                  COMPLEMENTS TO LI'S CRITERION
                   FOR THE RIEMANN HYPOTHESIS

                  Enrico Bombieri and Jeffrey C. Lagarias



                                       Abstract
   In aPrecent
          
               paper Xian-Jin
                         
                              Li showed that the Riemann Hypothesis holds if and only if
n =  1 (1 1=)n has n > 0 for n = 1; 2; 3; : : : where  runs over the complex
zeros of the Riemann zeta function. We show that Li's criterion follows as a consequence of
a general set of inequalities for an arbitrary multiset of complex numbers  and therefore
is not speci c to zeta functions. We also give an arithmetic formula for the numbers n
in Li's paper, via the Guinand-Weil explicit formula and relate the conjectural positivity
of n to Weil's criterion for the Riemann Hypothesis.
1990 Mathematics Subject Classi cation 11M26, 11R42.
x1. Introduction. In a recent paper Xian-Jin Li [3] obtained an interesting
criterion for the validity of the Riemann Hypothesis. His criterion can be stated in
terms of the Riemann  -function
                           (s) = 12 s(s 1) s=2 2s (s)
                                                    


and the sequence
                                        dn sn 1 log (s)
                         n = (n 1 1)! dsn
                                                              s=1
for n = 1; 2; 3; : : : , in the form that the Riemann Hypothesis is equivalent to the
statement that n > 0 for every positive integer n. He also showed that an iden-
tical result applies to the Riemann Hypothesis for the Dedekind zeta function of a
number eld.
   The number n can be written in terms of the complex zeros  of the Riemann
zeta function (or Dedekind zeta function) as
                                       X            n
                                  n =     1 1     1
                                       
where the sum* over  is understood as
                             X                   X
                                 = Tlim
                                     !1
                                                          :
                                              j=()jT
   * We denote by <(z) and =(z) the real and imaginary part of the complex number z
2                                BOMBIERI AND LAGARIAS
   Li's proof uses the positivity of certain numbers aj and bm (see [3, formulae (1.6)
and (3.5)] obtained from an appropriate integral representation of the zeta function
and one may wonder to what extent Li's criterion is speci c for zeta functions and
whether there is an arithmetic interpretation of it.
   In this paper, Theorem 1 proves a general criterion for a multiset* of complex
numbers to lie in the half-plane <(s)  21 . Then the Corollary of Theorem 1 is a
criterion for a multiset which is invariant by complex conjugation and the functional
equation map s 7! 1 s to lie on the critical line <(s) = 21 . In the special case of
the multiset of non-trivial zeros of a zeta function, we recover Li's theorem. These
results show that Li's criterion is a consequence of a general set of inequalities that
is not speci c to zeta functions.
   Theorem 2 and the nal considerations in this paper give an arithmetic formula
for the numbers n in Li's paper via the Guinand-Weil explicit formula. Their
conjectural positivity is then related to Weil's well-known criterion for the Riemann
Hypothesis.
x2. Li's criterion. In this and the following sections, convergence of a sum over
a multiset R of complex numbers  is understood as the existence of the limit
                                      X               X
                                           = Tlim
                                               !1
                                                                                                  ()
                                       R            j=()jT
where in the sum each  occurs according to its multiplicity. In this case, we say
that the sum is  -convergent.
   In what follows, for  = 0 the quantity (1 1=) n = (=( 1))n is interpreted
as 0 if n is a positive integer and as 1 if n is a negative integer. We have
Lemma 1. Let R be a multiset of complex numbers and suppose that 0 2= R if n
is a positive integer, 1 2= R if n is a negative integer, and
                                    X 1+   j<()j < +1:                                      (2:1)
                                     R (1 + jj)2
Then for all integers n the sum
                                    X                         
                                         < 1 (1 1=)n
                                     R
converges absolutely.
                P                               P                                      
    Moreover if R 1= is  -convergent then n =  1                       (1 1=)n is also
 -convergent.
Remark. If R is invariant by complex conjugation, 0 2= R and (2.1) holds, then
P
  1= is  -convergent to a real value.
    * A multiset is a set whose elements have positive integral multiplicities assigned to them
                LI'S CRITERION FOR THE RIEMANN HYPOTHESIS                           3
Proof: We leave the proof of these assertions to the reader, as an easy exercise.
  We are now ready to state our rst result.
Theorem 1. Let R be a multiset of complex numbers  such that
  (i) 1P2= R ;
  (ii)     (1 + j<()j)=(1 + jj) < +1 .
                                  2


