How to prove the Riemann Hypothesis

The aim of this paper is to prove the celebrated Riemann Hypothesis. I have already discovered a simple proof of the Riemann Hypothesis. The hypothesis states that the nontrivial zeros of the Riemann zeta function have real part equal to 0.5. I assume that any such zero is s = a+ bi . I use integral calculus in the first part of the proof. In the second part I employ variational calculus. Through equations (50) to (59) I consider (a) as a fixed exponent, and verify that a = 0.5 .From equation (60) onward I view (a) as a parameter (a <0.5 ) and arrive at a contradiction. At the end of the proof (from equation (73)) and through the assumption that (a) is a parameter, I verify again that a = 0.5


INTRODUCTION
The Riemann zeta function is the function of the complex variable s = a + bi (i = 1  ), defined in the half plane a >1 by the absolute convergent series The function ) (s  has zeros at the negative even integers -2, -4, … and one refers to them as the trivial zeros. The Riemann hypothesis states that the nontrivial zeros of ) (s  have real part equal to 0.5.

PROOF OF THE HYPOTHESIS
We begin with the equation Separating the real and imaginary parts we get In equation (11) replace the dummy variable x by the dummy variable y We form the product of the integrals (12)and (13).This is justified by the fact that both integrals (12) and (13) are absolutely convergent .As to integral (12) we notice that And as to integral (13) If we replace the dummy variable z by the dummy variable x, the integral takes the form

Journal of Progressive Research in Mathematics(JPRM) ISSN: 2395-0218
Volume 12, Issue 2 available at www.scitecresearch.com/journals/index.php/jprm 1857| Let p <0 be an arbitrary small positive number.We consider the following regions in the x -y plane.
Let us find the limit of F (x,y) as x   and y   . This limit is given by ((z)) is the fractional part of the number z ,0  ((z)) < 1 The above limit vanishes ,since all the functions [ -((y)) ] ,cos (blog xy ), -(( x 1 )), and ((x)) remain bounded as x   and y   Note that the function F (x,y) is defined and bounded in the region I 1. We can prove that the integral where t is a very small arbitrary positive. number. Since the integral ] dx is bounded.

I3 the integral
We denote the integrand of (47) by ( here we do not consider a as a parameter, rather we consider it as a given exponent) Using equation (50) we deduce that Since the value of [ F x  ] (at x = p)is bounded, we deduce from equation (54) that Thus we must have that First we compute We conclude from (56) that the product Assume that a =0.5 .( remember that we considered a as a given exponent )This value a =0.5 will guarantee that the quantity { (( x 1 ))-1 2  a x } will remain bounded in the limit as (x 0  ) .Therefore , in this case (a=0.5) (56) will approach zero as (x 0  ) and hence remain bounded . Now suppose that a< 0.5 .In this case we consider a as a parameter. Hence we have Substituting from (61) we get We return to equation (49) and write Fdx ( t is a very small positive number 0<t<p) Let us compute That is We conclude from this equation that ( since lim(x 0  ) x = 0 , which is the same thing as lim(t 0  )  x = 0) Since 2 2 F ( at p) is bounded , we deduce at once that 2 2 F must remain bounded in the limit as (t 0  ), which is the same thing as saying that F must remain bounded in the limit as (x ) 0  . Therefore .
It is evident that this last limit is unbounded. This contradicts our conclusion (71) that must remain bounded (for a< 0.5 ) Therefore the case a<0.5 is rejected .We verify here that ,for a = 0.5 (71)remains bounded as (x 0  ) .
We have that  Hence we get Therefore we must apply L 'Hospital ' rule with respect to x in the limiting process (75) We must apply L 'Hospital ' rule Thus we have verified here that ,for a = 0.5 (71) approaches zero as (x 0  ) and hence remains bounded.
We consider the case a >0.5 . This case is also rejected, since according to the functional equation, if ( ) (s  =0) (s = a+ bi) has a root with a>0.5 ,then it must have another root with another value of a < 0.5 . But we have already rejected this last case with a<0.5

Journal of Progressive Research in Mathematics(JPRM)
ISSN: 2395-0218 Volume 12, Issue 2 available at www.scitecresearch.com/journals/index.php/jprm 1866| Thus we are left with the only possible value of a which is a = 0.5 Therefore a = 0.5

Conclusion.
The Riemann Hypothesis is now proved.