An Investigation of the Non-Trivial Zeros of the Riemann Zeta Function 3

arXiv:1804.04700v15 [math.GM] 15 Jun 2020 h imn-o agltepii oml o h smttcexpan asymptotic the for formula explicit Mangoldt Riemann-von the eafnto nete ie fteciia ie Given line. critical the of t sides of zeros either the on between relationship function a zeta to leads functional zeta mann h rtclstrip critical the n ¯ and ieo acy h imn yohsssaigta l non-trivial produc all Euler that a line stating the analysis critical hypothesis the complex as Riemann on in known The developments Cauchy. formula recent of primes, more a time and ze with in function s linked the zeta was to is the Although function approach referred between power [14]. connection is inverse numbers the the and establishing prime of ago series of time inﬁnite distribution long an the is primar which of which 1 function numbers, study - complex (1826 to the function Riemann is zeta Bernhard the after of named extension an function zeta Riemann The Introduction 1 Keywords seog oso hr r ozrso ihrsdso h rtcllin critical the of sides either on zeros no strip are critical there the show to enough is eann ato h rtclstrip critical the of part remaining yohsssae htalnntiilzrsleo h rtclline critical the on lie zeros non-trivial all that states hypothesis rsn address: Present USA 30332, [email protected] GA Atlanta, E-mail: Technology, of Institute Georgia Heymann Yuri zeta line Riemann critical the of the zeros on many located While zeros. are trivial so-called the are Abstract date Accepted: / date Received: Heymann Yuri the of zeros function non-trivial zeta the Riemann of investigation An wl eisre yteeditor) the by inserted be (will No. manuscript Noname s t ope ojgt,if conjugate, complex its h eo fteReanzt ucinotieteciia strip critical the outside function zeta Riemann the of zeros The imn eafnto,Reanhypothesis Riemann function, zeta Riemann u hniu 22Gnv,Switzerland Geneva, 1202 Chandieu, rue 3 ℜ ℜ ℜ p p s s p s P q s P q s “ q 0 0 , , 1 1 1 { r r scuilfrtepienme hoy o only Not theory. prime-number the for crucial is 2 osyteReanhptei strue. is hypothesis Riemann the say to , hnw have we then , ℜ s p s ℜ sazr fteReanzt ucinin function zeta Riemann the of zero a is “ q p s s P q 1 { ,tenneitneo eo nthe in zeros of non-existence the 2, 0 , 1 ζ r p s ean ob rvn h Rie- The proven. be to remains “ q ζ p 1 ´ s s ¯ ope number complex a q steRiemann the As . ℜ p eRiemann he s Riemann’s “ q purpose y function within e eo lie zeros 6)is 866) inof sion tudied 1 { the t ,it 2, ta t 2 Yuri Heymann

the prime-counting function involves a sum over the non-trivial zeros of the Riemann zeta function, but also the Riemann hypothesis has implications for the accurate estimate of the error involved in the prime-number theorem, which is the description of the asymptotic behavior of the number of primes less than or equal to a real variable.

The Riemann zeta function is a holomorphic function deﬁned as follows:

8 1 ζpsq“ , (1) ns n“1 ÿ where s is a complex number and domain of congruence ℜpsqą 1. The domain of convergence of the Riemann zeta function is extended to the left of ℜpsq“ 1 by analytic continuation. The Dirichlet eta function, which is the product of 2 ℜ the factor 1 ´ 2s with the Riemann zeta function is convergent for psqą 0. This is an alternating series, expressed as follows: ` ˘ 2 8 p´1qn`1 ηpsq“ 1 ´ ζpsq“ , (2) 2s ns n“1 ˆ ˙ ÿ where ℜpsqą 0, and ηp1q“ lnp2q by continuity. 2 ℜ The function 1 ´ 2s has an inﬁnity of zeros on the line psq “ 1 given by 2kπi Z˚ 2 α´1 iβ ln 2 α iβ ln 2 sk “ 1 ` ln 2 where k P . As 1 ´ 2s “ 2 ˆ 2 e ´ 1 { 2 e , ` 2 ˘ the factor 1 ´ s has no poles nor zeros in the strip ℜpsq Ps0, 1r. It follows 2 ` ˘ ` ˘ that by analytic continuation, the Dirichlet eta function can be used as a proxy ` ˘ for zero ﬁnding of the Riemann zeta function in the critical strip ℜpsqPs0, 1r. The Riemann zeta function can be further extended to the left, i.e. ℜpsqď 0 by analytic continuation with the Riemann zeta functional. Whenever the Riemann zeta function is invoked for ℜpsqď 1, it is implicitly referring to its analytic continuation.

2 Elementary propositions

Before delving into core propositions, a prerequisite is the Riemann zeta functional, which is used explicitly in propositions 2, 4 and 9 and implicitly in 5, 6, 7 and 8. The Riemann zeta functional is expressed as follows: sπ ζpsq“ Πp´sqp2πqs´12 sin ζp1 ´ sq . (3) 2 ´ ¯ This equation was established by Riemann in 1859. The details of the derivation are provided in [2], p. 13. The standard formulation of (3) is as follows: sπ ζp1 ´ sq“ Γ psqp2πq´s2cos ζpsq , (4) 2 ´ ¯ An investigation of the non-trivial zeros of the Riemann zeta function 3

involves the substitution s Ñ 1 ´ s and the relationship Πps ´ 1q“ Γ psq. Let us proceed with key propositions.