  Then the following conditions are equivalent:
  (a) P<()  1=2 for every  ;
                            n  0 for n = 1; 2; 3; : : : ;
  (b)     < 1 (1 1=)
  (c) for every xed " > 0 there is a constant c(") such that
               X                      
                    < 1 (1 1=) n  c(")e"n ; n = 1; 2; 3; : : : :
                
Remarks. From a formal point of view if
                                                   
                                               1 z
                                           Y
                                 f(z) =                                        (2:3)
                                           2R
and if                                X                   
                             n =          1 (1 1=)n
                                    2R
then one has
                            
              d log f 1 = 1 f 0 1 = X
                                                         1
                                                          

              dz        1 z       (1 z)2 f 1 z                n 1z n :      (2:4)
                                                         n=0
By means of this formula, one could attack Theorem 1 by methods of conformal
mapping and complex function theory. In fact, there is a function-theoretic inter-
pretation of Theorem 1 when f(z) is well de ned. The mapping w = 1=(1 z)
is a conformal mapping of the open unit disk jz j < 1 onto the open half-plane
<(w) > 21 , hence condition (a) asserts that ddz logf(1=(1 z)) is holomorphic in
the unit disk, which is equivalent to lim sup j n 1j1=n  1. Conditions (b) and
(c) assert that this growth condition can be replaced by one-sided conditions on
<( n 1 ). In this paper we shall follow a direct approach to proving Theorem 1
which applies even if f(z) is not well de ned.
Proof: We have  6= 1 and
                         j1 1=j 2 = 1 + (2            1)=j1 j2
where  = <(), therefore
                             (a) implies j1 1=j 1  1 for every . A fortiori we
have < 1 (1 1=) n  0 for every positive integer n, hence (a) implies (b).
  It is clear that (b) implies (c).
  Now suppose that (a) does not hold, so that <() > 21 for at least one  2 R .
4                                 BOMBIERI AND LAGARIAS
   We use again the fact that j1 1=j 2 = 1+(2 1)=j1 j2 . Since (2 1)=j1 j2
tends to zero, the maximum over  of this quantity is attained and there are nitely
many elements k 2 R , k = 1; : : :; K , such that j1 1=j 1 = 1 + t = max. Note
that t > 0 because > 21 for at least one .
   For any other  we have j1 1=j 1  1 + t  for a xed small  > 0. Let k
be the argument of 1 1=k . Then
                        1 (1 1=k ) n = 1 (1 + t)n e ink :
                                                                 
For  6= k we have j1 1=j n = O (1 + t )n and also
                                                                         
                  < 1 (1 1=) n = O (n j<()j + n2 )=jj2
as soon as jj > n, as one easily veri es using
                                           n
                          1                     = 1 en(  + 22 +:::) :
                                                             1       1

                                       1
                                                         P
Hence the sum over jj > n is O(n2 ) because (1+ j<()j)=(1+ jj)2 is convergent.
The number of elements in the sequence with jj  n is O(n2), again because
P
   (1 + j<()j)=(1 + jj)2 is convergent.
                                P 
                                          Hence elements
                                                   
                                                         other than k contribute
at most O(n2(1 + t )n) to < 1 (1 1=) n , while the remaining elements
k contribute
                                          XK
                              K (1 + t)n cos(nk ):
                                                   k=1
We have shown that
     X                                               K
                                                       X                               
          < 1 (1 1=) n = K (1 + t)n                         cos(nk ) + O n2(1 + t )n :
                                                      k=1
    By Dirichlet's theorem on simultaneous Diophantine approximation
                                                                P 
                                                                       we can make
the sum of cosines arbitrarily close to K , making it plain that  < 1 (1 1=) n
is in nitely often negative and exponentially large in absolute value as n tends to
1 . Thus the negation of (a) implies the negation of (c), hence (c) implies (a),
concluding the proof.
Corollary 1. (Li's Criterion) Let R be a multiset of complex numbers  such that
    (i) 0; 1 2= R ;
    (ii) ifP 2 R then 1  and  are in R , with the same multiplicity as  ;
    (iii)     (1 + j<()j)=(1 + jj) < +1 .
                                     2


