Proof of the Riemann Hypothesis antzakouras Nikos(nikmantza@gmail.com) August 2015 - Athens Abstract: The Riemann zeta function is one of the most LeonhardEuler important and fascinating functions in mathematics. Analyzing the matterof conjecture of Riemann divide our analysis in the zeta function and in the proof of conjecture, which has consequences on the distribution of prime numbers. .Introduction #1.The Riemann Zeta Function Let C denote the complex numbers. They form a two dimensional real vector space spanned by 1i and where i is a fixed square rootof-1, that is, C = {x + iy : x,y e R} #2.Definition of Zeta. The Riemann zeta function is the function of a complex variable It is conventional to writeit+ = s and it- = swhere , t eR. As the equality on the second line follows from unique prime factorization, we can say that the equation

is an analytic statement of unique prime factorization." It is known as the Euler product. This gives us a first example of a connection between the zeta function and the primes.[3] #3. Definition of complex function,Inverse functions.Suppose A, B two subsets of the complex plane C. Each f tally the sets A and B f: A B with the restriction that each zA corresponds one and only one point w = f (z) B, called complex function with domain the set A and the range set B. The z called independent variable and the dependent variable w.In cases where the above restriction does not apply, ie where that each value of the variable zA match more than one value f (z), the f match is called multi-valued function in contrast to the previous case where you use the term single-valued function or simply function. If we set z = x + iy and w = u + iv, then the functionw = f (z) corresponds to each complex number zA with coordinates x and y two real numbers u and v.In other words, the set A set of two real functions u = u (x, y) and v = v (x, y) of two real variables x and y. So a complex function w = f (z) is equivalent to two real functions u = u (x, y) and v = v(x, y) of two real variables. Example.. The complex function:ixy y x iy x z 2 ) (2 2 2 2+ = + =It is equivalent to the actual functions: xy v y x u 2 ,2 2= =If a complex function w = f (z) is one to one (1-1) and on (i.e. z1 z2 f (z1) f (z2) or f (z1) = f (z2) z1 = z2 and f (A) = B), then the correlation that corresponds to each wV that zA such that f (z) = w, defines a new function, called reverse the function f (z) and is denoted by1f,that is) (w z1f= .So if we have Zeta(z1)=Zeta(z2) z1 = z2 in the domainof set A, if occurring earlier.In this case Zeta[z] is 1-1 which is acceptable. #4.Extensions of holomorphic functions One of the main properties of holomorphic functions is uniqueness in the sense that if two holomorphic functions f and g defined in a domain G are equal on a sequenceG zn e , G z G z Limnne = e 0 i.e.,) ( ) (n nz g z f = for n = 1, 2..,, then f = g in G, see Fig.1. Fig..1

Of course such a property is not true for functions in real calculus. In particular, if a holomorphic function f is de_ned in a domainC G c1and another holomorphic function g is defined in a domainC G c2 with02 1= G Gand f = g on the intersection, then g is determined uniquely by f; see Fig. 2. Fig..2 f cannot be holomorphically extended beyond this disc. A simple example: 1 , ) (0< == nnz z z f Obviously the series diverges for values1 s z . However, the function f can be holomorphically extended to the entire complex plane C except z = 1, by the formula A natural question appears: whether the Riemann zeta function can be holomo-rphically extended beyond the half-plane Re s > 1? The answer is yes, which we show in two steps. The first step is easy, the second more dificult. #5.Extension of (s) from {Re s > 1} to {Re s > 0} Let us calculate We obtained another formula for (s), Fig..3 so the alternating series converges in a bigger half-plane (see Fig. 3) than the originally defined function (s) in (3), but we have to remove s = 1 since the denominators 12 1 vanishes there. This rather easy extension of(s) from s with Re s > 1 to s with Re s > 0 is already signi_cant as it allows us to formulate the Riemann Hypothesis about the zeros of(s) in the critical strip. #6.Functional equations for the Riemann zeta function. The second step, which provides a holomorphic extension for (s) from {Re s > 0,1 = s } {Re s < 0}, see Fig. 4, was proved by Riemann in 1859.We do not give a proof here of the so-called functional equation, but the proof can be found, e.g. in the book by Titchmarsh . Alternatively one can first holomorphically extend (s) step by step to half-planes {Re s > k, 1 = s }, where k is any negative integer. For details of this method, see for example the papers [5] and [6]. There are few versions of the functional equation; here we formulate two of them: where Fig..4 Before we give more information about the function (s), we mention that each of the two before equations , give an extension of (s) on the entire plane C except s = 1, as is illustrated in Fig. 4. The gamma function was already known to Euler. It generalizes the factorial n!, namely Its basic properties are that (z) is holomorphic on the entire plane C except for the points z = 0,-1,-2,... . At these points there are simple singularities, called poles, where we have the limits

