3   The Riemann Zeta Function  Recall that the Riemann zeta function is defined by          s       n   1   1 n s     for Re   s     1   Remarkable properties    1        s   extends meromorphically to      2    A functional equation relating     s         1   s      3    All the  non trivial  zeroes appear to be on the line Re   s     1 2  Riemann hypothesis     4        s   closely relates to the distribution of primes   Notation  f   x     g   x   means   f   x     g   x     f   x        means that the constant in    can depend on     Lemma 3 1  Assuming that       1  t      Then            i t         1   Proof  By the Euler product formula               i t       p   1   p       i t     1       hence                i t         p   1   p       i t     1       By the triangle inequality       1   p         1   p       i t     1     p       i t     1   p           Hence         p   1   p         1             i t         p   1   p         1       Note that these products converge if and only if   p p     converges  and this sum converges by the comparison test     Lemma 3 2   Polynomial growth of e t a in half planes     i  f       extends to a meromorphic function on    with the only pole being a simple pole which is at s   1    ii  Let k   0 be an integer  Then  for Re   s       k and   s   1     1 1 0 we have        s       k   s   k   2   1   Proof  Let k   0 be an integer   We claim that there exist polynomials P k  Q k of degree   k   1 and such that for Re   s     1          s     1 s   1   Q k   s     s   s   1       s   k     1   P k     t     t s   k   1 d t           First assume      holds   Then  since  P k     t       k 1  for Re   s       k   1 2    s   1     1 1 0       gives        s       k   s   k   2   1   So  ii  follows   For  i   using analytic continuation and       it suffices to show that the RHS of      is meromorphic for Re   s       k  with the only pole a simple one at s   1   Suffices to show that        1   P k     t     t s   k   1 d t     is analytic for Re   s       k   This follows from the following criterion  If U     is open  f   U         is piecewise continuous  and if s   f   s   t   is analytic in U for any t      then     f   s   t   d t is analytic in U  provided that       f   s   t     d t is bounded on compact subsets of U   Applying this with  f   s   t     P k     t     t s   k   1     1       concludes the proof of  i  assuming        We are left with proving       We use induction on k   Case k   0  By Partial Summation          s       n   1   1 n s   s   1     u   u s   1 d u  apply partial summation to   n   x n   s and let x         s   1   1 u s d u     1     u   u s   1 d u   1 s   1   1   s   1     u   u s   1 d u     Take P 0   u     u  Q 0   u     1   Case k   1 assuming k  Let      c k     0 1 P k     t     d t       For Re   s     1  we have          s     1 s   1   Q k   s     c k s   s   1       s   k   1     s   s   1       s   k     1   P k     t       c k t s   k   1 d t       Let      P k   1   u         0 u   P k   t     c k   d t       This is a polynomial of degree   k   2  By integration by parts         1   P t     t       c k t s   k   1 d t     s   k   1     1   P k   1     u     u s   k   1 d u       Substituting this into the previous equation  we get that case k   1 follows     The Gamma function  Definition   Gamma function   For Re   s     0  let          s       0   t s   1 e   t d t       Lemma 3 3      s   is analytic for Re   s     0   Proof  Apply the same criterion for integral of analytic function being analytic as in the previous lemma  taking  f   s   t     t s   1 e   t     0         t     Note that for 0     1   Re   s       2            f   s   t     d t     0 1 t   1   1 e   t d t     1   t   2   1 e   t d t             Lemma   Functional equation for     The   function extends meromorphically to    with the only poles being simple poles at s   0     1     2       Moreover    i      s   1     s     s   for s        ii      s       1   s       sin     s   for all s      Euler reflection formula    Proof   i  For Re   s     0  by integration by parts         0   t s e   t d t   s   0   t s   1 e   t d t       This proves  i  for Re   s     0   Now for any k      for Re   s     0 we have          s         s   k     s   k   1       s   1   s       The RHS is analytic for Re   s       k  so can use analytic continuation to extend     s   meromorphically to Re   s       k  with the only poles being simple poles at s   0     1         k     1   Let k        ii  Since both sides are analytic in        by analytic continuation  it suffices to prove the formula for 0   s   1  Now  for any t   0      t s   1     s   1     t s   1   0   u   s e   u d u     0   v   s e   v t d t   u   v t       Multiply by e   t and integrate  and use Fubini to get         s       s   1       0     