    Then the following conditions are equivalent:
    (a) <() =P1=2 
                     for every  ; 
    (b) n =  1 (1 1=)n  0 for n = 1; 2; 3; : : : ;
                   LI'S CRITERION FOR THE RIEMANN HYPOTHESIS                                       5
   (c) for every xed " > 0 there is a constant c(") such that
                   X                      
                         1 (1 1=)n  c(")e"n ; n = 1; 2; 3; : : : :
                     
 Proof: Conditions (ii) and (iii) ensure that the sums in (b) and (c) are  -convergent
and real and n =  n for n = 1; 2; 3; ::: . The proof is completed by applying
Theorem 1 to R and to 1 R .
Remark. Condition (iii) can be relaxed to P 1=(1 + jj)2 < +1 ; we leave the
details to the reader.
x3. An arithmetic interpretation. In a well-known paper A. Weil [4] obtained a
general formulation of the so-called explicit formula of the theory of prime numbers.
An equivalent formula was obtained earlier for the case of the Riemann zeta function
by A.P. Guinand [2], under the assumption of the Riemann Hypothesis*. Both
papers [2] and [4] are formulated using the Fourier transform. As a historical
note, we mention here the fact that there is an earlier paper [1] by Guinand, in
which he states an explicit formula [1, (3.5), p.36] using the Mellin transform, with
an indication of proof. In this paper Guinand does not assume the validity of the
Riemann Hypothesis but says (referring to a formal proof of his summation formula)
\This cannot readily be done without a series of involved assumptions, so I shall
only derive a formal result which will indicate the type of formula to be expected.
Any particular case of the result can be investigated separately." However it should
be noted that Guinand's point of view, namely summation formulae arising from
self-reciprocal transforms, is quite di erent from Weil's, where the emphasis is on
the arithmetic.
   We restate Weil's formula as follows, in the special case of the Riemann zeta
function and in terms of the Mellin transform. Its formulation in the general case
does not involve additional diculties, so we leave it as an exercise for the interested
reader.
   Consider functions f(x) on the positive half-line (0; 1) such that:
   (A) f(x) is continuous and continuously di erentiable everywhere except at
 nitely many points ai , in which both f(x) and f 0 (x) have at most a disconti-
nuity of the rst kind, and in which one sets
                             f(ai ) = 21 f(ai + 0) + f(ai 0) ;
                                                           


   (B) there is  > 0 such that f(x) = O(x ) as x ! 0+ and f(x) = O(x 1  )
as x ! +1 .
   De ne an involution f ! fe by the formula
                                f(x)
                                 e    = x1 f( x1 )
   * It appears that the Riemann Hypothesis is used in [2] only to obtain the validity of the
Explicit Formula for a wider class of test functions. For example, one application of [2] is an exact
formula for N (T ), the number of complex zeros of  (s) with 0 < =(s)  T .
6                                     BOMBIERI AND LAGARIAS
and the Mellin transform of f by
                                                    Z 1
                                       f(s) =
                                       b                  f(x) xs 1 dx;
                                                     0


where, by assumption (B), the integral is absolutely convergent for  < <(s) <
1 +  . The inverse Mellin transform formula is
                                      Z
                                   1
                          f(x) = 2i          b x s ds;
                                              f(s)
                                       <(s)=c
valid for  < c < 1 +  .
   With this notation, we have
Explicit Formula. With f(x) satisfying (A), (B) above, we have
      X                 Z 1                 Z 1                 1
                                                                X          n           o
               f()
               b    =           f(x) dx +           f(x)
                                                    e    dx           (n) f(n) + f(n)
                                                                                  e
                       0                   0                   n=1
                                                         Z 1                         
                                                                                2        x dx
                                (log  + )f(1)                  f(x) + f(x)     x2 f(1) x2 1
                                                                       e
                                                          1


where the rst sum ranges over all complex zeros of the Riemann zeta function and
is understood as                      X
                                             lim
                                            T !1 j=()jT
                                                                f():
                                                                b