Proposition 1 A formula is introduced here for calculations further down, which is expressed as follows:

a2 ´ b2 ´2 ab ` i ra ` ibs“ a ´ i b , (5) a2 ` b2 a2 ` b2 „ ˆ ˙ where a and b are reals.

Proof By identifying the real and imaginary parts of px`iyqpa`biq“ a´ib, we obtain two equations a x ´ by “ a and b x ` ay “ ´b. Eq. (5) is the resolution of these two equations.

Proposition 2 Given s a complex number ands ¯ its complex conjugate, we have:

ζps¯q“ ζpsq , (6) where ζpsq is the complex conjugate of ζpsq.

Proof Let us say s “ α ` iβ where α and β are real numbers and i2 “ ´1. We have:

8 rcospβ ln nq´ i sinpβ ln nqs ζpsq“ , (7) nα n“1 ÿ where ℜpsqą 1.

1 1 1 We have ns “ nα exppβi ln nq “ nαpcospβ ln nq`i sinpβ ln nqq . We then multiply both the numerator and denominator by cospβ lnpnqq ´ i sinpβ lnpnqq. After sev- eral simpliﬁcations, we get (7). Note that in (7) when β changes its sign, the real part of ζpsq remains unchanged, while imaginary part changes sign. This means, the Riemann zeta function has mirror symmetry with respect to the real axis of the complex plane and ζps¯q“ ζpsq when ℜpsqą 1.

By the same process, the Dirichlet eta function is expressed as follows:

8 rcospβ ln nq´ i sinpβ ln nqs ηpsq“ p´1qn`1 , (8) nα n“1 ÿ where ℜpsqą 0, and ηp1q“ lnp2q by continuity.

Using the same reasoning as above, the Dirichlet eta function is symmetrical with respect to the real axis of the complex plane. Furthermore, the factor 1 ´ 21´s “ 1 ´ 21´α pcospβ ln 2q´ i sinpβ ln 2qq also has mirror symmetry with respect to the real axis. As the product of holomorphic functions which are` symmetrical˘ with respect to the real axis yields a holomorphic function 4 Yuri Heymann

which is also symmetrical with respect to the real axis, we get ζps¯q “ ζpsq when ℜpsq Ps0, 1r.

Let us consider the Riemann zeta functional (4), and introduce the function ξ : C Ñ C such that: sπ ξpsq“ 2Γ psqp2πq´s cos . (9) 2 ´ ¯ If ξpsq is symmetrical with respect to the real axis, then by analytic continuation, the Riemann zeta function is also symmetrical with respect to the real axis, when ℜpsqă 0. From formula 6.1.23 p. 256 in [1], we have Γ psq“ Γ ps¯q. ´s cospβ ln 2π´i sinpβ ln 2πqq ´s Furthermorre, p2πq “ p2πqα , hence p2πq is symmetrical with respect to the real axis. From the formula 4.3.56 p. 74 ins [1], we have sπ απ βπ απ βπ cos 2 “ cos 2 cosh 2 ´ i sin 2 sinh 2 . As coshp´xq“ coshpxq sπ and` sinh˘p´xq“´` sinh˘ pxq´, cos¯ 2 is` also˘ symmetrical´ ¯ with respect to the real axis. Hence, ξpsq has mirror symmetry with respect to the real axis. Therefore, ` ˘ ζps¯q“ ζpsq when ℜpsqă 0.

As the Riemann zeta function is a meromorphic function, by continuity ζps¯q“ ζpsq on the lines ℜpsq“ 0 and ℜpsq“ 1.

Proposition 3 Given s a complex number ands ¯ its complex conjugate, we deﬁne a holomorphic function ν : C Ñ C such that:

ζpsq“ νpsqζp1 ´ s¯q . (10) A property of νpsq is that it is equal to one if ℜpsq“ 1{2.

Proof Let us set s “ α ` iβ where α and β are reals. When α “ 1{2, we get 1 1 ζpsq“ ζ 2 ` iβ and ζp1 ´ s¯q“ ζ 2 ` iβ . Hence, ζpsq“ ζp1 ´ s¯q provided that ℜpsq“ 1 . For a given α ‰ 1 , we have to introduce a complex factor νpsq ` 2 ˘ 2 ` ˘ because we can no longer satisfy the two equations ℜpζpsqq “ ℜpζp1 ´ s¯qq and ℑpζpsqq “ ℑpζp1 ´ s¯qq with one degree of freedom given by β.

Proposition 4 Given s a complex number ands ¯ its complex conjugate, we have:

1 Γ ps¯q πs¯ u2 ´ v2 ´2u v “ 2 cos ` i , (11) νpsq p2πqs¯ 2 u2 ` v2 u2 ` v2 ´ ¯ „ ˆ ˙ where u “ ℜpζpsqq and v “ ℑpζpsqq.