0   v   s e   v t d v e   t d t     0     0   e     v   1   t d t v   s d v     0   v   s 1   v d v           e   1   s   x 1   e x d x   v   e x       Hence  the remaining task is to show              e   1   s   x 1   e x d x     sin     s         We will do this next time      Theorem 3 4   Functional equation for z e t a    Assuming that       s     1 2 s   s   1       s   2     s s       s   for s      Then    is an entire function and     s         1   s   for s      Hence          s   2     s 2       s         1   s 2     1   s 2       1   s       for s         0   1     Proof  Let Re   s     1  Then           s 2       0   t s 2   1 e   t d t       Make the change of variables f     n 2 u to get          s 2       s   2 n s   0   u s 2   1 e     n 2 u d u       Hence          s 2 n   s     s 2       0   u s 2   1 e     n 2 u d u       Summing over n     and using Fubini          s 2     s 2       s       n   1     0   u s 2   1 e     n 2 u d u   1 2   0   u s 2   1       u     1   d u     u       n         e     n 2 u   1 2   0 1 u s 2   1       u     1   d u   1 2   1   u s 2   1       u     1   d u           By the functional equation          u     1 u     1 u         we have       0 1 u s 2   1       u     1   d u     0 1 u s 2   3 2     1 u   d u     0 1 u s 2   1 d u     1   v   s   1 2     v   d v     0 1 u s 2   1 d u   u   1 v       1   v   s   1 2       v     1   d v     1   v   s   1 2 d v   1 s   1 2     0 1 u s 2   1 d u   1 s 2     Plugging this into       we get          s 2     s 2       s     1 2   1     u   s   1 2   u s 2   1         u     1   d u   1 s   s   1         Hence          s       1 2   1 4 s   s   1     1     u   s   1 2   u s 2   1         u     1   d u            Since        u     1     e     u  applying the criterion for integrals of analytic functions being analytic  we see that     s   is entire  So by analytic continuation          holds for all s       Moreover  the expression for     s   is symmetric with respect to s   1   s  so     s         1   s    s         Corollary 3 5   Zeroes and poles of z e t a    The   function extends to a meromorphic function in   and it has   i  Only one pole  which is a simple pole at s   1  residue 1    ii  Simple zeroes at s     2     4     6        iii  Any other zeroes satisfy 0   Re   s     1   Proof   ii     iii    Follows from the lemma on polynomial growth of   on vertical lines    ii     iii    We know     s     0 for Re   s     1  We want to show that if     s     0 and Re   s     0 then s       2     4     6       and s is a simple zero  Let Re   s     0  By the functional equation for            s       s   1 2     0     1   s 2       s 2       1   s         We claim that     s     0 for all s      By the Euler reflection formula           s       1   s   sin     s             for s       If     s     0  then   has a pole at 1   s  Hence  1   s     n for some n   0 integer  But then s   n   1  and     n   1     n     0   We conclude that for Re   s     0          s     0   s 2 is a pole of     s     2 n   n         Since the poles of   are simple    has a simple zero at s     2 n  n         3 1   Partial fraction approximation of    This is a formula for       s       s     For the proof we need a lemma  We write  for z      r   0      B   z 0   r       z         z   z 0     r   B   z 0   r         z         z   z 0     r     B   z 0   r       z         z   z 0     r       Lemma 3 6   Borel Caratheodory Theorem    Assuming that   0   r   R  f analytic in B   0   R      with f   0     0  Then      sup   z     r   f   z       2 r R   r sup   z     R Re   f   z           Proof  This is Exercise 10 on Example Sheet 2     Lemma 3 7   Landau    Assuming that   z 0     and r   0  f analytic in B   z 0   r    for some M   1 we have   f   z       e M   f   z 0     for all z   B   z 0   r    Then  for z   B   z 0   r   4          f     z   f   z           Z 1 z         9 6 M r       where Z is the set of zeroes of f in B   z 0   r   2      counted with multiplicities   Note   If f is a polynomial  we can factorise      f   z     a       z             and then     f     z   f   z       log f   z           log a       log   z                 1 z         Proof  Let      g   z     f   z         Z   z             Then g is analytic and non vanishing in B   z 0   r   2      Note that      g     z   g   z     f     z   f   z           Z 1 z            f 1   f n     f 1   f n     i   1 n f 1   f 1    Hence  it suffices to prove        g     z   g   z       9 6 M r     for z   B   z 0   r   4    Write      h   z     g   z 0   z   g   z 0         Then h is analytic and non vanishing in B   0   r   2      and h   0     1  We want to show        h     z   h   z       9 6 M r     for z   B   0   r   4     For all z 0     B   0   r   we have        h   z         f   z 0   z   f   z 0         Z z 