x s 1 e x dx, for σ > 1. If we replace x by nx in the integral then we obtain x s 1 e nx dx. x s 1

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1 Recall 9. The Riemann Zeta function II Γ(s) = x s e x dx, for σ >. If we replace x by nx in the integral then we obtain Now sum over n to get n s Γ(s) = x s e nx dx. x s ζ(s)γ(s) = e x dx. Note that as σ > the integral is absolutely convergent at both ends, x = and x = and so we can switch the order of integration and summation. Also we define x s (s ) log x = e unambiguously, in the usual way. Now we define two paths C and C n. For C we come in from positive infinity just above the real axis, describe most of a small circle centred at the origin and return to infinity just below the real axis. We don t care too much about the exact definition of C except that the circle has radius r less than. For C n we start at point of C describe most of a square encompassing ±ki, k n and end at point of C just below the x-axis. We then describe the bounded part of C to complete a full cycle. Theorem 9.. If σ > then Γ( s) ζ(s) = i C e z dz, where is defined on the complement of the positive real x-axis as e (s ) log( z) with < Im log( z) <. Proof. The integral obviously converges. By Cauchy s theorem the integral does not depend on C, as long as C does not go around any non-zero multiples of i. In particular we are free to let the radius of the circle go to zero. Consider the integral around the circular part of C. As r goes to zero the length of the path is proportional to r. As the denominator is also proportional to r and goes to zero the integral around the circular part goes to zero.

2 We are left with an integral along the positive real axis described both ways. On the upper edge and on the lower edge = x s e (s )i, = x s e (s )i. It follows that C e z dz = x s e (s )i dx + e x = i sin (s )ζ(s)γ(s). Now use the fact that x s e (s )i dx e x sin (s ) = sin s and Γ(s)Γ( s) = sin s. Corollary 9.. The ζ-function can be extended to a meromorphic function on the whole complex plane whose only pole is a simple pole at s = with residue. Proof. Consider the RHS of the equation in (9.). Γ( s) is meromorphic on C and the integral defines an entire function. Since the RHS is meromorphic on the whole complex plane we can use this equation to extend ζ(s) to a meromorphic function on the whole complex plane. Γ( s) has poles at s =, s =,.... But we already know that ζ(s) is holomorphic for σ > so the zeroes of the integral must cancel with the poles and the only pole is at the origin. As s =, Γ( s) has a simple pole with residue. On the other hand dz = i, C e z by the Residue theorem. Thus ζ(s) has residue one at s =. We can calculate ζ( n) where n N explicitly. We already know that e z = z + ( ) k B k (k)! zk. We have ζ( n) = ( ) n n! z) n i C e z dz. Thus ζ( n) is ( ) n n! times the coefficient of z n above. in the expansion

3 Thus ζ() = / ζ( m) = and ζ( m + ) = ( )m B m m. The points m are called the trivial zeroes of the ζ-function. Note that if σ > we have ζ(s) ζ(σ). Thus we have good control on the ζ-function for σ >. In fact we also have good control for σ < : Theorem 9.3. ζ(s) = s s sin s Γ( s)ζ( s). Proof. We use the path C n. We assume that square part is defined by the lines t = ±(n + ) and σ = ±(n + ). The cycle C n C has winding number one about the points ±mi with m =,,..., n. The poles at these points of are simple with residues Thus i C n C e z dz = e z ( mi) s. = = = [ ( mi) s + (mi) s ] m= (m) s (i s + ( i) s ) m= (m) s (i s ( i) s )/i m= (m) s sin s, m= where we used the fact that i = e i/. We divide C n into two parts, C n + C n, where C n is the square bit and C n is the rest. It is easy to see that e z is bounded below on C n by a fixed positive constant, independent of n, while is bounded by a multiple of n s. The length of C n is of the order of n and so e z dz Anσ, C n 3

4 for some constant A. If σ < then the integral over C n will tend to zero as n tends to infinity and the same is true for the integral over C n. Therefore the integral over C n C will tend to the integral over C and so the LHS tends to ζ(s) Γ( s). Under the same condition on σ the series m s converges to ζ( s). Thus the RHS is a multiple of ζ( s). Taking the limit gives the desired equation. A priori this is only valid for σ < but if two meromorphic functions are equal on an open set they are equal everywhere. One can rewrite the functional equation. For example, replacing s by s we have ζ( s) = s s cos s Γ(s)ζ(s). One can also derive this using the functional equation Γ(s)Γ( s) = sin s. We also have Corollary 9.4. The function ξ(s) = s( s) s/ Γ(s/)ζ(s), is entire and satisfies ξ(s) = ξ( s). Proof. ξ(s) is a meromorphic function on C. The pole of ζ(s) at s = cancels with s and the poles of Γ(s/) cancel with the trivial zeroes of ζ(s). Thus ξ(s) is an entire function. Note that from the functional equation Γ(s)Γ( s) = sin s we get Γ(( s)/)γ(( + s)/) = cos s/. Recall also Legendre s duplication formula Γ(z) = z Γ(z)Γ(z + /). 4

5 ξ( s) = ( s)s(s )/ Γ(( s)/)ζ( s) = s( s)(s )/ Γ(( s)/) s s cos s Γ(s)ζ(s) = s( s) s/ ζ(s) s / Γ(s)Γ(( s)/) cos s = s( s) s/ ζ(s) / Γ(s) s Γ(( + s)/) = s( s)ζ(s) s/ Γ(s/) = ξ(s). We know from the series development of ζ(s) that there are no zeroes in the region σ >. Using the functional equation it follows that ζ(s) has no zeroes in the region σ < apart from the trivial zeroes. So all of the zeroes belong to the strip σ. The Riemann hypothesis states that the only zeroes in the strip σ belong to the line σ = /. It is known that there are no zeroes on the lines σ = and σ =. It also known that at least one third of the zeroes lines on the line σ = /. 5