Proof This formula is derived from the Riemann zeta functional. By the def- 1 ζp1´s¯q inition in proposition 3, we have νpsq “ ζpsq . From (4), the Riemann zeta Γ psq πs functional is expressed as ζp1 ´ sq “ p2πqs 2cos 2 ζpsq. Hence, ζp1 ´ s¯q “ ` ˘ An investigation of the non-trivial zeros of the Riemann zeta function 5

Γ ps¯q πs¯ p2πqs¯ 2cos 2 ζps¯q. From proposition 2, we have ζps¯q“ ζpsq, where ζpsq is the complex conjugate of ζpsq. ` ˘ 2 2 u ´v ´2u v We use proposition 1 and get ζps¯q “ ζpsq u2`v2 ` i u2`v2 where u “ ℜpζpsqq and v “ ℑpζpsqq. Eq. (11) follows. ” ´ ¯ı

Proposition 5 Given s a complex number, the function νpsq as defined above is equal to zero only at the points s “ 0, ´2, ´4, ´6, ´8, ..., ń where n is an even integer. Proof We set s “ α ` iβ where α and β are reals. The function νpsq tends 1 to zero, when its reciprocal νpsq approaches ˘8. We use proposition 4 to find 1 the values when νpsq Ñ ˘8.

2 2 u ´v ´2u v (i) We note that the term u2`v2 ` i u2`v2 is bounded: we have ´1 ď 2 2 2 2 2 2 v u ´v u ” ´´2 maxp|¯ıu|,|v|q ´2uv 2 maxp|u|,|v|q ´ u2`v2 ď u2`v2 ď u2`v2 ď 1 and ´2 ď u2`v2 ď u2`v2 ď u2`v2 ď 2.

2 2 u ´v ´2u v (ii) We note that the expression u2`v2 ` i u2`v2 is never equal to zero as it is not possible to have both the” real and´ imaginary¯ı parts equal to zero at the same time. For the real part to be equal to zero either u “ v or u “ ´v. When u “ v (or u “ ´v), the imaginary part is equal to ´1 (or 1). For the imaginary part to be equal to zero, either u or v should be equal to zero. If u is equal zero (or v is equal to zero), then the real part is equal to ´1 (or 1).

2 2 u ´v ´2u v uí v Eq. (5) can also be expressed as u2`v2 ` i u2`v2 “ uì v . This formula is defined when u ` i v ‰ 0. ” ´ ¯ı

As implied by (i) and (ii), when u ` i v is in the neighborhood of 0, no matter u´i v how close to 0, the factor u`i v cannot take the value zero and is bounded.

πs¯ Consequently, of (i) and (ii), νpsq“ 0 at a point s0 if only lim cos 2 Γ ps¯q“ sÑs0 ˘8. ` ˘

πs¯ π πα ´πβ We check that cos 2 is bounded. We have cos 2 pα ´ iβq “ cos 2 cosh 2 ´ πα ´πβ ´ ¯ i sin 2 sinh 2` ˘. As the cosine, the sine,` the hyperbolic˘ cosine` ˘ and the πs¯ hyperbolic` ˘ sine´ are bounded¯ on s´8, 8r, cos 2 is bounded when β is ﬁnite. ` ˘ Hence, the condition νpsq “ 0 occurs only if Γ ps¯q Ñ ˘8 or when the re- 1 ciprocal Gamma Γ ps¯q “ 0. The Euler form of the reciprocal Gamma function 8 s 1 1` n (see [2] p. 8) is “ s 1 s . Thus, the only zeros of the reciprocal Γ psq 1` n“1 p n q Gamma are s “ 0, ´1, ´2, ´ś3, ..., ´n where n P N. These are the points where νpsq could potentially be equal to zero. 6 Yuri Heymann

πs¯ We check the points where the cosine term cos 2 is equal to zero. We have πs¯ π πα ´πβ πα ´πβ cos 2 “ cos 2 pα ´ iβq “ cos 2 cosh 2` ´˘ i sin 2 sinh 2 . As πs¯ the` hyperbolic˘ ` cosine is never˘ equal` to˘ zero,´ the¯ real part` of˘ cos 2´ is equal¯ to zero if and only if cos πα “ 0. As we cannot have both the sine and cosine 2 ` ˘ functions equal to zero simultaneously, when they share the same argument, ` πs¯˘ ´πβ the imaginary part of cos 2 equals zero if and only if sinh 2 “ 0, which πs¯ occurs only if β “ 0. Hence,` ˘ the term cos 2 can be equal´ to¯ zero on the real line at the points s “˘1, ˘3, ˘5, ˘7, .... Thus, νpsq is equal to zero with ˘ certainty at the points s “ 0, ´2, ´4, ´6, ´8`, ....

What about the points s “´1, ´3, ´5, ´7, ´9, ... ?