0     z 0   z           f   z 0   z   f   z 0       e M     since        z 0         r 2   r   r 2     z 0   z           for z     B   0   r     By the maximum modulus principle    h   z       e M for z   B   0   r   2      so     Re log h   z     log   h   z       M       By the Borel Caratheodory Theorem with radii 3 r 8  r 4 we have for for z   B   0   3 r   8          log h   z       2 r 4 3 r 8   r 4 M   4 M       Now  for z   B   0   r   4    Cauchy s theorem gives us       h     z   h   z         1 2   i     B   0   3 r   8   log h   w     z   w   2 d w     1 2     2   3 r 8 4 M   3 r 8   r 4     2   9 6 M r       Theorem 3 8   Partial Fraction approximation of z e t a     z e t a     i  Let s       i t with         1 0  s   1 and     s     0  Then            s       s       1 s   1           s     1 1 0 1 s       O   log     t     2           where the sum is over the zeroes   of   counted with multiplicity    ii  For any T   0  there are    log   T   2   many zeroes   of    counted with multiplicity  with   Im             T   T   1     Proof  We apply Landau with z 0   2   i t  r   5 0  with f   s       s   1       s     By the lemma on polynomial growth of   on vertical lines  for     i t   B   z 0   5 0    we have       f   s       C     t     5 2       t   2   5 0   C exp   5 1 log     t     5 2       C exp   5 1 log     t     5 2       f   z                z   i t       1    Let s   B   z 0   2 5 2    Then Landau gives us     f     s   f   s     1 s   1         s       s             z 0     2 S 1 s       O   log     t     2               Since B   z 0   2 5 2   contains all the points s       i t with         1 0  it suffices to show              z 0     2 5       s     1 1 0 1 s       O   log     t     2                Substituting s   z 0 in       we get             z 0     2 5 1 z 0       O           z 0       z 0       1   log     t     2       O   log     t     2         since             2   i t       2   i t           n   1       n   n   2   i t       n   1       n   n   2         2       2       Taking real parts               z 0     2 5 2   Re         z 0       2   O   log     t     2           Since Re         1              z 0     2 5 1   z 0       2   O   log     t     2           This proves part  ii   It gives also        since the sum there contains O   log     t     2     zeros     3 2   Zero free region  Proposition 3 9  Assuming that       1 and t      Then      3               4 Re               i t       Re               2 i t       0            Proof  Recall that              s         n   1       n   n s       where    is the von Mangoldt function function  and Re   s     1   Taking linear combinations  the LHS of        becomes          n   1       n   3   4 Re   n i t     Re   n   2 i t   n         n   1       n   3   4 cos   t log n     cos   2   log n   n        Re   n i u     cos   u log n      We are done by the inequality       3   4 cos     cos 2     2   1   cos     2   0     for           Theorem 3 10   Zero free region    There is a constant  c   0 such that         i t     0 whenever     1   c log     t     2    In particular      s     0 for Re   s     1   Proof  Let       1   2   and t      Suppose         i t    uwe know that     1  We know that   has no zeroes in some ball B   1   r   for some r   0  otherwise the entire function   s   1       s   would have an accumulation point for its zeros    Choosing c   0 small enough  we can assume that   t     r  By the key inequality for            3               4 Re               i t         Re               2 i t         0       Apply partial fraction decomposition of        So              s       1 s   1       s         1 1 0 1 s       O   log     t     2                 t   Im   s      Since Re         1 for any zero        Re 1     i u           Re             i u       2   0          1    Discarding terms  we get        3     1   4         C log     t     2         Take     1   s c log     t     2    and assume     1   c log     t     2   to get        3 log     t     2   5 c   4 log     t     2   6 c   C   log   t     2         Take c   1 1 6 C to get a contradiction     Theorem 3 11   Bounding z e t a     z e t a    Assuming that   c   0 sufficiently small  T   0  Re   s       1 0    Im   s         T   T   1    s is at least distance c log   T   2   away from any zero or pole  Then              s       s       c   log   T   2     2       Proof  If s       i t with     1 0  then             s       s           n   1       n   n   s       n   1       n   n       1     Assume then that Re   s         1 0   1 0    Apply         Each term satisfies     1   s         log   T   2   c 1   s   1     log   T   2   c     We know that there are O   log   T   2     zeros with multiplicity having imaginary part     T   2   T   2    The claim follows from triangle inequality