1 + ǫ ) / (1 + ǫ ) / − t − / ! . where d := (cid:18) π (cid:19) / ǫj / , d := (cid:18) π (cid:19) / j / . In the end we have (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ⌊ √ t π ⌋ X n = ⌈ jt / ⌉ n it (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ A t − / + A t − / + A t − / + A t − / + A (95)where A := d C (1 + ǫ ) / (1 + ǫ ) / − , A := d C (1 + ǫ ) / (1 + ǫ ) / − , A := d √ α (1 + ǫ ) / (1 + ǫ ) / − , A := d C (1 + ǫ ) / (1 + ǫ ) / − , A := d √ α (1 + ǫ ) / (1 + ǫ ) / − d C (1 + ǫ ) / (2 π ) / j / ((1 + ǫ ) / − . At this stage we bound the third sum in (72) trivially and thus we get the following bound validfor t > (2 /j ) | ζ (1 + it ) | ≤ γ + log j + 15 log t + 1 jt / − A t − / + A t − / + A t − / + A t − / + A + g ( t ) √ π + R (96)On the other hand, we also have the option of bounding the last sum in (72) using Lemma 3 . (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) N X n =1 n − it (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) N X n =1 n it (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) and following a similar intervalsplit and then steps taken to estimate S as in (80). However, the improvements obtained in thiscase is negligible unless t is astronomically large. Hence we omit the details of such a computationhere.Finally, we choose the following values for our parameters via numerical experimentation: ǫ = 0 . , j = 60 , η = (cid:18) √ π (cid:19) / , η = (cid:18) √ (cid:19) / , η = 2 . , t = 8 × Additionally, we use that W ≥ πj , λ ≤ (1 + ǫ ) , C ≤ C ( t ) , C ≤ C ( t ) , g ( t ) ≤ g ( t ) , R ≤ R ( t ) (97)and plugging them into (96) we get | ζ (1 + it ) | ≤ . t / − . t / + 0 . t / + 0 . t / + 2 . t / + 15 log t (98)Now from (98) we deduce that | ζ (1 + it ) | −
15 log t ≤ F ( t ) . (99)where F ( t ) := 43 . t / − . t / + 0 . t / + 0 . t / + 2 . t / Note that F ( t ) is decreasing in t and thus for t ≥ t and we have that F ( t ) ≤ F ( t ) ≤ . | ζ (1 + it ) | ≤
15 log t + 44 .
02 (100)Now for t ≥
47, we use (71) and get that | ζ (1 + it ) | ≤
12 log t + 1 . . (101)Lastly for t ≥ | ζ (1 + it ) | ≤ log t. (102)Ultimately combining (100) , (101) and (102) for t ≥ | ζ (1 + it ) | ≤ min (cid:18) log t,
12 log t + 1 . ,
15 log t + 44 . (cid:19) (103) N EXPLICIT UPPER BOUND FOR | ζ (1 + it ) | Concluding remarks
Note that choosing a different set of parameters, one can obtain a slight improvement on theconstant 44 .
02 in (9). However it seems that this constant cannot be improved beyond the Eulerconstant γ = 0 . . . . as the harmonic sums (73) are bounded above and below by at leastlog N + γ . Moreover the leading constant in (10) is the best that can be obtained if one insistson using Lemma 3 . / | ζ (1 + it ) | usinghigher explicit derivative tests are in preparation by the author. References
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