8 1 s 1 p1` n q The Euler form of the Gamma function is expressed as Γ psq“ s 1` s . n“1 n πs The Taylor series of cos 2 in ´m where m is an odd integer isś as follows: πs π mπ π 3 mπ 3 cos 2 “ 2 sin 2 ps`` m˘ q´ 2 sin 2 ps ` mq ` ... . If we multiply the first term of the Taylor series by the m-th product of the Euler form of ` ˘ ` ˘ ` ˘ ` ˘ s`m the Gamma function, we obtain a factor of order s`m , which is finite when s tends to ´m. The multiplication of the successive terms of the Taylor series with the m-th product of the Euler form of the Gamma function leads to a k ps`mq factor of order s`m where k “ 3, 5, 7, ..., which converges towards zero when s tends to ´m. Furtherore, the limit of the n-th product of the Euler form 1 s p1` n q of the Gamma function s when n tends to `8 is equal to 1. Therefore, 1` n πs¯ lim cos 2 Γ ps¯q is finite at the points s “ ´1, ´3, ´5, ´7, ´9, ... so we can sÑs0 say that`νps˘q‰ 0 at these points.

As a validation step, from proposition 3 we have ζpsq “ νpsqζp1 ´ s¯q, and from proposition 9 the only pole of ζpzq is at the point z “ 1. Furthermore, according to section 3, we have ζp1 ´ s¯q‰ 0 when ℜpsqă 0. Hence, νpsq cannot be equal to zero at the points s “ ´1, ´3, ´5, ´7, ´9, ...; otherwise, ζpsq would be equal to zero at these points, which contradicts the non-trivial zeros of the Riemann zeta function (see section 4). Although νp0q “ 0, ζp0q ‰ 0 because the function ζpzq has a pole at z “ 1.

Proposition 6 Given s a complex number, the function νpsq as deﬁned above has an inﬁnite number of poles at the points s “ 1, 3, 5, 7, 9, ..., n where n is an odd integer. Proof We use the expression of the reciprocal of νpsq introduced in proposition 4. 2 2 u ´v ´2u v We have shown in proof of proposition 5, that the factor u2`v2 ` i u2`v2 cannot take the value zero and is bounded. Hence, a point” s0 is a pole´ of the¯ı πs¯ function νpsq if and only if lim Γ ps¯q cos 2 “ 0. Such poles occur either sÑs0 ` ˘ An investigation of the non-trivial zeros of the Riemann zeta function 7

πs¯ πs¯ when Γ ps¯q or cos 2 are equal to zero. The term cos 2 is equal to zero at the points s “ ˘1, ˘3, ˘5, ˘7, .... The Euler form of the Gamma function ` ˘ 8 1 s ` ˘ 1 p1` n q N˚ 1 is expressed as Γ psq“ s 1` s . As @n P the term 1 ` n is not equal n“1 n to zero, we can say that theś Gamma function is never equal to zero. However, the Gamma function tends to infinity at the points s “ 0, ´1, ´2, ´3, ..., ń where n P N, hence the points s “ 1, 3, 5, 7, ... can be determined to be poles of νpsq with certainty. The points s “´1, ´3, ´5, ´7, ... are poles of νpsq only πs if the limit of Γ psq cos 2 is equal to zero when s tends to either of these points. The Taylor series of cos πs in ´m where m is an odd integer is as ` ˘ 2 πs π πm π 3 πm 3 follows: cos 2 “ 2 sin 2 ps` `˘mq´ 2 sin 2 ps ` mq ` ... . If we multiply the m-th product of the Euler form of the Gamma function by the ` ˘ ` ˘ ` ˘ ` πs˘ first term of the Taylor series we computed for cos 2 , we obtain a factor s`m of order s`m , which does not converge towards zero when s tends to ´m. The multiplication of the successive terms of the Taylor` ˘ series with the m-th k ps`mq product of the Euler form of Gamma function leads to a factor of order s`m where k “ 3, 5, 7, ..., which converges towards zero when s tends to ´m. In addition, the limit of the n-th product of the Euler form of the Gamma func- 1 s p1` n q πs tion 1 s when n tends to `8 is equal to 1. Hence, lim Γ psq cos 2 ‰ 0 ` n sÑs0 at the points s0 “´1, ´3, ´5, ´7, .... Thus, the points s “´1, ´3, ´`5, ´˘7, ... are not within the set of poles of νpsq.

Proposition 7 Given s a complex number ands ¯ its complex conjugate, if s is a complex zero of the Riemann zeta function in the strip ℜpsqPs0, 1r, then 1 ´ s¯ must also be a zero. Proof From proposition 5 we have shown that νpsq is only equal to zero at even negative integers including zero. Hence νpsq is never reaches zero when ℜpsqą 0. Furthermore, in proposition 6 we have shown that νpsq has no poles in the strip ℜpsqPs0, 1r. As ζpsq“ νpsqζp1 ´ s¯q from proposition 4 and given that νpsq has no poles nor zeros in the strip ℜpsqPs0, 1r, we can say that if s is a zero, then 1 ´ s¯ must also be a zero. Proposition 8 Given s a complex number ands ¯ its complex conjugate, if s is a complex zero of the Riemann zeta function in the strip ℜpsqPs0, 1r, then we have:

ζpsq“ ζp1 ´ s¯q . (12) Proof This is a corollary of proposition 7. If s and 1 ´ s¯ are zeros of the Riemann zeta function, both terms ζpsq and ζp1 ´ s¯q are equal to zero. Hence, both terms must match. The converse is not necessarily true, meaning that ζpsq“ ζp1 ´ s¯q does not imply that s and 1 ´ s¯ are zeros of the function.