Proof: This is obtained from Weil's result [4], p.261-262, taking F(x) = ex=2 f(ex )
and making the change of variable from ex to x. We obtain
       X                Z 1                 Z 1                  1
                                                                 X          n              o
               f()
               b    =           f(x) dx +           f(x)
                                                    e    dx             (n) f(n) + f(n)
                                                                                    e
                           0                   0                n=1                            (3:1)
                                                          Z 1
                                 (log2)f(1) PF f(x) max(1; x) dx
                                               0     j x 1=x j x
where PF is the limit
     Z 1                Z 1                                 
               dx
  PF a(x) x = lim                           dx
                             1 min(x ; x ) a(x) x 2a(1) log  : (3:2)
      0               !1 0
   By splitting the integral over (0; 1) as a sum of two integrals over (0; 1) and
(1; 1) and making the change of variable from x to 1=x in the integral over (0; 1),
we see that
 Z 1                                          Z 1
                           max(1;
      1 min(x ; x ) f(x) jx 1=xj x =  x) dx               
                                                  1 x  f(x) + f(x)
                                                                          x dx
                                                                          x2 1 :
                                                                     e
  0                                            1
                                                                              (3:3)
                LI'S CRITERION FOR THE RIEMANN HYPOTHESIS                         7
Now for  ! 1 we have the asymptotic formula
                Z 1
                      1 x  x12 xx2 dx1 = 12 log(=2) + + o(1);
                                                          
                                                                               (3:4)
                   1

which may be veri ed as follows. An easy step shows that we may assume that 
is an even integer 2N . Then the integral is
                       Z 1
                            1 x 2N dx = 1 + 1 + : : : + 1
                        1    1 x 2 x3 2 4                 2N
and the asymptotic formula follows from the de nition of Euler's constant as
                                                          
                         = Nlim     1   1          1
                                      + + : : : + N logN :
                              !1 1 2
   From (3.2), (3.3) and (3.4) we deduce
      Z 1
   PF f(x) max(1;       x) dx
       0        j x  1=x  j x      Z 1            
      = ( log 2 + )f(1) + lim           1   x  f(x) + e
                                                         f(x)    2 f(1) x dx
                               !1 1                              x2         x2 1
                              Z 1                       
      = ( log 2 + )f(1) +                   f(x) x22 f(1) xx2 dx1
                                     f(x) + e
                                1

because here we may interchange the limit and the integral. By (3.1), this completes
the proof of the Explicit Formula.
   Now we are able to give an arithmetic interpretation to Li's criterion.
Lemma 2. For n = 1; 2; 3; : :: the inverse Mellin transform of 1 (1 1=s)n is
                              8
                              < Pn(log x)
                              >                if 0 < x < 1
                     gn (x) = > n=2            if x = 1
                              :
                                0              if x > 1
where Pn(x) is the polynomial
                                      n n
                                      X        xj 1 :
                            Pn(x) =
                                      j =1 j (j 1)!
Proof: We have
   n
   X
        
        n   1 Z 1 (log x)j 1xs 1 dx = X n n      
                                                1 dj 1 Z 1 xs 1 dx
                                                       j 1 0
   j =1 j (j 1)! 0                    j =1 j (j 1)! ds
                                      X n n ( 1)j 1
                                    =           sj
                                      j =1 j
                                    = 1 (1 1=s)n:
8                                 BOMBIERI AND LAGARIAS
This completes the proof.
Theorem 2. For n = 1; 2; 3; : :: we have
    X                  
         1 (1 1=)n =
    
                                                    8                                   9
            n                
                                                    (m)(log m)j 1           1 (log 1=")j =
                 ( 1)j 1 nj (j 1 1)! "!
            X                               < X
                                      lim0+ :             m                  j            ;
            j =1                              m1="
                                     n              
          + 1 (log 4 + ) n2                ( 1)j 1 nj (1 2 j ) (j):
                                     X

                                     j =2

Remark. For n = 1 this gives the well-known evaluation*
                                = 1 + 12            1
                         X1
                                                    2 log 4 = 0:0230957 :
                                                                        +

                             

Proof: The function gn(x) does not satisfy condition (B), so we cannot apply the
Explicit Formula directly. Hence for 0 < " < 1 we replace gn(x) by its truncation
                                     8
                                     < gn (x)
                                     >                    if " < x  1
                         gn;"(x) =     1
                                     > 2
                                             gn(")        if x = "
                                     :
                                         0                if x < ":
The function gn;"(x) satis es (A) and (B). Then the Explicit Formula yields
                  Z 1           Z 1
    X                    dx
      gbn;"() = gn;"(x) x + gn;"(x) dx
                                                  X (m)
                                                          gn;" m1
                                                                  

               "                "               m1=" m
                                    Z 1="                            (3:5)
                                            1      1  2          x dx
                 (log + )gn;" (1)
                                     1      x gn;" x x2 gn;"(1) x2 1 :
   We want to compute the limit of this formula for " ! 0. For the right-hand
side, we have
                 Z 1                 n n         
                                           1 Z 1 (log x)j 1 dx
                        gn;"(x) dx
                                     X
                                x  =   j (j 1)!             x
                   "                     j =1                    "
                                             n n 1
                                             X
                                                                                       (3:6)
                                     =               j j!   (log ")j
                                             j =1