From the de_nition of the gamma functionit is not clear that it can be extended onto the entire plane except for z = 0,-1,-2,... , and that is non-vanishing. Fortunately there are other equivalent de_nitions of(z) from which these properties follow more easily; see[8,9]. Namely we have From the second formula for the gamma function we see that (z) is holomorphic and nonvanishing for Re z > 0. #7.The Riemann Hypothesis The Riemann Hypothesis is the most famous open problem in mathematics. Originally formulated by Riemann, David Hilbert then included the conjecture on his list of the most important problems during the Congress of Mathematicians in 1900, and recently the hypothesis found a place on the list of the Clay Institute's seven greatest unsolved problems in mathematics.It follows from the formula of (s) as the product that the function does not vanish for Re s > 1.Next, using the functional equationand the fact that0 (z) = for Re z > 0, we see that (s) vanishes in the half-plane Re s < 0 only at the points where the function sine is zero, namely we obtain... The above considerations do not tell us about the zeros of (s) in the strip0 +1,then N n kn Lnkt tk ke + =+, ,) (21t. Proof .. i) Here we formulate two of the functional equations We look at each one individually in order to identify and set of values that we want each time .. a. For the first equation and for real values with Re s>0 and if takelogarithm of 2 parts of equation then we have i k s s Cos LogsLog Log s s Log s s Cos s sst tt , , t t , ,2 )] ( ) 2 / ( [] 2 [ ] 2 [ )] ( / ) 1 ( [ ) ( ) 2 / ( ) 2 ( 2 ) ( / ) 1 (+ I+ = => I = but solving for sand if) ( ) 2 / ( ) ( s s Cos s f I = tand from Lemma 2 if 1 ) ( / ) 1 ( lim0= s ss s, ,or0 )] ( / ) 1 ( lim [0= s s Logs s, , we get ] 2 [2 )] ( [] 2 [] 2 [ttt Logi k s f LogLogLogs++ = with 0] 2 [2 )] ( [>=+ttLogi k s f Log Finally because we need real s we will have Re s 3771 . 0] 2 [] 2 [= >t LogLog.This is the lower bound which gives us, the first of Riemann Zeta Function of (s). b.For the second equation for real values with Re s I = but solving for sand if) 1 ( ) 2 / ( ) ( s s Sin s f I = tand from Lemma 2if1 ) 1 ( / ) ( lim0= s ss s, ,or0 )] 1 ( / ) ( lim [0= s s Logs s, , we get ] 2 [2 )] ( [] 2 [] [ttttLogi k s f LogLogLogs+ = with 0] 2 [2 )] ( [>=+ttLogi k s f Log In Follow because we need real s we will take Re s 62286 . 0] 2 [] [= sttLogLog.This is the upper bound which gives us, the second of Riemnn Zeta Function of (s). So we see that existsthe lower and upper bound for Re s and is well defined. ii) Assumingk low kit s + =o and k upperit s + = ' okwith ] 2 [] 2 [toLogLoglow = and ] 2 [] [ttoLogLogupper =thenapply lowo o =1anduppero o=2 in Next do the differences 1228 . 0 2 / 1 2 / 1 Re Re = = ' = ' Auppers s ok k 1228 . 0 2 / 1 Re 2 / 1 Re = = = Alow k ks s oThen this suggests our absolute symmetry aroundfrom the middle value is 2 / 1 Re = s , generally as shown in Figure 6 Fig. 6 iii) THE AVERAGE value of upper lower bound is Re s=1/2 because2 / 1 Re =s 2 / 1] 2 [] [ ] 2 [=+ttLog Log Log iv)If we accept the non-trivial zeroes of the Riemann Zeta Function (s) as 1 0 1it s + =o and 2 0 2it s + = o with 2 1s s > , suppose that the real coordinate 0oof each non-trivial zero of the Riemann Zeta, [2] function (s )corresponds with two imaginative coordinates t1 and t2 , then, we have folloing equation group: Taking the first equation minus the second equation, we obtain where =0Enable above expression equals to zero, we must have so, we obtain Thus, theorem 1 has been proved. Theorem 2. For real part of non-trivial zeroes Riemann Zeta function (s), the upper-lower bound converge in lineRe s=1/2, then solving the 2 equations ofthe Riemann zeta functions. Proof.. Using the Generalized theorem of Lagrange (GRLE),[7] to solve an equation attempt to bring the equation in such form, as to give us the limit of the roots, which converge on one number. Taking advantage of the method, we develop resolve in 2 parts to show the convergence as a number, using the two equations ofthe Riemann zeta functions . a)Solve of first equation For real values with Re s>0 and if takelogarithm of 2 parts of equation[7] then we have i k s s Sin LogLog s Log s s Log s s Sin s sst tt , , t t , ,2 )] 1 ( ) 2 / ( [] 2 [ ) 1 ( ] 2 [ )] 1 ( / ) ( [ ) 1 ( ) 2 / ( ) 2 ( 2 ) 1 ( / ) (1+ I+ + = => I = the correlation theory we ill have after using Lemma 2 ie..0 )] 1 ( / ) ( lim [0= s s Logs s, ,and relations 3 groups fields, and therefore for our case weget.. y s p = ) (1 this means thats y p y f = =) ( ) (11,but with an initial value] 2 [2ttLogy iks+=With yLog[] and total form Withq is the count of repetitions of the Sum(). We very simply with a language mathematica to calculate with data q=25,and we get the program. In():Clear[k,q,y] k := 1; q := 25; t := (Log[]); s = N[(2 I k + y)/(Log[2] + Log[])] +