Proposition 9 The only pole of the Riemann zeta function ζpsq is a simple pole at s “ 1. 8 Yuri Heymann

This is documented in [16], p.13. A sketch ofthe proof is provided below for the sake of completleness.

Proof For notation purposes, we set s “ α ` iβ where α and β are reals.

Using Abel’s lemma for summation by parts, see [8] at p. 58, we get:

m m´1 n´s “ n n´s ´ pn ` 1q´s ` m1´s , (13) n“1 n“1 ÿ ÿ ` ˘ where m P N.

When ℜpsqą 1, we have lim m1´s “ 0. Thus, we get: mÑ8

8 8 n´s “ n n´s ´ pn ` 1q´s n“1 n“1 ÿ ÿ8 ` n`1 ˘ “ s n x´s´1 dx n“1 żn (14) ÿ8 “ s txux´s´1 dx ż1 s 8 “ ´ s txux´s´1 dx , s ´ 1 ż1

where txu denotes the ﬂoor function of x and txu “ x ´ txu denotes the fractional part of x.

Hence, we have:

s 8 ζpsq“ ´ s txux´s´1 dx , (15) s ´ 1 ż1 where ℜpsqą 1.

To say that the only pole of ζpsq on the half-plane ℜpsqě 1 is s “ 1, we have 8 ´s´1 ℜ too show that the integral 1 txux dx in (15) is bounded when psqě 1. ş Given a sequence ui of complex numbers, we have the inequality | ui| ď i |ui|. As an integral is an inﬁnite sum, this inequality still holds andř we get: i ř An investigation of the non-trivial zeros of the Riemann zeta function 9

8 8 txux´s´1 dx ď |txux´s´1| dx ˇż1 ˇ ż1 ˇ ˇ 8 ˇ ˇ p´α´1q ´iβ ln x ˇ ˇ ď |txux e | dx 1 ż 8 (16) ď |txux´α´1| dx 1 ż 8 ď txux´α´1 dx . ż1 n`1 ´α´1 n`1 ´α´1 We have n txux dx ă n x dx. Thus, we get: ş ş 8 8 n`1 txux´s´1 dx ă x´α´1 dx ˇż1 ˇ n“1 żn ˇ ˇ ÿ (17) ˇ ˇ 1 8 ˇ ˇ ă n´α ´ pn ` 1q´α . α n“1 ÿ ` ˘ When α “ 1, we have:

8 8 txux´s´1 dx ă n´1 ´ pn ` 1q´1 ˇż1 ˇ n“1 ˇ ˇ ÿ ` ˘ (18) ˇ ˇ 8 1 ˇ ˇ ă . npn ` 1q n“1 ÿ 8 1 From the integral test [4] p. 132, we can show that the series npn`1q ă n“1 8 1 ℜ ř n2 converges. Hence, the only pole on the line psq“ 1 is at s “ 1, which n“1 isř a simple pole. 8 8 1 1 From the integral test, when α ą 1, the series nα and pn`1qα converge. n“1 n“1 8 ´s´1 Hence, 1 txux dx in (18) is bounded. Thus,ř we can sayř that ζpsq has no poles when ℜpsqą 1. ˇş ˇ ˇ ˇ According to the Riemann zeta functional (4), we have:

pα ` iβqπ ζp1 ´ α ´ βiq“ Γ pα ` iβqp2πq´pα`iβq2cos ζpα ` iβq . (19) 2 ˆ ˙ The Euler form of the Gamma function is expressed as follows: 10 Yuri Heymann

s 1 8 1 ` 1 Γ psq“ n . (20) s 1 ` s n“1 ` n˘ ź Hence, the Gamma function has no poles when ℜpsqě 1. The cosine term in (19) is bounded. Also, we have shown that the only pole of ζpsq in the half-plan ℜ 1 psqě 1 is at s “ 1. The Riemann zeta function evaluated at 0 is ζp0q“´ 2 , see [18] p. 135. Thus, considering (19) when α ě 1, we can say that ζpsq has no poles when ℜpsqď 0.

The Dirichlet eta function is used as an extension of the Riemann zeta function ℜ 2 to show that ζpsq has no poles in the strip psqPs0, 1r. As the factor 1 ´ 2s has no poles nor zeros in the strip ℜpsq Ps0, 1r, it is enough to show that ` ˘ the Dirichlet eta function is bounded, to say that ζpsq has no poles when ℜpsq Ps0, 1r. The Dirichlet eta function in its integral form is expressed as follows:

1 8 xs´1 ηpsq“ dx . (21) Γ psq ex ` 1 ż0 As the Gamma function is never equal to zero (see details in proposition 6), 8 xs´1 ηpsq is bounded if the integral 0 ex`1 dx is bounded. Let us consider the case when α Ps0, 1r, hence: ş

