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.
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
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
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
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