    * H. Davenport, Multiplicative Number Theory, Markham 1967, pp.83-84
                     LI'S CRITERION FOR THE RIEMANN HYPOTHESIS                                9
and
   Z 1                   n
                         Xn
                              
                                1 Z 1 (logx)j 1 dx
         gn;"(x) dx =
     "              j =1 j (j 1)! "
                    X n n 1                        Z "             
                 =                        j
                                     ( 1) (j 1)!
                                            1                   j
                                                          (logx) dx
                                                                  1

                    j =1 j (j 1)!                       0                  (3:7)
                      n           
                 = ( 1)j 1 nj + O "(log 1=")n 1
                    X                               

                    j =1
                                      
                 = 1 + O "(log 1=")n 1 :
Since gn;"(1) = n=2, from (3.5), (3.6) and (3.7) we obtain that the limit of the
right-hand side of (3.5) as " ! 0 is
                                                   8                                    9
         n              
                                                            (m)(log m)j 1
                ( 1)j 1 nj (j 1 1)! "!                                       1 (log 1=")j
         X                                         < X                                  =
                                     lim0+ :                        m        j            ;
         j =1                                       m1="
                                        Z 1
           + 1 (log  + ) n2       [gn(x) nx] 1 dxx2 :
                                 0

Here we have made the change of variable from x to 1=x in the last integral. We
make the change of variable x = e t and compute
 Z 1                       Z 1              n n 1                  
     [gn(x) nx] 1 dxx2 =         (n nx) +
                                           X
                                                           (log x)j 1 dx
  0                          0             j =2 j (j 1)!                1 x2
                                        n                  Z 1          t
                                                1 n    1            j 1 e
                                      X
                         = (log 2)n + ( 1)    j
                                      j =2        j (j 1)! 0      t    1 e 2t dt
                                        n                   1              
                                                               X (j 1)!
                         = (log 2)n + ( 1)j 1 nj (j 1 1)!
                                      X
                                                                            j
                                      j =2                     m=0 (2m + 1)
                                        n         
                                                1 n
                                      X
                         = (log 2)n + ( 1)    j
                                                  j  (1 2 j ) (j);
                                      j =2
obtaining the right-hand side of the formula of Theorem 2.
   Finally, we note that
                                                        !
                                         X                     X
                                lim
                               "!0+
                                                  gbn;"() =       gbn():                (3:8)
                                                              
This follows from the Prime Number Theorem with error term. We have (without
error term if n = 1)
                                        Z "
                   gbn (s) gbn;"(s) =            Pn (logx)xs 1 dx
                                         0
                                                 s     <(s)                   
                                   = Pn(log ") "s + O "jsj2 (log1=")n 2
10                                   BOMBIERI AND LAGARIAS
                                                                                            P
for 0P< <(s) < 1 and jsj  1. Thus to prove (3.8) we need to show that " =
and "<() =jj2 tend to 0, as " ! 0, faster than any negative power of log1=".
   By the De la Vallee-Poussin zero-free region we have
                             c       < ()  1         c
                       log(jj + 2)               log(jj + 2)
for some constant c > 0, therefore
                                                                                     
                   X "<()                                                X     1
                         jj2          max
                                         
                                            "c= log(jj+2) jj 1=2            jj3=2
                                                                                              (3:9)
                                            p
                                           c log 1=" 
                                  =O e      0




for some constant c0 > 0. This quantity tends to 0, as " ! 0, faster than any
negative power of log1=".
   In a similar way, one has
                             X "         p
                                        c log 1=" ;
                                  = O e                  0



                              
possibly by readjusting the constant c0 > 0. In fact,
           X "         X "1 

                 =     1  
                                               
                         X (1=")       X "<()
                    = "         + O  jj2
                         
                                                         p
                    = " 1" 1" log2 12 log(1 "2 ) + O e c log 1="
                                                                                 0



                           p        
                    = O e c log 1="
                              0




by the classical explicit formula for (x), the Prime Number Theorem with error
term* and by (3.9). This proves what we want.
x4. Concluding remarks. Let f(x) and g(x) be complex valued functions
satisfying conditions (A) and (B) of the preceding section. The multiplicative con-
volution of f(x) and g(x) is given by
                                                 Z 1
                                  (f  g)(x) =         f(x=y)g(y) dy
                                                                  y :
                                                  0