8 xs´1 8 xs´1 dx ď dx ex ` 1 ex ` 1 ˇż0 ˇ ż0 ˇ ˇ ˇ ˇ 8 ˇxα´1 ˇ ˇ ˇ ď ˇ dxˇ ˇ ˇ ˇex ` 1 ˇ ż0 (22) 1 xα´1 8 1 ă dx ` dx ex ` 1 ex ` 1 ż0 ż1 1 xα´1 8 ă dx ` e´x dx . ex ` 1 ż0 ż1 α´1 1 x 1 1 We solve 0 ex`1 dx using integration by parts, where u v “ru vs´ u v with 1 α´1 1 u “ x and v “ x . We get: ş e `1 ş ş

1 xα´1 xα 1 1 1 xα ex dx “ ` dx ex ` 1 α ex ` 1 α pex ` 1q2 ż0 „ 0 ż0 xα 1 1 1 1 ex ă ` dx (23) α ex ` 1 α pex ` 1q2 „ 0 ż0 1 1 1 1 ă ` ´ . αpe1 ` 1q α 2 e1 ` 1 ˆ ˙ Thus, we have: An investigation of the non-trivial zeros of the Riemann zeta function 11

8 xs´1 1 1 1 1 dx ă ` ´ ` e´1 , (24) ex ` 1 αpe1 ` 1q α 2 e1 ` 1 ˇż0 ˇ ˆ ˙ ˇ ˇ ˇ ˇ and ηpsq isˇ bounded in theˇ strip s0, 1r, so we can say that ζpsq has no poles in this strip.

3 Zeros when ℜpsq ą 1

Though this is a familiar result, in this section the Euler product is used to prove there are no zeros when ℜpsqą 1. We also need the integral form of the remainder of the Taylor expansion of ex.

The Taylor expansion of ex in 0 may be expressed as follows:

ex “ 1 ` x ` εpxq , (25) where εpxq is the remainder of the Taylor approximation.

The integral form of the remainder of the Taylor expansion of ex is as follows:

x εpxq“ px ´ uq eudu . (26) ż0 When u P r0, xs, we have px ´ uq eu ě 0. Hence, we have εpxq ě 0. Conse- quently, we get:

ex ě 1 ` x . (27) Let us say p is a prime number larger than one, and s “ α ` iβ a complex number where α and β are reals. We have:

|1 ´ p´s|ď 1 ` |p´s| ď 1 ` |p´α´iβ| (28) ď 1 ` p´α ˆ |epiβ ln pq| ď 1 ` p´α .

By setting x “ p´α in (27), we get:

1 ` p´α ď exp p´α . (29) From (28) and (29), we get: ` ˘

|1 ´ p´s|ď exp p´α . (30) The Euler product [2] p. 6 is expressed as: ` ˘ 12 Yuri Heymann

8 1 ζ s , p q“ ´s (31) p 1 ´ p ź where p is the sequence of prime numbers larger than 1 and ℜpsqą 1.

Hence, we have:

8 1 ζ s . | p q| “ ´s (32) p |1 ´ p | ź Using inequality (30) in (32), we get:

8 |ζpsq| ě expp´p´αq p ź 8 (33) ě exp ´ p´α ˜ p ¸ ÿ ą 0 .

8 8 8 We have 0 ă p´α ă n´α. From the integral test, the series n´α p n“1 n“1 ř ř 8 ř converges when α ą 1, hence exp ´ p´α ą 0. From (33), we get @α ą ˜ p ¸ 1, |ζpsq| ą 0. Hence, we can say the Riemannř zeta function has no zeros when ℜpsqą 1.

4 Zeros for ℜpsq ă 0

The below functional equation is used to obtain the zeros of the Riemann zeta function for ℜpsqă 0: s 1 ´ s π´s{2Γ ζpsq“ π´p1´sq{2Γ ζp1 ´ sq . (34) 2 2 ´ ¯ ˆ ˙ The derivation of this variant of the Riemann zeta functional is provided in [7] p. 8-12. By setting s “ α ` iβ where α and β are reals, (34) is expressed as follows:

1´αíβ Γ 2 ζpα ` iβq“ πpαìβ´1{2q ζp1 ´ α ´ iβq . (35) ´ αìβ ¯ Γ 2 Let us consider (35) when α ă 0. In the previous´ ¯ section, we have shown that the Riemann zeta function has no zeros when ℜpsqą 1, hence ζp1 ´ α ´ iβq is never equal to zero when α ă 0. The Gamma function is never equal to An investigation of the non-trivial zeros of the Riemann zeta function 13

zero (see insider of proof proposition 6). Thus, the zeros of the Riemann zeta ℜ s function when psqă 0 are found at the poles of Γ 2 provided ζp1 ´ sq has no poles when ℜpsqă 0. From proposition 9, the Riemann zeta function ζpzq ` ˘ only has one pole at the point z “ 1, hence ζp1 ´ sq has no poles when α ă 0. Thus, we obtain the trivial zeros of ζpsq at s “ ´2, ´4, ´6, ´8, ... Note that s “ 0 is not a zero of the Riemann zeta function due to the pole at ζp1q. We 1 have ζp0q“´ 2 , see [18].

5 Zeros on the lines ℜpsq “ 0, 1

The fact that the Riemann zeta function has no zeros on the line ℜpsq “ 1 was already established by both Hadamard and de la Vall´ee Poussin in their respective proofs of the prime-number theorem. A sketch of the proof of de la Vall´ee Poussin is available in [2] p. 79-80. We provide here a summary for the sake of completeness.