w 1q1wGammaw 1D 1 Log2LogSin 22 I k yLog2 LogLogGamma12 I k yLog2 Logw,y, w 1 FQ = N[s /. y -> t, 30]; N[1 - 2^(FQ)*()^(FQ - 1)*Sin[*FQ/2]*Gamma[1 - FQ]] Out(): 0.500006 +3.43623 I -3.71908*10-6-6.00526*10-6 I we see a good approximation of upper Re s=0.500006 of the order 10-6 Schematically illustrate get fig 7 with program mathematica Clear [z]; f=Log[1-z]/Log[z]-2*(2*)^(-z)*Cos[*z/2]*Gamma[z];ContourPlot[{Re[f/.z->x+I y]==0,Im[f/.z->x+I y]==0},{x,-15,15},{y,-10,10},ContourStyle->{Red,Blue}] 15 10 5 0 5 10 151050510 Fig 7 b)Solve of second equation For real values with Re s I = the correlation theory we ill have after using Lemma 2 ie..0 )] ( / ) 1 ( lim [0= s s Logs s, ,and relations 3 groups fields, and therefore for our case we get.. y s p = ) (1 this means thats y p y f = =) ( ) (11,but with an initial value] 2 [2ttLogy iks+ =With yLog[2] and total form Withq is the count of repetitions of the Sum(). Similarly forming the program a language mathematica to calculate with data q=25,we have. In():k := 1; q := 25; t := -Log[2]; S =-(2 I k + y)/Log[2] +w 1q1wGamma w 1yw 11 Log2 LogCos 22 I k yLog2 LogLogGamma2 I k yLog2 Logw FQ = N[der /. y -> t, 30] N[1 - 2*(2*)^(-FQ)*Cos[*FQ/2]*Gamma[FQ]] Out(): 0.499994 -3.43623 I, -3.71908*10-6-6.00526*10-6 I a good approximation of lower Re s=0.499994 of the order 10-6Schematically illustrate get fig 8 with program mathematica Clear [z];f=Log[z]/Log[1-z]-2^z*()^(z-1)*Sin[*z/2]*Gamma[1-z];ContourPlot[{Re[f/.z->x+I y]==0, Im[f/.z->x+I y]==0},{x,-15,15},{y,-10,10},ContourStyle->{Red,Blue}] 15 10 5 0 5 10 151050510

Fig.8With this analysis as we see the lower and upper limit converge in value Re s=1/2, what we want results which proves the theorem2. Theorem 3. The Riemann Hypothesis states that the nontrivial zeros of (s) have real part equal to 1/2. Proof Constant Hypothesis (s)=0. In this case we use the 2 equations ofthe Riemann zeta functions, so if they apply what they represent the (s) and (1-s) to equality. BEFORE We developed the method make three assumptions: i) a)ForC z e1.{ 0 Re 0 0 > . = = z a az} 2.{ 0 0 1 . = z a az} which It refers to the inherent function similar in two Riemann zeta functions as0 ) 2 ( =ztor0 ) 2 (1= zt and it seems they do not have roots in C-Z because 0 ) 2 ( = t . b)For allZ C z etrue the form z Sinz Gamma z Gammatt= ) 1 ( ) ( . From this form resulting that Gamma(z)=0 or Gamma(1-z)=0 dont has roots in C-Z. c)Sin(/2z)=0 or Cos(/2z)=0 More specifically if z=x+yi then 1.0 y)/2] Sinh[( x)/2] Cos[( I x)/2] Sin[( y)/2] Cosh[( 0 ) 2 / ( = + = t t t t t z SinHas solution of the generalized solution seems that the pairs (x, y) will always arise integers which makes impossible the case is 0