The Mellin transform of f  g is f(s)
                                    b gb(s), whence, denoting by f the complex
conjugate of f , the Mellin transform of f  fe is f(s)
                                                   b f(1b
                                                            s), which is real and
     * H. Davenport, loc. cit., Ch.17,18
                 LI'S CRITERION FOR THE RIEMANN HYPOTHESIS                                             11
positive on the critical line <(s) = 12 . Thus the positivity of the Explicit Formula
on functions of the type f  fe is a necessary condition for the validity of the Riemann
Hypothesis. As shown by Weil, this is also a sucient condition.
   Li's criterion can be interpreted quite easily in this light. Let gn (x) be the
function de ned in the preceding section. We have the identity
                                                                                 
  1 (1 1=s)n + 1 (1 1=(1 s))n = 1 (1 1=s)n  1 (1 1=(1 s))n
and taking the inverse Mellin transform we nd
                              gn (x) + gen (x) = (gn  gen)(x):
Since the right-hand side of the Explicit Formula is invariant by changing f(x) into
f(x),
 e     the positivity in Li's criterion has the same meaning as in Weil's criterion.
The interesting point is that Li's criterion requires an explicit and rather simple set
of test functions for its veri cation.
   The numbers                   8                                          9
                  (  1) n        <X
                                         (m)(log  m) n       1             =
             n = n! xlim                                        (log x)n+1         (4:1)
                           !1 :mx             m            n+1             ;

which appear in Theorem 2 are analogous to the Stieltjes constants
                                      8                                                  9
                     ( 1)n            <X
                                              (log m)n           1          n+1
                                                                                         =
                n=      n! xlim
                            !1 :mx                m           n + 1 (log x) ;                       (4:2)

which appear in the Laurent expansion*
                                             1
                           (s + 1) = 1s +
                                            X
                                                n sn :
                                            n=0
In fact, we have                              1
                             0 (s + 1) = 1 + X  n sn :                                             (4:3)
                                         s             n=0
Hence                                                   1           n+1
                                                              n ns + 1
                                                       X
                               log s (s + 1) =
                                                        n=0
and also
                       
                                         1
                                          X
                                                        
        log s (s + 1) = log 1 +               n sn+1
                                        n=0
                               1 (
                               X        h
                                      1) 1 X
                                             1                    h
                           =          h                 n sn+1
                               h=1                n=0
                           = 0s +             1 2 s2 +                       + 31 03 s3 + : : : :
                                                                                     
                                       1
                                              2 0              2        0 1


  * A. Erdelyi, W Magnus, F. Oberhettinger, F.G. Tricomi, Higher Transcendental Functions,
McGraw-Hill 1953, vol. I, p.34, errata p.1
12                             BOMBIERI AND LAGARIAS
This expresses the constants n as polynomials in the constants n and in view of
Theorem 2 one obtains a formula for n in terms of the constants log4, h for
h  0 and (h) for h = 2; 3; : : : . A direct proof of such a formula can also be
given without recourse to the explicit formula, via power series expansions relating
(2.4) to (4.3) using the change of variable s = 1=(1 z).

                                    References

   [1] A.P. Guinand, Summation Formulae and Self-reciprocal Functions (III),
Quarterly J. Math., 13 (1942), 30-39.
   [2] A.P. Guinand, A Summation Formula in the Theory of Prime Numbers,
Proc. London Math. Soc., (2) 50 (1948), 107-119.
   [3] X.-J. Li, The Positivity of a Sequence of Numbers and the Riemann Hypoth-
esis, J. Number Theory, 65 (1997), 325-333.
   [4] A. Weil, Sur les \formules explicites" de la theorie des nombres premiers,
Meddelanden Fran Lunds Univ. Mat. Sem. (dedie a M. Riesz) (1952), 252-265.
Also uvres Scienti ques - Collected Papers, Springer-Verlag, Corrected Second
Printing 1980, Vol.II, 48-61.

     Enrico Bombieri                          Je rey C. Lagarias
     Institute for Advanced Study             Information Sciences Research
     School of Mathematics                    AT&T Labs
     Princeton, NJ 08540, USA                 Florham Park, NJ 07932-0971, USA
     eb@math.ias.edu                          jcl@research.att.com