From the Euler product, we have:

1 8 “ p1 ´ p´sq , (36) ζpsq p ź where p is the sequence of prime numbers larger than 1. Eq. (36) is deﬁned for ℜpsqą 1.

By expanding (36), we get:

1 8 8 8 “ 1 ´ pp´sq` ppqq´s ´ ppqrq´s ` ... , (37) ζ s p q p păq păqăr ÿ ÿ ÿ where p, q, r, ... are prime numbers. Therefore, we get an inﬁnite sum of integers, which are the product of unique primes. For such an integer n, the coeﬃcient of n´s is `1 if the number of prime factors of n is even, and ´1 if the number of prime factors of n is odd. Thus, we get:

1 8 “ µpnq n´s , (38) ζpsq n“1 ÿ where µpnq is the M¨obius function.

8 8 1 ´s lnpnq We have ζpsq“ ns “ e . Hence, we get: n“1 n“1 ř ř 8 ζ1psq“´ lnpnq n´s . (39) n“1 ÿ By multiplying (38) with (39) we get the logarithmic derivative of ζpsq, which is deﬁned for ℜpsqą 1. The expansion yields: 14 Yuri Heymann

ζ1psq 8 “´ Λpnq n´s , (40) ζpsq n“2 ÿ where Λpnq is the Mangoldt function, which is equal to lnppq if n “ pk for some prime p and integer k ě 1 and 0 otherwise.

Let us say we have a function f which is holomorphic in the neighborhood of a point a. Suppose there is a zero in a, hence we can write fpzq “ pz ´aqn gpzq 1 f pzq where gpaq ‰ 0. By computing the derivative of f, we get: pz ´ aq fpzq “ 1 1 1 g pzq f pa`εq g pa`εq n ` pz ´ aq gpzq . We set ε “ z ´ a, hence we get ε fpa`εq “ n ` ε gpa`εq . 1 g paq Suppose a is not a pole of f, hence we get gpaq is equal to a constant. If we 1 f pa`εq take the limit of the real part of ε fpa`εq when ε tends to zero, we get: f 1pa ` εq wpf,aq“ lim ℜ ε “ n , (41) εÑ0 fpa ` εq „  which is the multiplicity of the zero in a.

Let us set s “ 1 ` ε ` iβ where β is a real number and ε ą 0. We get:

ζ1psq 8 ℜ ε “´ε Λpnq n´1´ε cospβ lnpnqq . (42) ζpsq n“2 „  ÿ The proof uses the Mertens’ trick, which is based on the below inequality:

0 ď p1 ` cos θq2 “ 1 ` 2cos θ ` cos2 θ 1 ` cosp2θq “ 1 ` 2cos θ ` 2 (43) 3 1 “ ` 2cos θ ` cosp2θq . 2 2 Hence, we get:

3 ` 4cos θ ` cosp2θqě 0 . (44) From (42) and (43), we get:

ζ1 ζ1 ζ1 3 ℜ ε p1 ` εq ` 4 ℜ ε p1 ` ε ` iβq ` ℜ ε p1 ` ε ` 2 iβq ď 0 , (45) ζ ζ ζ „  „  „  where ε ą 0. By taking the limit of (45) when ε tends to zero from the right (ε ą 0), we get:

3 wpζ, 1q` 4 wpζ, 1 ` iβq` wpζ, 1 ` 2 iβqď 0 . (46) An investigation of the non-trivial zeros of the Riemann zeta function 15

From proposition 9, ζpsq only has a simple pole in s “ 1. Hence, when β ‰ 0, we have wpζ, 1 ` 2 iβqě 0 and wpζ, 1q“´1. Thus, we get:

3 0 ď wpζ, 1 ` iβqď´ wpζ, 1q 4 (47) 3 ď . 4

As the Riemann zeta function is holomorphic, the multiplicity of a zero in a denoted wpζ,aq must be an integer. Thus, according to (47), @β ‰ 0, wpζ, 1 ` iβq “ 0. Hence, we can say that the Riemann zeta function has no zeros on the line ℜpsq“ 1.

The fact that ζpsq has no zeros on the line ℜpsq“ 0 is obtained by reﬂecting the line ℜpsq“ 1 with the Riemann zeta functional. By considering (35) when α “ 0, we get:

1´iβ Γ 2 ζpiβq“ πpiβ´1{2q ζp1 ´ iβq . (48) ´ iβ ¯ Γ 2 ´ ¯ As the Gamma function is never equal to zero, the zeros on the line ℜpsq “ iβ iβ 0require that Γ 2 is a pole. The only pole of Γ 2 occurs when β “ 0. However, as ζps´q has¯ a pole in s “ 1, we cannot say´ ¯ that the point s “ 0 1 1 is a zero. In fact, we have ζp0q“´ 2 . To show that ζp0q“´ 2 , we use the Riemann zeta functional (3), which is expressed as follows:

sπ ζpsq“ πs´12s sin Γ p1 ´ sqζp1 ´ sq . (49) 2 ´ ¯ The Gamma functional equation, which is derived using integration by part on the integral form of the Gamma function, see formula 6.1.15 p. 256 in [1] is expressed as follows:

sΓ psq“ Γ ps ` 1q . (50)

s´1 s sπ By multiplying (49) with p1 ´ sq, we get p1 ´ sqζpsq “ π 2 sin 2 p1 ´ sqΓ p1 ´ sqζp1 ´ sq. From (50), we get p1 ´ sqΓ p1 ´ sq“ Γ p2 ´ sq. As ζpsq has ` ˘ a simple pole in s “ 1 (see proposition 9 ), we get limp1 ´ sqΓ psq“´1. Also, sÑ1 we have lim πs´12s sin sπ Γ p2 ´ sqζp1 ´ sq“ 2ζp0q. Hence, ζp0q“´ 1 . sÑ1 2 2 Therefore, we can say` the˘ Riemann zeta function has no zeros on the line ℜpsq“ 0. 16 Yuri Heymann

6 Zeros in the critical strip ℜpsqPs0, 1r

Let us set the complex number s “ α ` iβ where α and β are reals and its complex conjugates ¯ “ α´iβ. From proposition 8, if s is a zero of the Riemann zeta function in the strip ℜpsqPs0, 1r, then ζpsq“ ζp1 ´ s¯q.

Let us rewrite the latter equation in terms of α and β. We get:

ζpα ` iβq“ ζp1 ´ α ` iβq . (51)

1 R A solution set of (51) is α “ 2 for any β P , where zeros of the Riemann zeta function are found.

The Riemann hypothsis states that all non-trivial zeros, i.e. in the critical strip ℜpsq Ps0, 1r lie on the critical line ℜpsq “ 1{2. By reﬂexion of the zeros ℜ 1 of the Riemann zeta function with respect to the critical line psq “ 2 (see proposition 8 ), it is enough to show there are no zeros on either of the intervals ℜ 1 1 psq Ps0, 2 r or s 2 , 1r, to say all non-trivial zeros of the Riemann zeta function lie on the critical line ℜpsq“ 1{2.

7 Discussion

The Riemann hypothesis is an old problem of number theory related to the prime-number theorem, which statement is that all non-trivial zeros lie on the critical line ℜpsq“ 1{2. While the Riemann zeta functional leads to proposition 8, this criterion alone is not enough to say whether the Riemann hypothesis is true or false. Since this criterion establishes a symmetry between the zeros of the Riemann zeta function on either sides of the critical line ℜpsq“ 1{2, it is enough to show that the Riemann zeta function has no zeros on either of the ℜ 1 1 strips psq Ps0, 2 r or s 2 , 1r, to say the Riemann hypothesis is true.

References

1. Abramowitz, M., Stegun, I.: Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables. National Bureau of Standards Applied Mathematics Series 55 (1964) 2. Edwards, H.M.: Riemann’s Zeta Function. Dover Publications Inc. (2003) 3. Firk, F.W.K., Miller, S.J.: Nuclei, primes and the random matrix connection. arXiv:0909.4914 pp. 1–54 (2009) 4. Grigorieva, E.: Methods of Solving Sequence and Series Problems. Birkhäuser (2016) 5. Hadamard, J.: Etude´ sur les propriétés des fonctions entières et en particulier d’une fonction considérée par Riemann. J. de Math. Pures Appl. 9, 171–215 (1893) 6. Hadamard, J.: Sur la distribution des zéros de la fonction ζpsq et ses conséquences arithmétiques. Bull. Soc. Math. France 24, 199–220 (1896) 7. Ivic, A.: The Riemann zeta-function: theory and applications. Dover Publications, Inc (2003) 8. Krantz, S.G.: Foundations of Analysis. CRC Press (2015) An investigation of the non-trivial zeros of the Riemann zeta function 17

9. Lapidus, M.L., Maier, H.: The Riemann hypothesis and inverse spectral problems for fractal strings. J. London Math. Soc. 52, 15–34 (1995) 10. Mazur, B., Stein, W.: Prime Numbers and the Riemann Hypothesis. Cambridge Uni- versity Press (2016) 11. Mehta, M.L.: Random Matrices. Academic Press; 2nd edition (1991) 12. Milinovich, M.B., Ng, N.: Simple zeros of modular L-functions. Proc. London Math. Soc. 109, 1465–1506 (2014) 13. Organick, E.I.: A Fortran IV primer. Addison-Wesley (1966) 14. Riemann, G.F.B.: Uber¨ die Anzahl der Primzahlen unter einer gegebenen Grösse. Monatsber. Königl. Preuss. Akad. Wiss. Berlin Nov., 671–680 (1859) 15. Selberg, A.: An elementary proof of the prime-number theorem. Annals of Mathematics, Second Series 50, 305–313 (1949) 16. Titchmarsh, E.C., Heath-Brown, D.R.: The Theory of the Riemann Zeta-function. Ox- ford University Press, 2nd ed. reprinted in 2007 (1986) 17. de la Vallée Poussin, C.J.: Recherches analytiques sur la théorie des nombres premiers. Ann. Soc. Sci. Bruxelles 20, 183–256 (1896) 18. van der Veen, R., van des Craat, J.: The Riemann hypothesis. MAA Press (2015)